tag:blogger.com,1999:blog-37414104180967168272019-12-03T10:16:20.744-08:00Unizor - Creative Mind through Art of MathematicsUnizor is a site where students can learn the high school math (and in the future some other subjects) in a thorough and rigorous way. It allows parents to enroll their children in educational programs and to control the learning process.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.comBlogger387125tag:blogger.com,1999:blog-3741410418096716827.post-36938997835457674002019-12-03T10:16:00.001-08:002019-12-03T10:16:20.607-08:00Unizor - Physics4Teens - Energy - Light as Quants<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/pYSCh1p3Zh0" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Quants of Light</u><br /><br />By the end of 19th century the <i>electrons</i>, as carriers of electricity, were discovered by Sir Joseph Thomson in 1897, and many scientists experimented with electricity.<br /><br />At that time the <i>wave theory of light</i> was dominant. It could explain most of experimental facts and was shared by most physicists.<br />However, there were some new interesting experimental facts that could not be easily explained in the framework of the <b>classical</b> <i>wave theory of light</i> as oscillation of electromagnetic field.<br /><br />The <i>photoelectric effect</i> was one of such experimental facts that physicists could not fit into classical wave theory.<br /><br />Consider a simple electrical experiment when two poles, positive and negative, positioned close to each other, are gradually charged. After a charge reaches certain value, a spark between these poles causes the discharge of electricity.<br /><br />The electric charge was attributed to electrons with a negatively charged pole having more electrons than in an electrically neutral state and a positively charged one having less electrons than in an electrically neutral state.<br />The electric spark was the flow of excess electrons from a negative pole to a positive one, thus bringing them both to an electrically neutral state.<br /><br />It was observed that, if the negative pole is lit by light, the discharge occurs earlier, with less amount of charge accumulated in the poles.<br />This was so-called <i>photoelectric effect</i>.<br /><br />The experimental characteristics of the <i>photoelectric effect</i> were:<br />(a) if <i>photoelectric effect</i> is observed with specific frequency of the light, the number of electrons leaving the negative pole in a unit of time is proportional to intensity of the light;<br />(b) the speed of photo-electrons and, therefore, their kinetic energy do not depend on intensity of light, but on its frequency; higher intensity light produces more electrons, but their speed remains the same, while higher frequency light produces faster photo-electrons;<br />(c) for each material, used as a negative pole, there is minimal frequency of light necessary to initiate the photoelectric effect; high intensity or prolong time exposure to light of a lesser frequency do not produce photoelectric effect.<br /><br />The explanation coming to mind within a framework of the <i>wave theory of light</i> would be as follows.<br />Light is oscillations of the electromagnetic field. Electrons, accumulated during the charging process, are vibrating more intensely as a result of the oscillations of the electromagnetic field of the light, whose energy is transformed to electrons, so photo-electrons leave the negative pole easier, thus facilitating an earlier discharge.<br /><br />This would be a great explanation if not for a couple of contradictory facts.<br />The first contradiction is the property (b) of the <i>photoelectric effect</i>. According to the classical wave theory, the speed of photo-electrons must be dependent on the intensity of the light (amplitude of electromagnetic waves), which was not observed. And the (c) property is also unexplainable by classical wave theory, because, again, within a framework of the classical wave theory for any frequency we can find an intensity of light sufficient to "knock" out the electrons from the negative pole or keep the light of lesser intensity long enough time to transfer to electrons sufficient amount of energy to fly off the surface of the pole, which was not observed.<br /><br />The explanation of these phenomena came with introduction of <i>quants of light</i> - a hypothesis offered by Planck and used by Einstein to explain the properties of photoelectric effect.<br /><br />According to the explanation of photoelectric effect offered by Einstein, light propagates in space not as a continuous stream of waves of electromagnetic oscillations, but in small indivisible packets (<i>quants of light</i> or <i>photons</i>), separated in space and traveling along the same path, thus resurrecting the <i>corpuscular theory</i>, but without rejecting the electromagnetic origin of light.<br /><br />The energy of each <i>photon</i> proportionally depends on the frequency of oscillation of electromagnetic field that carries the light, not on intensity of light, with intensity of light being just a measure of the number of <i>photons</i> passing through a point in space in a unit of time.<br /><br />The energy of light is absorbed by electrons also in these <i>photons</i>. To break away the electron needs certain minimal energy.<br />If the energy of a single photon is sufficient to overcome the atomic forces that keep the electron inside the negative pole, this electron becomes a photo-electron and flies away to a positive pole.<br />If the energy of a photon is less than this minimal amount necessary to overcome the atomic forces, this energy is dissipated as heat, and no photo-electrons are produced.<br /><br />Within the framework of this new <i><b>quantum theory of light</b></i> all the characteristics of the <i>photoelectric effect</i> can find their explanation.<br />Let's analyze them.<br /><br />(a) If <i>photoelectric effect</i> is observed with specific frequency of the light, the number of electrons leaving the negative pole in a unit of time is proportional to intensity of the light;<br /><b>Explanation</b>: each photon has sufficient amount of energy to "knock" out an electron, and intensity is the number of such <i>photons</i> per unit of time.<br /><br />(b) The speed of photo-electrons and, therefore, their kinetic energy do not depend on intensity of light, but on its frequency; higher intensity light produces more electrons, but their speed remains the same, while higher frequency light produces faster photo-electrons;<br /><b>Explanation</b>: speed of electrons and, therefore, their kinetic energy depend on the energy of a <i>photon</i> that "knocked" these electrons out, which, in turn, is proportional to the frequency of light, while intensity of light (the number of <i>photons</i> per unit of time) affects the number of photo-electrons produced by light, not their individual energy.<br /><br />(c) For each material, used as a negative pole, there is minimal frequency of light necessary to initiate the photoelectric effect; high intensity or prolong time exposure to light of a lesser frequency do not produce photoelectric effect.<br /><b>Explanation</b>: the energy, needed by an electron to break away from atomic forces that keep it inside the material, obviously depend on the material; as soon as the light frequency is sufficient for one <i>photon</i> to carry that amount of energy, the photoelectric effect can start; <i>photons</i> of lesser level of frequency cannot "knock" out the electrons from the surrounding material, and the energy of the light is just dissipated as heat.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-47607749201597744192019-11-14T19:10:00.001-08:002019-11-14T19:10:37.857-08:00Unizor - Physics4Teens - Energy of Light - Light as Waves<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/ZPbN9tLUjXY" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Light as Waves</u><br /><br /><i>Corpuscular theory</i> of light, suggested by Newton, while adequately explaining such properties of light as traveling along a straight line and reflection off the mirror, has not been able to explain certain other properties of light observed by experimentators. In particular, <i>interference</i>, <i>diffraction</i> and <i>polarization</i> of light remained unexplainable in the framework of <i>corpuscular theory</i>.<br /><br />Attempts to model the light as oscillation of waves were made by several scientists, but the main formulation and explanation of wave-like properties of light is mostly related to a brilliant English physician and physicist Thomas Young.<br />His famous double-slit experiment, that showed the interference of light passing through two small close to each other slits, convincingly proved that light has wave-like properties.<br /><br />This experiment is extremely simple, anybody can reproduce it at home, its schema is below.<br /><img src="http://www.unizor.com/Pictures/DoubleSlit.jpg" style="height: 200px; width: 200px;" /><br /><br />The light from a single monochromatic source equidistant from two slits goes through these slits and is projected on a screen. The picture on the screen is a sequence of light and dark stripes parallel to the slits. This is a picture of <i>interference</i> that can only be explained within a framework of waves.<br />Two light waves coming through two slits to a surface of a screen in phase (both at its top wave state or both at the bottom) increase each other, making bright stripes. Those that come in opposite phases (one at its top state, while another at the bottom) nullify each other, making dark stripes.<br /><br /><img src="http://www.unizor.com/Pictures/DoubleSlitStripes.png" style="height: 200px; width: 200px;" /><br /><br />Let's interpret the <i>interference</i> of light from the wave theory viewpoint.<br /><br />First of all, if light is a wave, we have to determine the carrier of these waves. Until late 19th century physicists accepted a hypothesis that certain substance called <i>ether</i> (or <i>aether</i>) fills all the space and is a material carrier of waves of light and other waves of electromagnetic oscillations.<br />This hypothesis of <i>ether</i> was later on rejected, but for understanding of wave-like properties of light we can still think about <i>ether</i> as a substance, through which light travels as oscillations of this substance.<br />It might be considered similar to sound spreading inside a metal object, when you hit this metal object with another one. The crystal structure of the metal will start vibrating at the point of impact (the source of the sound) and these vibrations will spread throughout the whole object in all directions in a form of compression waves.<br />Notice that in this model the amplitude of vibrations is a characteristic of the strength of the sound and speed of propagation of sound waves depends on properties of material the object is made of.<br /><br />If light is the waves of a carrier substance, we can talk about this wave's length (distance between two adjacent crests or two adjacent troughs), amplitude (maximum deviation from the neutral position) and speed of propagation (how far a crest moves in a unit of time).<br /><br />If the speed of light propagation in the <i>ether</i> is <i><b>c</b>(m/sec)</i> and the wave length is <i><b>λ</b></i>, it means that in a <i><b>1</b> sec</i> the crest of a wave moves by a distance <i><b>c</b> meters</i>. With a wave length <i><b>λ</b> meters</i> there will be <i><b>f=c/λ</b> crests</i> on this distance. So, the frequency of oscillation <i><b>f</b></i>, that is the number of oscillations per second, the wave length <i><b>λ</b></i> and the speed of propagation <i><b>c</b></i> are related by the equation<br /><i><b>f = c/λ</b></i><br />or<br /><i><b>λ = c/f</b></i><br />or<br /><i><b>f·λ = c</b></i><br /><br />Let's consider the double slit experiment and analyze what will be observed on the screen at the point equidistant from both slits. Two slits can be considered as two independent sources of light and, considering they are letting through the light emitted by a single source equidistant from them, we can assume that the waves of light are coming <i>in-phase</i> to these slits and go through them also <i>in-phase</i>. Then the two beams of light travel the same distance to a point on a screen that we have chosen as equidistant from the slits, arriving to this point also <i>in-phase</i>. At that point they are combined and intensity of the light is increasing at there. That's why the stripe on a screen in the middle of a picture is bright, it consists of points equidistant from both slits.<br /><br />If, however, the distance these two beams of light have to travel towards the screen is different, the compounded effect of them, observed on a screen, will be dependent on whether the beams come to a screen <i>in-phase</i> or not.<br />More precisely, if the difference in the distance these beams travel towards a screen is a multiple of the length of a wave, they will come <i>in-phase</i>, and the overlapping beams will strengthen each other, there will be a bright spot on a screen.<br />If the difference in the distance equals to a multiple of wave length plus half a wave length, they will come <i>out-of-phase</i>, and the overlapping beams will nullify each other, there will be a dark spot on a screen.<br />In other cases the effect depends on how close the difference in distance is to a multiple of wave length or a multiple of wave length plus half a wave length with intermediary results.<br /><br />So, the wave theory can explain the effect of interference. Other wave-related properties of light that can be explained within the framework of the wave theory will be presented in the Optics chapter of this course.<br /><br />While the corpuscular properties of light do exist, they also can be explained within a framework of the wave theory and the energy of waves. That makes the wave theory of light more universal and more acceptable among physicists, though the auxiliary concept of <i>ether</i> has been totally rejected.<br /><br />The concept of <i>ether</i> was rejected as there was no experiment that could prove its existence. Moreover, a famous experiment of Michelson-Morley has proven that the speed of light does not depend on the speed of the source of light, which cannot be explained from the position of <i>ether</i> as a medium, through which the light waves propagate.<br /><br />There were some interesting observations of electric and magnetic properties, which related to an experimental fact that the speed of propagation of electric signals along wires is practically the same as the speed of light. This and other experiments prompted physicists to seek the explanation of the nature of light in the domain of electricity and magnetism.<br /><br />James Maxwell came up with his famous equations that describe the <i>electromagnetic field</i> and concluded that light is the oscillations of <i>electromagnetic field</i>. There is no special medium, like <i>ether</i>, that carries the light as the sound is carried by the waves of compression of the medium. Instead, the <i>electromagnetic field</i> propagates through space, thus carrying the light.<br />The properties of <i>electromagnetic field</i> will be presented in the parts of this course dedicated to <i>Electricity</i> and <i>Magnetism</i>.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-16735381918438972062019-11-05T11:32:00.001-08:002019-11-05T11:32:43.538-08:00Unizor - Physics4Teens - Energy - Energy of Light - Light as Corpuscles<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/v3PrPxQBJ1o" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Light as Corpuscles</u><br /><br />The nature of light was one of the topics of Newton's research. He experimented with light, made instruments to research the nature of light and formulated the first theory of light - <i>corpuscular theory</i>.<br /><br />According to <i>corpuscular theory</i>, light is a bunch of very small particles emitted by a source of light and flying with a very high speed. When they enter our eye, we see the light as a result of bombarding of inner surface of an eye with these tiny particles of light that Newton called <i>corpuscles</i>.<br />Each <i>light corpuscle</i> has mass, speed, kinetic energy and trajectory, like any other material object.<br /><br />Furthermore, Newton decided that there are different kinds of <i>light corpuscles</i> that cause our perception as different colors. He assumed that different colors of <i>light corpuscles</i> are due to their different sizes.<br /><br />He also realized that the white color is a combination of different kinds of <i>light corpuscles</i> that can be separated into different individual colors. Thus, white light after going through a green glass becomes green, because a green glass separates different kinds of <i>light corpuscles</i>, letting through only those that are perceived by our eye as green.<br /><br />According to Newton, <i>light corpuscles</i> are elastic, which explains perfectly their reflection from the mirror.<br /><br />The effect of refraction of light, when it changes the direction going from one medium, like air, into another, like water, was explained by Newton as a result of changing the speed of propagation of <i>light corpuscles</i>, when they go from one medium into another.<br /><br />The <i>corpuscular theory</i> of light was unable to explain the effect of <i>diffraction</i>, when the light seems bending around an edge of an obstacles or aperture.<br />Here is the picture formed on a screen by red light going through a small round hole.<br /><img src="http://www.unizor.com/Pictures/Diffraction.png" style="height: 200px; width: 200px;" /><br /><br />The <i>corpuscular theory</i> of light was unable to explain the effect of <i>interference</i>, when the light going through two small holes positioned near each other forms a complex wave-like picture on the screen.<br /><img src="http://www.unizor.com/Pictures/Interference.png" style="height: 150px; width: 200px;" /><br /><br />The <i>corpuscular theory</i> of light was unable to explain the effect of <i>polarization</i>, when the light consecutively going through two crystals of tourmaline changes its intensity from maximum to zero, depending on orientation of these crystals relatively to each other, when we rotate them around the axis coinciding with the direction of light.<br /><img src="http://www.unizor.com/Pictures/Polarization.png" style="height: 130px; width: 200px;" /><br /><br />These and some other difficulties in explanation of observed properties of light led to another theory - the <i>wave theory of light</i>.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-26592601450851128222019-10-31T10:41:00.001-07:002019-10-31T10:41:09.678-07:00Unizor - Physics4Teens - Energy - Light as Energy Carrier<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/lWQLQFzZw24" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Light as Energy Carrier</u><br /><br />Sun is a source of light and a source of heat. Just looking at light, we don't see energy it carries, but, touching a surface that was under Sun's light for some time, we feel its temperature, which is a measure of inner energy of the molecules inside the object. The object warms up under the Sun's light, which can only be explained by energy the light carries.<br /><br />Let's conduct an experiment with light to prove that it carries energy.<br />On YouTube there are a few videos that show it, one of them demonstrates a <a href="https://www.youtube.com/watch?v=j7UtjEjh7k4">radiometer</a>.<br /><br />The radiometer ("solar mill") looks like this:<br /><img src="http://www.unizor.com/Pictures/Radiometer.png" style="height: 240px; width: 200px;" /><br />Four small plates are arranged on a spinning wheel in the vacuum to avoid interference of air motion. Each plate has two surfaces, one silver and another black.<br />When light is directed on this "solar mill", black sides absorb light and are heated more than silver ones that reflect light.<br />As a result, the molecules near the black surface are moving more intensely than those near the surface of the silver side, thus pushing the silver side more and causing the rotation of the "solar mill".<br /><br />So, there is no doubt that light carries energy. An important question is, how it does it.<br />We used to think about heat as the intensity of molecular motion. In case of light there is no such motion, light travels from Sun to Earth through vacuum.<br />Apparently, the situation is similar to gravity in a sense that gravity carries energy, but does not require any medium, like molecules, to carry it. Recall that we have introduced a concept of a <i>field</i> as a certain domain of space were forces exist and energy is present without any material substance. Somewhat similar situation is with light. Its nature is the waves of <i>electromagnetic field</i>.<br /><br /><i>Electromagnetic field</i> is a completely different substance than <i>gravitational field</i>, but both are capable to carry energy without any material presence.<br /><br />Classifying light as the waves of an <i>electromagnetic field</i>, we are opening the door to using this wave model for explanation of different kinds of light.<br />First of all, let's consider the light we see with a naked eye. The vision itself is possible only if the light carries some energy, that agitates some cells inside our eye, that, in turn, send an electric signal to a brain - one more argument toward a light as a carrier of energy.<br /><br />Light that can be seen by a naked eye is called <i>visible</i>. But what about different colors that we can differentiate by an eye and different intensity of light that we view as "bright" or "faint"? The only explanation within a wave theory of light is that different intensities and colors that our eye sees are attributable to different kinds of waves of an electromagnetic field.<br /><br />Any wave has two major characteristics: amplitude and frequency. This is similar to a pendulum, where amplitude is the maximum angle of deviation from a vertical and frequency is measured as a number of oscillations per unit of time. Light, as a wave, also has these two characteristics. The <i>amplitude</i> is an intensity of light, while <i>frequency</i> of the visible light is viewed as its color.<br /><br />We all know that the photo laboratories, developing old fashioned films, are using rather faint red light during the developing process in order not to overexpose the film to light. The obvious reason is that red light carries less energy than white one, that is known to be a combination of many differently colored kinds of light. So, the color-defining frequency, as well as an intensity-defining amplitude of electromagnetic waves, determine the amount of energy carried by light.<br /><br />Not only visible light is a manifestation of electromagnetic waves. From cosmic radiation called <i>gamma rays</i> to <i>x-rays</i> to <i>ultra-violet light</i> to <i>visible light</i> to <i>infra-red light</i> to <i>microwave</i> to <i>radio waves</i> - all are electromagnetic waves of different frequencies.<br /><br />The unit of measurement of frequency is called <i>Hertz</i>, abbreviated as <i><b>Hz</b></i> with <i><b>1 Hz</b></i> meaning 1 oscillation per second.<br /><br />The range of frequencies of electromagnetic waves is from a few oscillations per second for very low frequency radio waves to 10<sup>24</sup> oscillations per second for very high frequency gamma rays. The visible spectrum of frequencies is close to 10<sup>15</sup> oscillations per second with the light perceived as red having a smaller frequency around 0.4·10<sup>15</sup> Hz, followed in increasing frequency order by orange, yellow, green, blue and violet with a frequency around 0.7·10<sup>15</sup> Hz.<br /><br />Amount of energy carried by light depends on the amplitude and frequency of electromagnetic waves that constitute this light. Generally speaking, the higher the amplitude - the higher the energy is carried by light in a unit of time and, similarly, the higher the frequency - the higher the energy is carried by light in a unit of time.<br /><br />Dependency of the energy carried by light on the frequency of electromagnetic waves that constitute this light is a more complex problem, that was solved in the framework of the Quantum Theory of light. According to Quantum Theory, electromagnetic waves propagate in packets called <i>photons</i>. Each <i>photon</i> carries an energy proportional to the frequency of waves that constitute this <i>photon</i>, and the amplitude of electromagnetic waves is, simply, a measure of the number of photons participating in these electromagnetic waves. That's why the picture of a light as a sinusoidal wave is a very simplified view on the nature of light as it is understood by contemporary physics.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-85391035758300597142019-10-15T20:18:00.001-07:002019-10-15T20:18:42.738-07:00Unizor - Physics4Teens - Energy - Gravitational Potential<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/fWXogAfvR30" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Gravity Integration 4 -<br />Solid Sphere</u><br /><br /><i>1. Determine the potential of the gravitational field of a uniform solid sphere at any point outside it.</i><br /><br />Let's establish a system of coordinates with a sphere's center at the origin of coordinates and X-axis going through a point of interest <i><b>P</b></i>, where we have to determine the gravitational potential.<br /><br />Assume that the sphere's radius is <i><b>R</b></i> and the mass is <i><b>M</b></i>. Then its volume is <i><b>4πR³/3;</b></i> and the mass density per unit of volume is <i><b>ρ=3M<span style="font-size: medium;">/</span>(4πR³)</b></i>.<br />Assume further that X-coordinate of a point <i><b>P</b></i>, where we want to calculate the gravitational potential, is <i><b>H</b></i>, which is greater than the radius of a sphere <i><b>R</b></i>.<br />If, instead of a sphere, we had a point mass <i><b>M</b></i> concentrated in its center at point <i><b>O(0,0,0)</b></i>, its gravitational potential at a point <i><b>P</b></i> would be<br /><i><b>V<sub>0</sub>(H) = −G·M/H</b></i><br />(remember that the gravitational potential is negative, as it equals to amount of work needed to bring a unit of mass from infinity to point <i><b>P</b></i>, and the field performs this work for us, so we perform negative work).<br /><br />As we know, gravitational potential is additive, that is the potential of a combined gravitational field produced by two or more sources of gravitation at some point equals to sum of their individual gravitational potentials at this point.<br /><br />To calculate a gravitational potential of a solid sphere at point <i><b>P</b></i> on the X-axis, let's divide it into infinite number of infinitesimally thin concentric spherical shells, all centered at the origin of coordinates and use the results presented in the previous lecture about spherical shell.<br /><br />As the variable of integration we will chose a radius of a spherical shell <i><b>r</b></i> that varies from <i>0</i> to <i>R</i>. Its outside surface area is <i><b>4πr²</b></i>, its thickness is <i>d<b>r</b></i> and, therefore, its volume is <i><b>4πr²·</b>d<b>r</b></i>.<br />This allows us to calculate the mass of this spherical shell using the volume and mass density calculated above.<br /><i>d<b>m = ρ·4πr²·</b>d<b>r =<br />= 3M·4πr²·</b>d<b>r<span style="font-size: medium;">/</span>(4πR³) =<br />= 3M·r²·</b>d<b>r<span style="font-size: medium;">/</span>R³</b></i><br />The formula for gravitational potential of a spherical shell, derived in the previous lecture was <i>V=−G·M<span style="font-size: medium;">/</span>H</i>, where <i>G</i> is a gravitational constant, <i>M</i> is a mass of a spherical shell and <i>H</i> is a distance from a center of a shell to a point of interest.<br />In case of a solid sphere divided into infinite number of infinitesimally thin concentric spherical shells the distance <i>H</i> remains the same. So, all we have to do is to substitute the mass in the formula for a shell with the variable mass of a shells we divided our solid sphere and to integrate by variable radius <i>r</i>:<br /><i><b>V(H) = −(G/H)<span style="font-size: large;">∫</span><sub>[0;R]</sub></b>d<b>m =<br />= −(G/H)<span style="font-size: large;">∫</span><sub>[0;R]</sub>3M·r²·</b>d<b>r<span style="font-size: medium;">/</span>R³ =<br />= −</b></i>[<i><b>3M·G/(H·R³)</b></i>]<i><b><span style="font-size: large;">∫</span><sub>[0;R]</sub>r²·</b>d<b>r</b></i><br />The indefinite integral of <i>r²</i> is <i>r³/3</i>, which gives the value of the integral<br /><i><b><span style="font-size: large;">∫</span><sub>[0;R]</sub>r²·</b>d<b>r = R³/3 − 0 = R³/3</b></i><br />Therefore, finally,<br /><i><b>V(H) = −G·M/H</b></i><br /><br /><b>Remarkably, the formula is exactly the same as if the whole mass was concentrated in the center of a sphere, the same as in case of a spherical shell</b>.<br /><br />It's a justification for considering gravitational potential of space objects in many cases to be approximately equivalent to a potential of point-masses of the same mass concentrated in one point, the center of mass of an object.<br /><br />In theory, this result was easily predictable. The gravitational potential of each spherical shell is the same as if its mass is concentrated at its center. All shells are concentric, therefore the masses of all of them are concentrated in the origin of coordinates and can be added together, since the gravitational potential is additive. Thus, we come to the same value of gravitational potential of a sphere, as if its total mass is concentrated in one point - its center.<br /><br />Let's analyze the force of gravity, acting on a probe object of a mass <i><b>m</b></i> at a point of interest on the distance <i><b>H</b></i> from a center of a solid sphere. This is a mass of a prove object multiplied by a derivative of the gravitational potential:<br /><i><b>F(H) = m·</b>d<b>V(H)/</b>d<b>H =<br />= G·m·M/H²</b></i><br />which is a well known Newton's Law of Gravitation.<br /><br /><i>2. Determine the potential of the gravitational field of a uniform solid sphere of radius <i><b>R</b></i> at a point inside it at distance <i><b>r</b></i> from a center.</i><br /><br />For this problem, as in the <i>Problem 1</i> above, we will need a mass density per unit of volume <i><b>ρ=3M<span style="font-size: medium;">/</span>(4πR³)</b></i>.<br /><br />Assume that a probe object is a distance <i><b>r</b></i> from a center of sphere, which is less than the radius of a sphere <i><b>R</b></i>. Let's calculate the gravitational potential <i><b>V(r)</b></i> of the combination of two separate sources - the solid sphere of radius <i><b>r</b></i> with a probe object on its surface and a thick empty spherical object between a surface of a sphere of the radius <i><b>r</b></i> and surface of a sphere of the radius <i><b>R</b></i>.<br /><br />The gravitational potential of a uniform solid sphere of radius <i><b>r</b></i> on its surface is discussed above as a <i>Problem 1</i>. To use the results of this problem, we need a mass <i><b>M<sub>1</sub>(r)</b></i> of a source of gravity and the distance of a point of interest from a center <i><b>H</b></i>.<br />The mass is<br /><i><b>M<sub>1</sub>(r) = ρ·4πr³<span style="font-size: medium;">/</span>3 = M·r³<span style="font-size: medium;">/</span>R³</b></i><br />The distance form a center is<br /><i><b>H = r</b></i><br />The gravitational potential on the surface of this solid sphere of the radius <i><b>r</b></i> equals to<br /><i><b>V<sub>1</sub>(r) = −G·M·r²/R³</b></i><br /><br />Consider now the second source of gravity - a thick empty sphere between the radiuses <i><b>r</b></i> and <i><b>R</b></i>.<br />As in <i>Problem 1</i> above, we will divide a thick empty sphere into an infinite number of concentric infinitesimally thin spherical shells of a variable radius <i><b>x</b></i> and thickness <i>d<b>x</b></i>.<br />The mass of each shell is<br /><i>d<b>m(x) = ρ·4πx²·</b>d<b>x =<br />= 3M·4πx²·</b>d<b>x<span style="font-size: medium;">/</span>(4πR³) =<br />= 3M·x²·</b>d<b>x<span style="font-size: medium;">/</span>R³</b></i><br />The potential inside such an infinitesimally thin spherical shell of radius <i><b>x</b></i> is, as we know from a previous lecture, constant and equals to<br /><i>d<b>V(x) = −G·</b>d<b>m(x)<span style="font-size: medium;">/</span>x =<br />= −3G·M·x·</b>d<b>x<span style="font-size: medium;">/</span>R³</b></i><br />To get a full potential inside such a thick empty sphere we have to perform integration of this expression from <i>x=r</i> to <i>x=R</i>.<br /><i><b>V<sub>2</sub>(r) = (−3G·M<span style="font-size: medium;">/</span>R³)<span style="font-size: large;">∫</span><sub>[r;R]</sub>x·</b>d<b>x =<br />= −3G·M·(R²−r²)/(2R³)</b></i><br /><br />The total potential at distance <i><b>r</b></i> from a center equal to sum of two potentials calculated above<br /><i><b>V(r) = V<sub>1</sub>(r) + V<sub>2</sub>(r) =<br />= −G·M·r²/R³ −<br />−3G·M·(R²−r²)/(2R³) =<br />= −G·M·(3R²−r²)/(2R³)</b></i><br /><br />On the outer surface of this sphere, when <i><b>r=R</b></i>, the above formula converts into the one derived in <i>Problem 1</i>:<br /><i><b>V(R) = −G·M·/R</b></i><br /><br />In the center of a solid sphere, when <i><b>r=0</b></i>, the potential is<br /><i><b>V(R) = −(3/2)·G·M·/R</b></i><br /><br />Let's analyze the force of gravity, acting on a probe object of a mass <i><b>m</b></i> at a point of interest on the distance <i><b>r</b></i> from a center of a solid sphere. This is a mass of a prove object multiplied by a derivative of the gravitational potential:<br /><i>Case inside a solid sphere</i>:<br /><i><b>F(r) = m·</b>d<b>V(r)/</b>d<b>r =<br />= G·m·M·r/R³</b></i><br />So, as we move from a center of a solid sphere (<i><b>r=0</b></i>) towards its outer surface (<i><b>r=R</b></i>), the force is linearly growing from zero at the center to <i><b>G·m·M/R²</b></i> at the end on the surface.<br /><i>Case outside a solid sphere (using the results of Problem 1 above for V(H)=−G·M/H, where H=r is greater than R)</i>:<br /><i><b>F(r) = m·</b>d<b>V(r)/</b>d<b>r =<br />= G·m·M/r²</b></i><br />So, as we move from a surface of a solid sphere (<i><b>r=R</b></i>) outwards to infinity, increasing <i><b>r</b></i>, the force is decreasing inversely to a square of a distance from the center from <i><b>G·m·M/R²</b></i> to zero at infinity.<br /><br />It's quite interesting to graph the force of gravitation as a function of a distance of a probe object from a center of a solid sphere. We have two different functions that represent this force, one inside and one outside the surface of a sphere.<br />The graph looks like this:<br /><img src="http://www.unizor.com/Pictures/GravitySolidSphere.png" style="height: 150px; width: 200px;" /><br /><br /><br /><br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-78215585584398073532019-10-07T12:37:00.001-07:002019-10-07T12:37:05.587-07:00Unizor - Physics4Teens - Energy - Gravitational Potential - Thin Spheric...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/xI86jwVxmQk" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Gravity Integration 3 -<br />Thin Spherical Shell</u><br /><br /><i>1. Determine the potential of the gravitational field of a uniform infinitesimally thin spherical shell at any point outside it.</i><br /><br />Let's establish a system of coordinates with a spherical shell's center at the origin of coordinates and X-axis going through a point of interest <i><b>P</b></i>, where we have to determine the gravitational potential.<br /><br />Assume that the sphere's radius is <i><b>R</b></i> and the mass is <i><b>M</b></i>. Then its surface is <i><b>4πR²</b></i> and the mass density per unit of surface area is <i><b>ρ=M<span style="font-size: medium;">/(4πR²)</span></b></i><span style="font-size: medium;">.<br />Assume further that X-coordinate of a point <i><b>P</b></i>, where we want to calculate the gravitational potential, is <i><b>H</b></i>, which is greater than the radius of a spherical shell <i><b>R</b></i>.<br /><img src="http://www.unizor.com/Pictures/SphericalShell.png" style="height: 100px; width: 200px;" /><br /><br />If, instead of a spherical shell, we had a point mass <i><b>M</b></i> concentrated in its center at point <i><b>O(0,0,0)</b></i>, its gravitational potential at a point <i><b>P</b></i> would be<br /><i><b>V<sub>0</sub> = −G·M/H</b></i><br />(remember that the gravitational potential is negative, as it equals to amount of work needed to bring a unit of mass from infinity to point <i><b>P</b></i>, and the field performs this work for us, so we perform negative work).<br /><br />As we know, gravitational potential is additive, that is the potential of a combined gravitational field produced by two or more sources of gravitation at some point equals to sum of their individual gravitational potentials at this point.<br /><br />To calculate a gravitational potential of an infinitesimally thin spherical shell at point <i><b>P</b></i> on the X-axis, let's divide a spherical shell into infinite number of infinitesimally thin rings that are parallel to the YZ-plane and, therefore, perpendicular to X-axis, that goes through a center of each ring.<br /><br />The angle <i><b>φ</b></i> from X-axis (that is, from <i><b>OP</b></i>) to a radius from an origin of coordinates to any point on a ring will be our variable of integration.<br />Then the radius of a ring will be<br /><i><b>r(φ) = R·sin(φ)</b></i><br />The distance from the origin of coordinates to a center of a ring is <i><b>R·cos(φ)</b></i>.<br />The area of a ring between angles <i><b>φ</b></i> and <i><b>φ+</b>d<b>φ</b></i> will be equal to the product of infinitesimal width of a ring <i><b>R·</b>d<b>φ</b></i> and its circumference <i><b>2πR·sin(φ)</b></i><br />Therefore, the mass of a ring will be<br /><i>d<b>m(φ) = ρ·2πR²·sin(φ)·</b>d<b>φ =<br />= M·2πsin(φ)·</b>d<b>φ<span style="font-size: medium;">/</span>(4π) =<br />= M·sin(φ)·</b>d<b>φ<span style="font-size: medium;">/</span>2</b></i><br /><br />Knowing the mass of a ring <i>d<b>m(φ)</b></i>, its radius <i><b>r(φ)</b></i> and the distance from the ring's center to point of interest <i><b>P</b></i>, that is equal to <i><b>H−R·cos(φ)</b></i>, we can use the formula of the ring's potential from a previous lecture<br /><i>V = −G·M <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">R²+H²</span></i><br />substituting<br /><i>d<b>V(φ)</b></i> instead of <i>V</i><br /><i>d<b>m(φ)</b></i> instead of <i>M</i><br /><i><b>H−R·cos(φ)</b></i> instead of <i>H</i><br /><i><b>r(φ)</b></i> instead of <i>R</i><br /><br />Therefore,<br /><i>d<b>V(φ) = −G·</b>d<b>m(φ) <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">r²(φ)+[H−R·cos(φ)]²</span> =<br />= −G·M·sin(φ)·</b>d<b>φ <span style="font-size: medium;">/</span>2√<span style="text-decoration-line: overline;">R²+H²−2R·H·cos(φ)</span></b></i><br /><br />Now all we need is to integrate this by <i><b>φ</b></i> in limits from <i>0</i> to <i>π</i>.<br />Substitute<br /><i><b>y = √<span style="text-decoration-line: overline;">R²+H²−2R·H·cos(φ)</span></b></i><br />Incidentally, the geometric meaning of this value is the distance from point of interest <i><b>P</b></i> to any point on a ring for a particular angle <i><b>φ</b></i>.<br />Then<br /><i>d<b>y = R·H·sin(φ)·</b>d<b>φ <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">R²+H²−2R·H·cos(φ)</span></b></i><br /><br />The limits of integration for <i><b>φ</b></i> from <i>0</i> to <i>π</i> in terms of <i><b>y</b></i> are from |<i><b>H−R</b></i>| (which, for our case of point <i><b>P</b></i> being outside the sphere, equals to <i><b>H−R</b></i>) to <i><b>H+R</b></i>.<br /><br />In terms of <i><b>y</b></i><br /><i>d<b>V(y) = −G·M·</b>d<b>y <span style="font-size: medium;">/</span>(2R·H)</b></i><br />which we have to integrate by <i><b>y</b></i> from <i><b>H−R</b></i> to <i><b>H+R</b></i>.<br /><br />Simple integration of this function by <i><b>y</b></i> on a segment [<i><b>H−R;H+R</b></i>] produces <i><b>−G·M·y<span style="font-size: medium;">/</span>(2R·H)</b></i> in limits from <i><b>H−R</b></i> to <i><b>H+R</b></i>:<br /><i><b>V = −G·M·(H+R)<span style="font-size: medium;">/</span>(2R·H) +<br />+ G·M·(H−R)<span style="font-size: medium;">/</span>(2R·H) =<br />= −G·M<span style="font-size: medium;">/</span>H</b></i><br /><br /><b>Remarkably, it's exactly the same gravitational potential, as if the whole mass was concentrated in a center of a spherical shell, as noted above as <i><b>V<sub>0</sub></b></i></b>.<br /><br />It's a justification for considering gravitational potential of space objects in many cases to be approximately equivalent to a potential of point-masses of the same mass concentrated in one point, the center of mass of an object.<br /><br /><i>2. Determine the potential of the gravitational field of a uniform infinitesimally thin spherical shell at any point inside it.</i><br /><br />Using the same notation as in the previous case, this problem requires the distance from a point of interest <i><b>P</b></i> to a center of a spherical shell <i><b>O</b></i> to be less than the radius <i><b>R</b></i> of a spherical shell.<br />Doing exactly the same manipulation and substitution<br /><i><b>y = √<span style="text-decoration-line: overline;">R²+H²−2R·H·cos(φ)</span></b></i><br />we see that the only difference from the previous case is in the limits of integration in terms of <i><b>y</b></i>.<br />The limits of integration for <i><b>φ</b></i> from <i>0</i> to <i>π</i> in terms of <i><b>y</b></i> are from |<i><b>H−R</b></i>| (which, in this case of point <i><b>P</b></i> being inside the sphere, equals to <i><b>R−H</b></i>) to <i><b>H+R</b></i>.<br /><br />Integration by <i><b>y</b></i> on a segment [<i><b>R−H;H+R</b></i>] produces <i><b>−G·M·y<span style="font-size: medium;">/</span>(2R·H)</b></i> in limits from <i><b>R−H</b></i> to <i><b>H+R</b></i>:<br /><i><b>V = −G·M·(H+R)<span style="font-size: medium;">/</span>(2R·H) + G·M·(R−H)<span style="font-size: medium;">/</span>(2R·H) =<br />= −G·M<span style="font-size: medium;">/</span>R</b></i><br /><br />Remarkably, it's constant and is independent of the position of point <i><b>P</b></i> inside a spherical shell.<br /><br />We have mentioned in the earlier lecture on gravitational field that in one dimensional case the gravitational force is a derivative of gravitational potential by distance from the source of gravity times mass of a probe object:<br /><b><i>F(r)=G·M·m <span style="font-size: medium;">/</span>r²=m·</i></b><i>d<b>V(r)/</b>d<b>r</b></i><br /><br />The fact that the gravitational potential is constant and, therefore, its derivative is zero, signifies that there is no force of gravity inside a spherical shell. The forces of gravity from all directions nullify each other.<br /><br />An intuitive explanation of this is in the fact that, if you consider any conical surface with a vertex at point <i><b>P</b></i> inside a sphere, cutting pieces of spherical shell's surface in both directions, the areas of the pieces will be proportional to a square of a distance from point <i><b>P</b></i>, while the gravitational forces produced by these pieces of surface are inversely proportional to a square of a distance from point <i><b>P</b></i>, thus both forces from opposite ends of a cone are equal in magnitude and opposite in direction, thus nullify each other.</span><br /><br /><span style="font-size: medium;"><br /></span>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-2146035884629930812019-10-04T19:26:00.001-07:002019-10-04T19:26:02.124-07:00Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Gravit...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/nbQAFewy5Y4" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Gravity Integration 2</u><br /><br /><i>1. Determine the potential of the gravitational field of an infinitely thin uniform solid ring at any point on the line perpendicular to a plane of the ring and going through its center.</i><br /><br />Let's establish a system of coordinates with a ring in the XY-plane with its center coincided with the origin of coordinates. Then the Z-axis is the line, on which we have to determine the gravitational potential of the ring.<br />Assume that the ring's radius is <i><b>R</b></i> and the mass is <i><b>M</b></i>, so the density of mass per unit of length is <i><b>ρ=M/(2πR)</b></i>.<br />Assume further that the Z-coordinate of a point <i><b>P</b></i>, where we want to calculate the gravitational potential, is <i><b>H</b></i>.<br /><br />If, instead of a ring, we had a point mass <i><b>M</b></i> concentrated in its center at point <i><b>(0,0,0)</b></i>, its gravitational potential at a point <i><b>P</b></i> would be<br /><i><b>V<sub>0</sub> = −G·M/H</b></i><br />(remember that the gravitational potential is negative, as it equals to amount of work needed to bring a unit of mass from infinity to point <i><b>P</b></i>, and the field performs this work for us, so we perform negative work).<br /><br />Since the mass in our case is distributed along the circumference of a ring, and every point on a ring is on a distance <i><b>r=√<span style="text-decoration-line: overline;">R²+H²</span></b></i> from point <i><b>P</b></i>, which is further from this point than the center of a ring, the gravitational potential of a ring at point <i><b>P</b></i> will be smaller.<br /><br />As we know, gravitational potential is additive, that is the potential of a combined gravitational field produced by two or more sources of gravitation at some point equals to sum of their individual gravitational potentials at this point.<br />Therefore, to calculate a gravitational potential of a ring, we can divide it into infinite number of infinitesimal pieces, calculate the gravitational potential of each piece at the point of interest <i><b>P</b></i> and integrate all these potentials.<br /><br />Let's choose an angle from the positive direction of the X-axis to a point on a ring as the main integration variable <i><b>φ</b></i>∈[<i><b>0;2π</b></i>]. Its increment <i>d<b>φ</b></i> gives an increment of the circumference of a ring<br /><i>d<b>l = R·</b>d<b>φ</b></i><br />The mass of this infinitesimal segment of a ring is<br /><i>d<b>m = ρ·</b>d<b>l = M·R·</b>d<b>φ <span style="font-size: medium;">/</span>(2πR) = M·</b>d<b>φ <span style="font-size: medium;">/</span>(2π)</b></i><br /><br />The distance from this infinitecimal segment of a ring to a point of interest <i><b>P</b></i> is independent of variable <i><b>φ</b></i> and is equal to constant <i><b>r=√<span style="text-decoration-line: overline;">R²+H²</span></b></i>.<br /><br />Therefore, gravitational potential of an infinitecimal segment of a ring is<br /><i>d<b>V = −G·</b>d<b>m <span style="font-size: medium;">/</span>r = −G·M·</b>d<b>φ <span style="font-size: medium;">/</span>(2π√<span style="text-decoration-line: overline;">R²+H²</span>)</b></i><br /><br />Integrating this by variable <i><b>φ</b></i> on [<i><b>0;2φ</b></i>], we obtain the total gravitational potential of a ring at point <i><b>P</b></i>:<br /><i><b>V = <span style="font-size: large;">∫</span><sub>[0;2π]</sub></b>d<b>V = −<span style="font-size: large;">∫</span><sub>[0;2π]</sub>G·M·</b>d<b>φ <span style="font-size: medium;">/</span>(2π√<span style="text-decoration-line: overline;">R²+H²</span>)</b></i><br />Finally,<br /><i><b>V = −G·M <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">R²+H²</span></b></i><br /><br /><i>2. Determine the potential of the gravitational field of an infinitely thin uniform solid disc at any point on the line perpendicular to a plane of the disc and going through its center.</i><br /><br />Let's establish a system of coordinates with a disc in the XY-plane with its center coincided with the origin of coordinates. Then the Z-axis is the line, on which we have to determine the gravitational potential of the ring.<br />Assume that the ring's radius is <i><b>R</b></i> and the mass is <i><b>M</b></i>, so the density of mass per unit of surface is <i><b>ρ=M/(πR²)</b></i>.<br />Assume further that the Z-coordinate of a point <i><b>P</b></i>, where we want to calculate the gravitational potential, is <i><b>H</b></i>.<br /><br />Let's split our disc into infinite number of infinitely thin concentric rings of radius from <i><b>x=0</b></i> to <i><b>x=R</b></i> of width <i>d<b>x</b></i> each and use the previous problem to determine the potential of each ring.<br /><br />The mass of each ring is<br /><i>d<b>m(x) = ρ·2πx·</b>d<b>x</b></i><br />This gravitational potential of this ring at point <i><b>P</b></i>, according to the previous problem, is<br /><i>d<b>V(x) = −G·</b>d<b>m(x) <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">x²+H²</span> =<br />= −G·ρ·2πx·</b>d<b>x <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">x²+H²</span> =<br />= −G·M·2πx·</b>d<b>x <span style="font-size: medium;">/</span>(πR²√<span style="text-decoration-line: overline;">x²+H²</span>) =<br />= −G·M·2x·</b>d<b>x <span style="font-size: medium;">/</span>(R²√<span style="text-decoration-line: overline;">x²+H²</span>)</b></i><br /><br />To determine gravitational potential of an entire disc, we have to integrate this expression in limits from <i><b>x=0</b></i> to <i><b>x=R</b></i>.<br /><i><b>V = <span style="font-size: large;">∫</span><sub>[0;R]</sub></b>d<b>V(x) = −k·<span style="font-size: large;">∫</span><sub>[0;R]</sub>2x·</b>d<b>x <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">x²+H²</span></b></i><br />where <i><b>k = G·M <span style="font-size: medium;">/</span>R²</b></i><br /><br />Substituting <i><b>y=x²+H²</b></i> and noticing the <i>d<b>y=2x·</b>d<b>x</b></i>, we get<br /><i><b>V = −k·<span style="font-size: large;">∫</span><sub>[H²;H²+R²]</sub> </b>d<b>y <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">y</span></b></i><br />The derivative of <i><b>√<span style="text-decoration-line: overline;">y</span></b></i> is <i><b>1 <span style="font-size: medium;">/</span>(2√<span style="text-decoration-line: overline;">y</span>)</b></i> Therefore, the indefinite integral of <i><b>1 <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">y</span></b></i> is<br /><i><b>2√<span style="text-decoration-line: overline;">y</span> + C</b></i><br /><br />Finally,<br /><i><b>V = −k·(2√<span style="text-decoration-line: overline;">H²+R²</span>−2H) = −2G·M·(√<span style="text-decoration-line: overline;">H²+R²</span>−H) <span style="font-size: medium;">/</span>R²</b></i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-15361333087752802492019-09-24T19:31:00.001-07:002019-09-24T19:31:01.396-07:00Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Gravit...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/siZCP7D-Huo" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Gravity Integration 1</u><br /><br /><i>Determine the potential of the gravitational field of an infinitely thin solid rod at any point outside of it.</i><br /><br />Let's establish a system of coordinates with a rod and a point mass lying in the XY-plane with the rod on the X-axis with one end at point <i><b>A(a,0)</b></i> and another at point <i><b>B(b,0)</b></i>.<br />Assume that the rod's length is <i><b>L=b−a</b></i> and the mass is <i><b>M</b></i>, so the density of mass per unit of length is <i><b>ρ=M/L</b></i>.<br />Assume further that the coordinates of a point <i><b>P</b></i>, where we want to calculate the gravitational potential, are <i><b>(p,q)</b></i>.<br /><br />If, instead of a rod, we had a point mass <i><b>M</b></i> concentrated in the midpoint of a rod at point <i><b>((a+b)/2,0)</b></i>, its gravitational potential at a point <i><b>(p,q)</b></i> would be<br /><i><b>V<sub>0</sub>=G·M/r</b></i><br />where <i><b>r</b></i> is the distance between the midpoint of a rod and a point of measurement of gravitational potential <i><b>P</b></i>:<br /><i><b>r = </b></i>{<i><b></b></i>[<i><b>(p−(a+b)/2</b></i>]<i><b><sup>2</sup> + q<sup>2</sup></b></i>}<i><b><sup>1/2</sup></b></i><br /><br />Since the mass in our case is distributed along the rod, the gravitational potential will be different.<br /><br />As we know, gravitational potential is additive, that is the potential of a combined gravitational field produced by two or more sources of gravitation at some point equals to sum of their individual gravitational potentials at this point.<br />Therefore, to calculate a gravitational potential of a rod, we can divide it into infinite number of infinitesimal pieces, calculate the gravitational potential of each piece at the point of interest <i><b>P</b></i> and integrate all these potentials.<br /><br />Consider a picture below (we recommend to save it locally to see in the bigger format).<br /><img src="http://www.unizor.com/Pictures/GravityRod.png" style="height: 120px; width: 200px;" /><br />As a variable, we will use an X-coordinate of a point on a rod <i><b>Q</b></i> and calculate the gravitational potential at point of interest <i><b>P(p,q)</b></i> from an infinitesimal segment of a rod of the length <i>d<b>x</b></i> around point <i><b>Q(x,0)</b></i>.<br />Knowing that, we will integrate the result by <i><b>x</b></i> on a segment [<i><b>a;b</b></i>] to get the gravitational potential of the rod.<br /><br />The infinitesimal segment of a rod <i>d<b>x</b></i>, positioned around a point <i><b>Q(x,0)</b></i>, has an infinitesimal mass <i>d<b>m</b></i> that can be calculated based on the total mass of a rod <i><b>M</b></i> and its length <i><b>L=b−a</b></i> as<br /><i>d<b>m = M·</b>d<b>x <span style="font-size: medium;">/</span>L</b></i><br /><br />The gravitational potential of this segment depends on its mass <i>d<b>m</b></i> and its distance <i><b>r(x)</b></i> to a point of interest <i><b>P(p,q)</b></i>.<br /><i>d<b>V = G·</b>d<b>m <span style="font-size: medium;">/</span>r(x)</b></i><br />Obviously,<br /><i><b>r(x) = </b></i>[<i><b>(p−x)<sup>2</sup>+q<sup>2</sup></b></i>]<i><b><sup>1/2</sup></b></i><br />Combining all this, the full gravitational potential of a rod [<i><b>a;b</b></i>] of mass <i><b>M</b></i> at point <i><b>P(p,q)</b></i> will then be<br /><i><b>V(p,q) = <span style="font-size: large;">∫</span><sub>a</sub><sup><sup>b</sup></sup>G·</b>d<b>m <span style="font-size: medium;">/</span>r(x) = <span style="font-size: large;">∫</span><sub>a</sub><sup><sup>b</sup></sup>G·M·</b>d<b>x<span style="font-size: medium;">/</span></b></i>{<i><b>L·</b></i>[<i><b>(p−x)<sup>2</sup>+q<sup>2</sup></b></i>]<i><b><sup>1/2</sup></b></i>}<br /><br />We can use the known indefinite integral<br /><i><b><span style="font-size: large;">∫</span></b>d<b>t <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">(t²+c²)</span> = ln|t+√<span style="text-decoration-line: overline;">(t²+c²)</span>|</b></i><br /><br />Let's substitute in the integral for gravitational potential <i><b>t=x−p</b></i>.<br />Then<br /><i><b>V(p,q) = <span style="font-size: large;">∫</span>G·M·</b>d<b>t <span style="font-size: medium;">/</span>L·</b></i>[<i><b>t<sup>2</sup>+q<sup>2</sup></b></i>]<i><b><sup>1/2</sup></b></i><br />where integration is from <i><b>t=a−p</b></i> to <i><b>t=b−p</b></i>.<br /><i><b>V(p,q) = (G·M/L)·</b></i>[<i><b>ln|b−p+√<span style="text-decoration-line: overline;">(b−p)²+q²</span>| − ln|a−p+√<span style="text-decoration-line: overline;">(a−p)²+q²</span>|</b></i>]<br />where <i><b>L = b−a</b></i><br /><br />Since the difference of logarithms is a logarithm of the result of division,<br /><i><b>V(p,q) = G·M·ln(R) <span style="font-size: medium;">/</span>L</b></i><br />where<br /><i><b>L = b−a</b></i> and<br /><i><b>R = |b−p+√<span style="text-decoration-line: overline;">(b−p)²+q²</span>| <span style="font-size: medium;">/</span> |a−p+√<span style="text-decoration-line: overline;">(a−p)²+q²</span>|</b></i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-69288424898840422132019-09-20T13:21:00.001-07:002019-09-20T13:21:27.523-07:00Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Energy...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/HGOK0nYEe2w" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Gravitational Energy Conservation</u><br /><br />While moving an object from a distance <i><b>r<sub>1</sub></b></i> to a distance <i><b>r<sub>2</sub></b></i> from the center of gravity, the gravitational field has performed certain work <i><b>W<sub>[r<sub>1</sub>,r<sub>2</sub>]</sub></b></i>, spending certain amount of energy. Since energy must be conserved, it should materialize in some other way.<br /><br />Indeed, the kinetic energy of a probe object at the end of its movement from point {<i>r<sub>1</sub>,0,0</i>} to {<i>r<sub>2</sub>,0,0</i>} must be equal to the work performed by the field.<br /><br />We have positioned our probe object at point {<i>r<sub>1</sub>,0,0</i>} without any initial speed, that is <i><b>V<sub>r<sub>1</sub></sub>=0</b></i>. Therefore, the kinetic energy <i><b>K<sub>r<sub>1</sub></sub></b></i> at this initial point is zero.<br />At the end of a motion at point {<i>r<sub>2</sub>,0,0</i>} the speed <i><b>V<sub>r<sub>2</sub></sub></b></i> must have such a value that the kinetic energy <i><b>K<sub>r<sub>2</sub></sub></b></i> would be equal to work <i><b>W<sub>[r<sub>1</sub>,r<sub>2</sub>]</sub></b></i> performed by the field.<br /><i><b>K<sub>r<sub>2</sub></sub> = m·V²<sub>r<sub>2</sub></sub> <span style="font-size: medium;">/</span>2 = W<sub>[r<sub>1</sub>,r<sub>2</sub>]</sub></b></i><br /><br />From this equation, knowing how to calculate the work performed by a gravitational field (see the previous lecture), we can find a speed of a probe object at the end of its motion from point {<i>r<sub>1</sub>,0,0</i>}, where it was at rest, to point {<i>r<sub>2</sub>,0,0</i>}:<br /><i><b>m·V²<sub>r<sub>2</sub></sub> <span style="font-size: medium;">/</span>2 =<br />= <i>(</i>G·M <span style="font-size: medium;">/</span>r<sub>2</sub> − G·M <span style="font-size: medium;">/</span>r<sub>1</sub></b></i><b>)<i>·m</i></b><br /><i><b>V²<sub>r<sub>2</sub></sub> = 2·<i></i>G·M·(1<span style="font-size: medium;">/</span>r<sub>2</sub> − 1<span style="font-size: medium;">/</span>r<sub>1</sub></b></i><b>)<i></i></b><br /><br />In a particular case, when <i><b>r<sub>1</sub>=∞</b></i> and <i><b>r<sub>2</sub>=r</b></i>, that is a probe object falls with no initial speed from the infinitely long distance from a source of gravity to a point at distance <i><b>r</b></i> from it, the formula is simplified:<br /><i><b>V²<sub>r</sub> = 2·<i></i>G·M <span style="font-size: medium;">/</span>r</b></i><br /><br />We would like to warn against falling into a point-mass that is a source of gravity, when the final distance from it is zero, that is <i><b>r=0</b></i> in the above equation. It obviously produces infinite speed and infinite kinetic energy, which does not correspond to reality. The most important reason for this deviation from the reality is our assumption about a source of gravity to be a point-mass. Real objects have certain non-zero dimensions. For example, in case of a gravitational field around our planet should not be analyzed by this formula for values of <i><b>r</b></i> less than the radius of Earth.<br /><br />Back to <i>energy conservation</i>.<br />The <i>potential energy</i> of an object is a measure of work that it can do, if left alone, that depends on a position of an object relative to other objects and such properties as its mass. Actually, these two parameters are the only ones needed to calculate the potential energy of a probe object in a gravitational field, provided we know everything about the field.<br /><br />As we know (see the previous lecture), amount of work we need to move a probe object of mass <i><b>m</b></i> from an infinite distance to a distance <i><b>r</b></i> from a source of gravity equals to<br /><i><b>W<sub>r</sub> = −G·M·m <span style="font-size: medium;">/</span>r</b></i><br />It's negative from our external to the gravitational field viewpoint, because we don't actually perform work, the field performs it for us. So, from the external viewpoint, the field gives certain energy to external object by performing some work on it, similar to a person, pushing the cart, spends energy, transferring it to a cart.<br /><br />In this expression, skipping over the universal gravitational constant <i><b>G</b></i>, components <i><b>M</b></i> (mass of a source of gravitational field) and <i><b>r</b></i> (distance from the center of the gravitational field) characterize the gravitational field, while <i><b>m</b></i> (mass of a probe object) characterizes the object, whose potential energy we measure.<br /><br />This energy is transferred to a probe object as its <i>potential energy</i>. If an object is not moving from this position, because some force holds it there, it retains this potential energy. As soon as there is no force holding it there, it will start moving towards the source of gravity, losing its <i>potential energy</i> and gaining the <i>kinetic energy</i> because it will move faster and faster.<br /><br />As an example, let's calculate the kinetic energy and final speed of a small asteroid, free falling on the surface of the Moon, assuming the Moon is the only source of gravity in the Universe.<br />The Universal Gravitational Constant is<br /><i><b>G=6.67408·10<sup>−11</sup></b></i>,<br />its units are <i>m<sup>3</sup>·kg<sup>−1</sup>·sec<sup>−2</sup></i>.<br />The mass of the Moon is <i><b>M=7.34767309·10<sup>22</sup></b> kg</i>.<br />The radius of the Moon is <i><b>r=1.7371·10<sup>6</sup></b> m</i>.<br />Let's assume that an asteroid falling on the Moon is relatively small one, say, <i><b>m=50</b> kg</i>.<br /><br />According to the formula above, the gravitational field of the Moon did the work that equals to<br /><i><b>W ≅ 6.67408·10<sup>−11</sup>·7.34767309·10<sup>22</sup>·50 <span style="font-size: medium;">/</span>(1.7371·10<sup>6</sup>) ≅ 141,151,800 </b>(joules)</i><br />let's check the units to make sure we get joules, the units of work<br /><i>m<sup>3</sup>·kg<sup>−1</sup>·sec<sup>−2</sup>·kg·kg·m<sup>−1</sup> = kg·m<sup>2</sup>·sec<sup>−2</sup> = N·m = J</i><br /><br />The final speed <i><b>V</b></i> can be calculated by equating this amount of work and kinetic energy of an asteroid:<br /><i><b>V<sup>2</sup> ≅ 2·6.67408·10<sup>−11</sup>·7.34767309·10<sup>22</sup> <span style="font-size: medium;">/</span>(1.7371·10<sup>6</sup>) ≅ 5646072</b></i><br />let's check the units to make sure we get the square of speed units<br /><i>m<sup>3</sup>·kg<sup>−1</sup>·sec<sup>−2</sup>·kg·m<sup>−1</sup> = m<sup>2</sup>·sec<sup>−2</sup> = (m/sec)<sup>2</sup></i><br /><br />From this the speed of an asteroid falling from infinity onto Moon's surface is<br /><i><b>V ≅ √<span style="text-decoration-line: overline;">5646072</span> ≅ 2376 </b>(m/sec)</i><br />or about <i><b>2.4 km/sec</b></i>.<br /><br />Incidentally, this is the so-called <i>escape speed</i> from the Moon, the initial speed needed for an object to leave the gravitational field of the Moon. A stone, thrown perpendicularly to the surface of the Moon with an initial speed less than that will go for certain distance away from the Moon, but then it will be brought back by the Moon's gravitation. Only if the initial speed is equal or exceeds the one above, the distance an object will go will be infinite, that is the object will leave the gravitational field of the Moon.<br /><br />Let's do similar calculations for the Earth, using the same assumptions, the same asteroid and the same units of measurement.<br /><i><b>M = 5.972·10<sup>24</sup></b> kg</i><br /><i><b>r = 6.371·10<sup>6</sup></b> m</i><br /><i><b>W ≅ 6.67408·10<sup>−11</sup>·5.972·10<sup>24</sup>·50 <span style="font-size: medium;">/</span>(6.371·10<sup>6</sup>) ≅ 3,128,049,424 </b>(joules)</i><br /><br /><i><b>V<sup>2</sup> ≅ 2·6.67408·10<sup>−11</sup>·5.972·10<sup>24</sup> <span style="font-size: medium;">/</span>(6.371·10<sup>6</sup>) ≅ 125121977</b> (m/sec)<sup>2</sup></i><br />From this the speed of an asteroid falling from infinity onto Earth's surface is<br /><i><b>V≅√<span style="text-decoration-line: overline;">125121977</span>≅11186</b> m/sec</i><br />or about <i><b>11.2</b> km/sec</i>.<br /><br />This is also the <i>escape speed</i> needed to fly away from Earth's gravitational field.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-64974552626984366942019-09-16T19:28:00.001-07:002019-09-16T19:28:28.295-07:00Unizor - Physics4Teens - Energy - Gravitational Field - Problems<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/37caTtBahno" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Problems on Gravity</u><br /><br /><i>Problem 1</i><br />Gravitational potential of a spherical gravitational field around a point-mass <i><b>M</b></i> at a distance <i><b>r</b></i> from it is defined as the work performed by gravity to bring a probe object of a unit mass from infinity to this point and is expressed as<br /><i><b>V<sub>r</sub> = −G·M <span style="font-size: medium;">/</span>r</b></i><br />Why is this formula independent of trajectory of a probe object or its exact final position relative to the point-mass <i><b>M</b></i>, but only on a distance itself from the source of gravity?<br /><br /><i>Solution</i><br />Any movement can be represented as infinitely many infinitesimal displacements, combined together into a trajectory.<br />In our three-dimensional world the force and an infinitesimal displacement of a probe object are vectors, so the infinitesimal work <i>d<b>W</b></i> performed by the force of gravity <i><b><span style="text-decoration-line: overline;">F</span></b></i> during the movement of a probe object, described by the infinitesimal displacement <i><span style="text-decoration-line: overline;">d<b>S</b></span></i>, is a scalar product of these two vectors:<br /><i>d<b>W = <span style="text-decoration-line: overline;">F</span>·</b><span style="text-decoration-line: overline;">d<b>S</b></span></i><br />Note that the vector of gravitational force <i><b><span style="text-decoration-line: overline;">F</span></b></i> is always directed towards the source of gravity.<br />Since a displacement vector <i><span style="text-decoration-line: overline;">d<b>S</b></span></i> can be represented as a sum of radial (towards the source of gravity) <i><span style="text-decoration-line: overline;">d<b>S</b></span><b><sub>r</sub></b></i> and tangential (perpendicular to radius) <i><span style="text-decoration-line: overline;">d<b>S</b></span><b><sub>t</sub></b></i> components, the above expression for a differential of work can be written as<br /><i>d<b>W = <span style="text-decoration-line: overline;">F</span>·</b></i><b>(<i></i></b><i><span style="text-decoration-line: overline;">d<b>S</b></span><b><sub>r</sub> + </b><span style="text-decoration-line: overline;">d<b>S</b></span><b><sub>t</sub></b></i><b>)<i> =<br />= <span style="text-decoration-line: overline;">F</span>·</i></b><i><span style="text-decoration-line: overline;">d<b>S</b></span><b><sub>r</sub> + <span style="text-decoration-line: overline;">F</span>·</b><span style="text-decoration-line: overline;">d<b>S</b></span><b><sub>t</sub></b></i><br />The second component in the above expression is a scalar product of two perpendicular vectors and is equal to zero. That's why we can completely ignore tangential movements, when calculating the work done by a central gravitational field, as not contributing to the amount of work. The total amount of work will be the same as if our probe object moved along a straight line towards the source of gravity and stopped at a distance <i><b>r</b></i> from it.<br /><br /><i>Problem 2</i><br />Given two point-masses of mass <i><b>M</b></i> each, fixed at a distance <i><b>2R</b></i> from each other.<br />Prove that the gravitational potential of a gravitational field produced by both of them at each point on a perpendicular bisector between them equals to a sum of individual gravitational potentials of these point-masses at this point, as if they were the only source of gravitation. In other words, prove that gravitational potential is additive in this case.<br /><br /><i>Solution</i><br />Let's draw a diagram of this problem (you can download it to display in a bigger format).<br /><img src="http://www.unizor.com/Pictures/Gravity2Masses.png" style="height: 100px; width: 200px;" /><br />Our two point-masses are at points <i>A</i> and <i>B</i>, the probe object is at point <i>D</i> on a perpendicular bisector of a segment <i>AB</i> going through point <i>C</i>.<br />The force of gravity towards point <i>A</i> is a segment <i>DE</i>, the force of gravity towards point <i>B</i> is a segment <i>DF</i>.<br />We will calculate the potential of a combined gravitational field of two point-masses at point <i>D</i>, where the probe object is located.<br />Let's assume that the segment <i>CD</i> equals to <i><b>h</b></i>.<br />The magnitude of each gravitational force equals to<br /><i><b>F = G·M·m <span style="font-size: medium;">/</span>(h<sup>2</sup>+r<sup>2</sup>)</b></i><br />Represent each of these forces as a sum of two vectors, one (green on a drawing) going vertically along the bisector <i>CD</i>, another (red) going horizontally parallel to <i>AB</i>.<br />Vertical components of these two forces will add to each other, as equal in magnitude and similarly directed downwards on a drawing, while horizontal ones will cancel each other, as equal in magnitude and opposite in direction to each other. So, the combined force acting on a probe object is a sum of vertical components of gravitational forces with a magnitude<br /><i><b>F<sub>tot</sub> = 2·G·M·m·sin(φ)<span style="font-size: medium;">/</span>(h<sup>2</sup>+r<sup>2</sup>)</b></i><br />Since <i><b>sin(φ) = CD/AD</b></i>,<br /><i><b>sin(φ) = h <span style="font-size: medium;">/</span></b></i>[(<i><b>h<sup>2</sup>+r<sup>2</sup></b></i>)<i><b><sup>1/2</sup></b></i>]<br /><i><b>F<sub>tot</sub> = 2·G·M·m·h <span style="font-size: medium;">/</span>(h<sup>2</sup>+r<sup>2</sup>)<sup>3/2</sup></b></i><br />If the gravitational field pulls a probe object along the perpendicular bisector of a segment <i>AB</i> from infinity to a distance <i><b>h</b></i> from the segment, the magnitude of a combined force of gravity, as a function of a distance from the segment <i><b>x</b></i> is changing, according to a similar formula:<br /><i><b>F<sub>tot</sub>(x) = 2·G·M·m·x <span style="font-size: medium;">/</span>(x<sup>2</sup>+r<sup>2</sup>)<sup>3/2</sup></b></i><br />To calculate work performed by a gravitational field pulling a probe object from infinity to height <i><b>h</b></i> above the segment <i>AB</i>, we have to integrate<br /><i><b>W<sub>tot</sub> = <span style="font-size: large;">∫</span><sub>[∞;h]</sub>F<sub>tot</sub>(x)·</b>d<b>x</b></i><br />It's supposed to be negative, since the direction of a force is opposite to a positive direction of the coordinate axis, we will take it into account later.<br /><i><b>W<sub>tot</sub> = <span style="font-size: large;">∫</span>2GMm·x·</b>d<b>x <span style="font-size: medium;">/</span>(x<sup>2</sup>+r<sup>2</sup>)<sup>3/2</sup></b></i><br />(within the same limits of integration [<i><b>∞;h</b></i>])<br />This integral can be easily calculated by substituting<br /><i><b>y=x<sup>2</sup>+r<sup>2</sup></b></i>,<br /><i><b>2·x·</b>d<b>x = </b>d<b>y</b></i>,<br />infinite limit of integration remaining infinite and the <i><b>x=h</b></i> limit transforming into <i><b>y=h<sup>2</sup>+r<sup>2</sup></b></i>. Now the work expression is<br /><i><b>W<sub>tot</sub> = <span style="font-size: large;">∫</span>G·M·m·y<sup>−3/2</sup>·</b>d<b>y</b></i><br />with limits from <i><b>y=∞</b></i> to <i><b>y=h<sup>2</sup>+r<sup>2</sup></b></i>.<br />The indefinite integral (anti-derivative) of <i><b>y<sup>−3/2</sup></b></i> is <i><b>−2·y<sup>−1/2</sup></b></i>.<br />Therefore, the value of integral and the work are<br /><i><b>W<sub>tot</sub> = −2·G·M·m·(h<sup>2</sup>+r<sup>2</sup>)<sup>−1/2</sup></b></i><br />For a unit mass <i><b>m=1</b></i> this work is a gravitational potential of a combined gravitational field produced by two point-masses on a distance <i><b>h</b></i> from a midpoint between them along a perpendicular bisector<br /><i><b>V<sub>tot</sub> = −2·G·M·(h<sup>2</sup>+r<sup>2</sup>)<sup>−1/2</sup></b></i><br />At the same time, the gravitational potential of a field produced by each one of the point-masses, considered separately, equals to<br /><i><b>V<sub>single</sub> = −G·M·(h<sup>2</sup>+r<sup>2</sup>)<sup>−1/2</sup></b></i><br />As we see, the gravitational potential of two point-masses equals to a sum of gravitational potential of each of them, considered separately.<br /><b>IMPORTANT NOTE</b><br />With more cumbersome calculations this principle can be proven for any two (not necessarily equal) point-masses at any point in space (not necessarily along the perpendicular bisector). This principle means that <b>gravitational potential is additive</b>, that is the <u>gravitational potential of any set of objects at any point in space equals to sum of their individual gravitational potentials</u>.<br /><br /><i>Problem 3</i><br />Express mass <i><b>M</b></i> of a spherical planet in terms of its radius <i><b>R</b></i> and a free fall acceleration <i><b>g</b></i> on its surface.<br /><br /><i>Solution</i><br />Let <i><b>m</b></i> be a mass of a probe object lying on a planet's surface.<br />According to the Newton's 2nd Law, its weight is<br /><i><b>P = m·g</b></i><br />According to the Universal Law of Gravitation, the force of gravitation between a planet and a probe object is<br /><i><b>F<sub>gravity</sub> = G·M·m <span style="font-size: medium;">/</span>R<sup>2</sup></b></i><br />Since the force of gravitation is the weight <i><b>F<sub>gravity</sub> = P</b></i>,<br /><i><b>m·g = G·M·m <span style="font-size: medium;">/</span>R<sup>2</sup></b></i><br />from which<br /><i><b>M = g·R<sup>2</sup> <span style="font-size: medium;">/</span>G</b></i><br /><br /><i>Problem 4</i><br />Express gravitational potential <i><b>V<sub>R</sub></b></i> of a spherical planet on its surface in terms of its radius <i><b>R</b></i> and a free fall acceleration <i><b>g</b></i> on its surface.<br /><br /><i>Solution</i><br />From the definition of a gravitational potential on a distance <i><b>R</b></i> from a source of gravity<br /><i><b>V<sub>R</sub> = −G·M <span style="font-size: medium;">/</span>R</b></i><br />Using the expression of the planet's mass in terms of its radius <i><b>R</b></i> and a free fall acceleration <i><b>g</b></i> on its surface (see above),<br /><i><b>M = g·R<sup>2</sup> <span style="font-size: medium;">/</span>G</b></i><br />Substituting this mass into a formula for potential,<br /><i><b>V<sub>R</sub> = −G·g·R<sup>2</sup> <span style="font-size: medium;">/</span>(G·R) = −g·R</b></i><br /><br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-68786613128947835842019-09-10T04:52:00.002-07:002019-09-10T04:53:55.892-07:00Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Gravitational Field<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/XFuuNzOnO_Y" width="480"></iframe><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Gravitational Field</u><br /><br />Studying <i>forces</i>, we have paid attention to a force of attraction, that exists between any material objects, the <i>force of gravity</i>.<br/>For example, if a comet from outer space flies not far from a Sun, it is attracted by Sun and changes its straight line trajectory.<br/><br/>In Mechanics we used to see the force as something between the objects touching each other, like a man pushing a wagon. In case of gravity the force obviously exists, but it acts on a distance, in "empty" space.<br/>In Physics this concept of force acting on a distance is described by a term <i>field</i>. Basically, <i>field</i> is the area in space where some force acts on all objects or only on objects that have specific property. The force in this case depends on a point in space and an object that experiences this force and, as a result of the action of force, changes its movement.<br/><br/><i>Gravitational field</i> exists around any material object (the source object of a field) and acts as an attraction towards this source object, experienced by any other material object (probe object) positioned in this field.<br/>As described in the "Gravity, Weight" chapter of "Mechanics" part of this course, the magnitude of the <i>gravitational force</i> <i><b>F</b></i> is proportional to a product of masses of a source object and a probe object, <b><i>M</i></b> and <b><i>m</i></b>, and it is inversely proportional to a square of a distance <b><i>r</i></b> between these objects:<br/><b><i>F = G·M·m <font size=4>/</font>r²</i></b><br/>where <b><i>G</i></b> - a constant of proportionality, since the units of force (N - newtons) have been defined already, and we want to measure the gravitational force in the same units as any other force.<br/><br/>The direction of the <i>gravitational force</i> acting on a probe object is towards the source object.<br/><br/>Let's return to our example of a comet flying not far from the Sun and, being attracted to the Sun, changing its trajectory. Obviously, to change the trajectory, some energy must be spent. So, we conclude that <i>gravitational field</i> has certain amount of energy at each point that it spends by applying the <i>force</i> onto a probe object.<br/><br/>To quantify this, assume that the source of gravity is a point mass <i><b>M</b></i> fixed at the origin of coordinates. Position a probe object of mass <i><b>m</b></i> at coordinates {<i>r<sub>1</sub>,0,0</i>} and let it go. The force of gravity will cause the motion of this probe object towards the center of gravity, the origin of coordinates, so the movement will be along the X-axis. Let the ending position of the probe object be {<i>r<sub>2</sub>,0,0</i>}, where <i>r<sub>2</sub></i> is smaller then <i>r<sub>1</sub></i>. Let <i><b>x</b></i> be a variable X-coordinate (distance to the origin).<br/><br/>According to the Universal Law of Gravitation, the force of attraction of a probe object towards the source of a gravitational field at distance <i><b>x</b></i> from the origin equals to<br/><i><b>F = −G·M·m <font size=4>/</font>x²</b></i><br/>where minus in front of it signifies that this force is directed opposite to increasing the X-coordinate.<br/>This force causes the motion and, therefore, does some work, moving a probe object from point {<i>r<sub>1</sub>,0,0</i>} to point {<i>r<sub>2</sub>,0,0</i>} along the X-axis. To calculate the work done by this variable force, we can integrate <i><b>F·</b>d<b>x</b></i> from <i><b>x=r<sub>1</sub></b></i> to <i><b>x=r<sub>2</sub></b></i>:<br/><i><b>W<sub>[r<sub>1</sub>,r<sub>2</sub>]</sub> = <font size=5>∫</font><sub>[r<sub>1</sub>,r<sub>2</sub>]</sub>F·</b>d<b>x =<br/>= −<font size=5>∫</font><sub>[r<sub>1</sub>,r<sub>2</sub>]</sub>G·M·m·</b>d<b>x <font size=4>/</font>x² =<br/>= G·M·m <font size=4>/</font>x<font size=5></b>|<b></font><sub>[r<sub>1</sub>,r<sub>2</sub>]</sub> =<br/>= G·M·m <font size=4>/</font>r<sub>2</sub> − G·M·m <font size=4>/</font>r<sub>1</sub> =<br/>= <i>(</i>G·M <font size=4>/</font>r<sub>2</sub> − G·M <font size=4>/</font>r<sub>1</sub></i>)<i>·m</b></i><br/><br/>The expression<br/><i><b>V(r) = −G·M<font size=4>/</font>r</b></i><br/>is called <i>gravitational potential</i>.<br/>It's a characteristic of a <i>gravitational field</i> sourced by a point mass <i><b>M</b></i> at a distance <i><b>r</b></i> from a source.<br/>It equals to work needed by <b>external forces</b> to bring a probe object of mass <i><b>m=1</b></i> to a point at distance <i><b>r</b></i> from a source of the field <b>from infinity</b>.<br/>Indeed, set <i><b>m=1</b></i>, <i><b>r<sub>1</sub>=∞</b></i> and <i><b>r<sub>1</sub>=r</b></i> in the above formula for work <i><b>W<sub>[r<sub>1</sub>,r<sub>2</sub>]</sub></b></i> and take into consideration that gravitational field "helps" external forces to move a probe object, so the external forces spend negative amount of energy.<br/><br/>Using this concept of <i>gravitational potential</i> <i><b>V(r)</b></i>, we can state that, to move a probe object of a unit mass from distance <i><b>r<sub>1</sub></b></i> relative to a source of gravitational field to a distance <i><b>r<sub>2</sub></b></i> relative to its source in the gravitational field with <i>gravitational potential</i> <i><b>V(r)</b></i>, we have to spend the amount of energy equal to <i><b>V(r<sub>1</sub>)−V(r<sub>2</sub>)</b></i>.<br/>For a probe object of any mass <i><b>m</b></i> this amount should be multiplied by <i><b>m</b></i>.<br/>If <i><b>r<sub>2</sub></b></i> is greater than <i><b>r<sub>1</sub></b></i>, that is we move a probe object further from the source of gravity, working against the gravitational force, this expression is positive, we have to apply effort against the force of gravity. In an opposite case, when <i><b>r<sub>2</sub></b></i> is smaller than <i><b>r<sub>1</sub></b></i>, that is we move closer to a source of gravity, the gravitational force "helps" us, we don't have to apply any efforts, and our work is negative.<br/><br/>Therefore, an expression <i><b>E<sub>P</sub>=m·V(r)</b></i> represents <i>potential energy</i> of a probe object of mass <i><b>m</b></i> at a distance <i><b>r</b></i> from a source of a gravitational field with <i>gravitational potential</i> <i><b>V(r)</b></i>.<br/><br/>A useful consequence from a concept of a <i>gravitational potential</i> is that the <i>force of gravity</i> can be expressed as the derivative of the <i>gravitational potential</i>.<br/><b><i>F = G·M·m <font size=4>/</font>r² = m·</b>d<b>V(r)/</b>d<b>r</i></b><br/>which emphasizes the statement that the <i>gravitational potential</i> is a characteristic of a field itself, not its source.<br/>We, therefore, can discuss <i>gravitational field</i> as an abstract concept defined only by the function called <i>gravitational potential</i>.<br/><br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-54619554922461790012019-08-18T06:19:00.001-07:002019-09-10T04:42:09.654-07:00Unizor - Physics4Teens - Energy - Energy of Nucleus - Fusion<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/XcmHR2eISpc" width="480"></iframe><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Nucleus Fusion</u><br /><br /><br /><br /><i>Fusion</i> is a <i>nuclear reaction</i>, when light nuclei are brought together and combined into a heavier ones.<br /><br />The reason for this reaction to release the energy is the difference <br />between amount of energy needed to overcome the repulsion between nuclei<br /> because they have the same positive electric charge (this energy is <br />consumed by <i>fusion</i>) and the potential energy released by <i>strong forces</i>, when the formation of a combined nucleus occurs (this energy is released by <i>fusion</i>).<br /><br /><u>The former is less than the latter</u>.<br /><br /><br /><br />When the light nuclei are fused into a heavier one, the excess of potential energy of <i>strong forces</i>, released in the process of <i>fusion</i>,<br /> over the energy needed to squeeze together protons against their <br />repulsion is converted into thermal and electromagnetic field energy.<br /><br /><br /><br />Analogy to this process can be two magnets separated by a spring.<br /><img src="http://www.unizor.com/Pictures/FusionAnalogy.png" style="height: 400px; width: 200px;" /><br /><br />The magnets represent two separate protons, the magnetic force of attraction between them represents the <i>strong force</i><br /> that is supposed to hold the nucleus together, when these particles are<br /> close to each other, the spring represents the electrical repulsive <br />force between them, acting on a larger distance, as both are positively <br />charged.<br /><br />It's known that magnetic force is inversely proportional to a square of a<br /> distance between objects, while the resistance of a spring against <br />contraction obeys the Hooke's Law and is proportional to the length of <br />contraction.<br /><br />On the picture magnets are separated. To bring them together, we have to<br /> spend certain amount of energy to move against a spring that resists <br />contraction. But the magnetic attraction grows faster then the <br />resistance of the spring, so, at some moment this attraction will be <br />greater than the resistance of a spring. At this moment nothing would <br />prevent magnets to fuse.<br /><br /><br /><br />As is in the above analogy, if we want to fuse two protons, we have to bring them together sufficiently close for <i>strong forces</i> to overtake the repulsion of their positive charges.<br /><br /><br /><br />Consider the following nuclear reaction of <i>fusion</i>.<br /><br />One nucleus of hydrogen isotope <i>deuterium</i> <b><sup>1</sup>H<sup>2</sup></b> with atomic mass 2 contains one proton and one neutron.<br /><br />One nucleus of hydrogen isotope <i>tritium</i> <b><sup>1</sup>H<sup>3</sup></b> with atomic mass 3 contains one proton and two neutrons.<br /><br />If we force these two nuclei to fuse, they will form a nucleus of <i>helium</i> <b><sup>2</sup>He<sup>4</sup></b> and releasing certain amount of energy:<br /><br /><b><sup>1</sup>H<sup>2</sup> + <sup>1</sup>H<sup>3</sup> = <sup>2</sup>He<sup>4</sup> + <sup>0</sup>n<sup>1</sup></b><br /><br /><br /><br />It's not easy to overcome the repulsion of protons. High temperature and<br /> pressure, like in the core of our Sun, are conditions where it happens.<br /> On Earth these conditions are created in the nuclear bomb, using the <br />atomic bomd to achieve proper amount of heat and pressure, thus creating<br /> an uncontrlled <i>fusion</i>.<br /><br />Controlled nuclear reaction of <i>fusion</i> is what scientists are working on right now. So far, it's still in the experimental stage.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-43415492098950650422019-08-12T18:50:00.001-07:002019-08-12T18:50:19.438-07:00Unizor - Physics4Teens - Energy - Energy of a Nucleus - Fission<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/6WdOzCFKxxw" width="480"></iframe><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Nucleus Fission</u><br /><br /><i>Fission</i>, first of all, is a <i>nuclear reaction</i>, when heavier nuclei are split into lighter ones.<br/>The reason for this reaction to release the energy is the difference between amount of energy needed to break <i>strong forces</i> that hold the nucleus together (this energy is consumed by <i>fission</i>) and amount of potential energy in positively charged and repelling protons inside nucleus (this energy is released by <i>fission</i>).<br/><u>The former is less than the latter</u>.<br/><br/>When the heavy nucleus is broken into parts, the excess of potential energy of squeezed together protons against their repelling force over the energy of strong forces that keep nucleus together is converted into thermal and electromagnetic field energy.<br/><br/>Analogy of this is a spring squeezed tightly and held in this position by a thread. A thread plays the role of <i>strong forces</i>, while a potential energy of a squeezed spring plays the role of protons kept close to each other by a this force. When you cut a thread, the spring will release the potential energy, similarly to protons repelling from each other.<br/><br/>Electrically positively charged protons repel each other and, at the same time, are bonded together by <i>strong forces</i> inside a nucleus. At the same time neutrons are also bonded by <i>strong forces</i> among themselves and with protons without any repulsion.<br/>So, the more neutrons the nucleus has - the stronger it is. Neutrons only add "bonding material" to a nucleus without adding any repelling forces that work against the nucleus' stability.<br/><br/>Uranium-238 with 92 protons and 146 neutrons (<sup>92</sup>U<sup>238</sup>) naturally occurs on Earth and is relatively stable.<br/>Uranium-235 with the same 92 protons and 143 neutrons (<sup>92</sup>U<sup>235</sup>) has less "bonding material" (less neutrons) and is more susceptible to fission.<br/><br/>All it takes to break the nucleus of <sup>92</sup>U<sup>235</sup> is a little "push" from outside, which can be accomplished by bombarding it with neutrons. In the process of <i>fission</i>, caused by hitting a nucleus of <sup>92</sup>U<sup>235</sup> with a neutron, it can transforms into Barium-141 with 56 protons and 85 neutrons <sup>56</sup>Ba<sup>141</sup>, Krypton-92 with 36 protons and 56 neutrons <sup>36</sup>Kr<sup>92</sup> and 3 free neutrons.<br/>As we see, the numbers of protons is balanced (input: 92, output: 56 and 36), as well as a number of neutrons (input: 1 free hitting neutron and 143 in a nucleus of <sup>92</sup>U<sup>235</sup> total 144, output: 85 in a nucleus <sup>56</sup>Ba<sup>141</sup>, 56 in a nucleus of <sup>36</sup>Kr<sup>92</sup> and 3 new free neutrons total 144).<br/><br/>Let's express this reaction in a formula (letter <i><b>n</b></i> denotes a neutron):<br/><i><b><sup>0</sup>n<sup>1</sup> + <sup>92</sup>U<sup>235</sup> =<br/><sup>56</sup>Ba<sup>141</sup> + <sup>36</sup>Kr<sup>92</sup> + 3·<sup>0</sup>n<sup>1</sup></b></i><br/><br/>What's interesting in this reaction is that it not only produces energy because we break a heavy nucleus into lighter ones, but also that it produces 2 new neutrons that can bombard other atoms, causing a <b>chain reaction</b> and, potentially, an explosion (atomic bomb). However, if we absorb extra neutrons, it will allow to slowly release of nuclear energy (nuclear power stations).<br/><br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-69734146366792166512019-08-05T18:46:00.001-07:002019-08-05T18:46:22.756-07:00Unizor - Physics4Teens - Energy - Energy of a Nucleus<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/zi6gTkhDIds" width="480"></iframe><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Energy of Nucleus</u><br /><br /><br /><br />In this lecture we will analyze the energy aspect of nucleus - the central part of an atom.<br /><br /><br /><br />By now we have built a pyramid of energy types, related to the depth of our view inside the matter.<br /><br /><br /><br />First, we analyzed the <i>mechanical energy</i> - the energy of moving macro-objects.<br /><br /><br /><br />Our next view deep into the world of macro-objects uncovered the <i>molecules</i> - the smallest parts of macro-objects that retain their characteristics. The movement of these molecules was the source of <i>thermal energy</i>, which we often call the <i>heat</i>.<br /><br /><br /><br />Next step inside the molecules uncovered <i>atoms</i>, as the molecules'<br /> components. There are about 100 types of atoms and their composition <br />inside the molecules creates all the thousands of different molecules. <i>Chemical reactions</i><br /> change the composition of atoms in molecules, thereby creating new <br />molecules from the atoms of old molecules. This process broke some <br />inter-atomic bonds and created the new ones and is the source of <i>chemical energy</i>.<br /><br /><br /><br />Now we look deep inside the atoms and find there 3 major <b>elementary particles</b> - electrically positively charged <i>protons</i> and electrically neutral <i>neutrons</i> inside a small but heavy <i>nucleus</i> and electrically negatively charged <i>electrons</i>,<br /> circulating around nucleus on different orbits. For electrically <br />neutral atoms the numbers of protons and electrons are equal. <i>Nuclear energy</i> is hidden inside the nucleus and is the subject of this lecture.<br /><br /><br /><br />The first question we would like to answer is "What holds nucleus, its <br />protons and neutrons, together, considering protons, as electrically <br />positively charged particles must repel each other?"<br /><br /><br /><br />The answer is simple. There are other forces in the Universe, not only <br />electrostatic ones, that act in this case. These intra-nucleus forces <br />that hold the nucleus together are called <i><b>strong forces</b></i>. They are <b>strong</b><br /> because they are the source of attraction between the protons that is <br />stronger than electrostatic repelling. However, these strong forces act <br />only on a very small distance, comparable to the size of a nucleus <br />inside an atom. For example, at a distance <i>10<sup>−15</sup>m</i> the strong force is more than 100 times stronger than electrostatic one.<br /><br /><br /><br />If, by regrouping <i>protons</i> and <i>neutrons</i>, we will be able to create different <i>atoms</i> (inasmuch as regrouping <i>atoms</i> in <b>chemical reaction</b> we create new <i>molecules</i>), a new source of energy, based on <b>strong forces</b>, the <i>nuclear energy</i>, can be uncovered in the course of <b>nuclear reaction</b>.<br /><br /><br /><br />There is another form of <i>nuclear reaction</i> related to <br />transformation of elementary particles. Under certain circumstance a <br />neutron inside a nucleus can transform into proton and, to keep the <br />total electrical charge in balance, it emits an electron. This reaction <br />is called <i>beta-decay</i> and it also produces energy in the form of electromagnetic waves of very high frequency (<i>gamma-rays</i>).<br /><br /><br /><br />Nuclear reactions are a very powerful source of nuclear energy, which is<br /> so much more powerful than other types of energy, that, if misused, it <br />might represent a danger for life on our planet.<br /><br /><br /><br />There is a clear analogy between nuclear and chemical reactions.<br /><br />What happens with atoms in the chemical reaction, happens with protons <br />and neutrons in nuclear reaction. Some atomic bonds break in a chemical <br />reaction, some are created. Some nuclear bonds between protons and <br />neutrons break in a nuclear reaction, some are created.<br /><br /><br /><br />Sometimes the chemical reaction happens by itself, as long as <br />participating substances are close together, but sometimes we have to <br />initiate it, like lighting methane gas with a spark or a flame of a <br />match to initiate continuous burning.<br /><br />Similar approach is valid for nuclear reaction. Sometimes it happens by <br />itself, but sometimes it should be started, like bombarding the nucleus <br />with neutrons, after which it continues by itself.<br /><br /><br /><br />Here is an interesting fact.<br /><br />Physicists have measured the masses of protons, neutrons and many <br />different nuclei that contain these protons and neutrons and have <br />discovered that the sum of masses of individual protons and neutrons is <br />greater than the mass of a nucleus that contain these exact particles.<br /><br />For example,<br /><br />mass of proton is 1.0072766 atomic mass units or 1.6726·10<sup>-27</sup>kg,<br /><br />mass of neutron is 1.0086654 atomic mass units or 1.6749·10<sup>-27</sup>kg.<br /><br />At the same time, mass of deuterium nucleus, that contains 1 proton and 1<br /> neutron is 2.0135532 atomic mass units, which is smaller than the sum <br />of masses of proton and neutron (1.0072766 + 1.0086654 = 2.015942).<br /><br />This so-called "mass defect" is directly related to nuclear energy - the energy of <i>strong forces</i> that hold the nucleus together.<br /><br /><br /><br />A simplified explanation of this effect is based on the law of energy <br />conservation. Consider the force of gravity between a planet and an <br />object above its surface. The object has certain potential energy and, <br />if dropped to the ground, this potential energy transforms into other <br />forms, like kinetic, thermal etc.<br /><br /><br /><br />Similarly, if we consider two independent neutrons (or neutron and <br />proton, or two protons) on a very small distance from each other, but <br />not forming a nucleus, there is a potential energy of the <i>strong forces</i><br /> acting between them. If we let these two particles to form a nucleus, <br />analogously to an object falling towards the surface of a planet, this <br />potential energy should be transformed into other forms, like thermal.<br /><br /><br /><br />Now the Theory of Relativity comes to play, that has established the equivalence of <b>mass</b> and <b>energy</b> by a famous formula <i><b>E=m·c²</b></i>.<br /> According to this equivalence, if some energy is released during the <br />formation of a nucleus from individual protons and neutrons, there must <br />be certain amount of mass released associated with this energy. That is <br />the explanation of "mass defect".<br /><br /><br /><br />It should be noted that to form a nucleus of deuterium from 1 proton and<br /> 1 neutron is easier than to form a nucleus that contains more than one <br />proton, because electrostatic repulsion between positively charged <br />protons prevents their bonding. So, to bring protons sufficiently close <br />to each other for <i>strong forces</i> to overcome the electrostatic <br />repulsion, we have to spend some energy. The net energy released by <br />forming a nucleus from protons and neutrons is the difference between <br />the energy released from <i>strong forces</i> taking hold of these particles inside a nucleus and the energy consumed to overcome repulsion of protons.<br /><br /><br /><br />Actually, as we attempt to form bigger nuclei, the energy we have to <br />spend to overcome electrostatic repulsion forces become greater than <br />amount of energy released by forming a nucleus. This border line is <br />approximately around the nucleus of iron <i><b>Fe</b></i>. Forming iron <br />and heavier elements from protons and electrons is a process that <br />consumes more energy than releases. These heavier nuclei will produce <br />energy, if we reverse the procedure, breaking them into individual <br />protons and neutrons.<br /><br /><br /><br />The mechanisms described above are used in nuclear reactors and atomic <br />bomb, where heavier elements are broken into lighter ones (fission), <br />releasing energy, and in hydrogen bomb, where lighter elements are <br />bonded together to release the energy (fusion).Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-20467534891672103982019-07-29T14:09:00.001-07:002019-07-29T14:09:21.377-07:00Unizor - Physics4Teens - Energy - Atoms and Chemical Reactions - Interat...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Interatomic Bonds</u><br /><br /><br /><br />Atoms in a molecule are bonded together to form a stable chemical substance or compound.<br /><br />The mechanism of bonding is quite complex and different for different <br />molecules. In fact, the complexity of these bonds is outside of the <br />scope of this course. However, certain basic knowledge about molecular <br />bonding and molecular structure is necessary to understand the following<br /> lecture, where we will make certain calculations related to energy <br />produced or consumed in chemical reactions.<br /><br /><br /><br />The key to a mechanism of bonding atoms into molecules lies in an internal structure of atoms.<br /><br />For our purposes we can consider the orbital model of atom as consisting<br /> of electrically positive nucleus and electrically negative electrons <br />circulating on different orbits around a nucleus. <b>This is only a model</b>,<br /> not an exact representation of what's really happening inside the atom,<br /> but this model gives relatively good results that correspond to some <br />simple experiments.<br /><br /><br /><br />Two different particles can be found in a nucleus - positively charged <br />protons and electrically neutral neutrons. The number of protons inside a<br /> nucleus and electrons circulating on different orbits around a nucleus <br />should be the same for electrically neutral atoms in their most common <br />state.<br /><br /><br /><br />For reasons not well understood by many physicists, each orbit can have <br />certain maximum number of electrons that can circulate on it without <br />"bumping" into each other. The higher the orbit - the more electrons it <br />can hold. The lowest orbit can hold no more than 2 electrons, the next -<br /> no more than 8, the next - no more than 14 etc.<br /><br /><br /><br />Consider a few examples.<br /><br /><br /><br />1. Let's consider the structure of a simplest molecule - the molecule of<br /> hydrogen, formed by two atoms of hydrogen. Each hydrogen atom has one <br />electron on the lowest orbit around a nucleus. The maximum number of <br />electrons on this orbit is two, in which case the compound becomes much <br />more stable. So, two atoms of hydrogen grab each other and the two <br />electrons, each from its own atom, are shared by a couple of atoms, thus<br /> creating a stable molecule of hydrogen with symbol <i><b>H<sub>2</sub></b></i>. The bond between two atoms of hydrogen is formed by one pair of shared electrons, so structurally the molecule of hydrogen <i><b>H<sub>2</sub></b></i> can be pictured as<br /><br /><i><b>H−H</b></i>.<br /><br /><br /><br />2. Atom of oxygen has 8 electrons - 2 on the lowest orbit and 6 on the <br />next higher one. The next higher orbit is stable when it has 8 <br />electrons. So, two atoms of oxygen are grabbing each other and share 2 <br />out of 6 electrons on the outer orbit with another atom. So, each atom <br />has 4 "personal" electrons, 2 electrons that it shares with another atom<br /> and 2 electrons that the other atom shares with it. Thus, the orbit <br />becomes full, all 8 spots are filled. The bond between two atoms of <br />oxygen is formed by two pairs of shared electrons, so structurally the <br />molecule of oxygen <i><b>O<sub>2</sub></b></i> can be pictured as<br /><br /><i><b>O=O</b></i><br /><br />(notice double link between the atoms).<br /><br /><br /><br />3. Our next example is gas methane. Its molecule consists of one atom of<br /> carbon (6 electrons, 2 of them on the lowest orbit, 4 - on the next <br />one) and 4 atoms of hydrogen (1 electron on the lowest orbit of each <br />atom). Obviously, having only 4 electrons on the second orbit, carbon is<br /> actively looking for electrons to fill the orbit. It needs 4 of them to<br /> complete an orbit of 8 electrons. Exactly this it finds in 4 atoms of <br />hydrogen that need to complete their own lowest orbit. Sharing <br />electrons, one atom of carbon and 4 atoms of hydrogen fill their <br />corresponding orbits, thus creating a molecule of methane <i><b>CH<sub>4</sub></b></i> with can be pictured as<br /><br /><i><b> H<br /><br /> |<br /><br /> H−C−H<br /><br /> |<br /><br /> H</b></i><br /><br /><br /><br />4. Carbon dioxide molecule contains 1 atom of carbon, that needs 4 <br />electrons to complete its orbit, and 2 atoms of oxygen, each needs 2 <br />electrons to complete its orbit: <i><b>CO<sub>2</sub></b></i>. By <br />sharing 2 electrons from each atom of oxygen with 4 electrons from atom <br />of carbon they all fill up their outer orbit of electrons and become a <br />stable molecule, pictured as<br /><br /><i><b>O=C=O</b></i><br /><br />(notice double link between the atoms).<br /><br /><br /><br />5. Ethanol molecule contains 2 atoms of carbon, 1 atom of oxygen and 6 atoms of hydrogen connected as follows<br /><br /><i><b> H H<br /><br /> | |<br /><br /> H−C−C−O−H<br /><br /> | |<br /><br /> H H</b></i><br /><br />(notice single bond between atoms of carbon and oxygen in ethanol, while<br /> the bond between them in carbon dioxide has double link)<br /><br /><br /><br />6. Hydrogen peroxide molecule contains 2 atoms of hydrogen and 2 atoms of oxygen connected as follows<br /><br /><i><b>H−O−O−H</b></i><br /><br />(notice single bond between atoms of oxygen, not like in a molecule of oxygen)<br /><br /><br /><br />Numerous examples above illustrate that bonds between atoms can be <br />different, even between the same atoms in different molecules. That's <br />why it is important to understand the structure of molecules, how <br />exactly the atoms are linked and what kind of links exist between them. <br />This is the basis for calculation of the amount of energy produced or <br />consumed by chemical reactions that rearrange the atoms from one set of <br />molecules to another.<br /><br /><br /><br />Obviously, bonds <i><b>O−O</b></i> and <i><b>O=O</b></i> are different. <br />The first one is facilitated by one shared electron, the second one - by<br /> two. The amounts of energy, needed to break these bonds, are different <br />too. Therefore, when calculating the energy of chemical reaction, it's <br />important to understand the kind of bond between atoms in each separate <br />case.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-66298229959949839832019-07-23T12:21:00.001-07:002019-07-23T12:21:16.716-07:00Unizor - Physics4Teens - Energy - Chemical Energy of Atomic Bonds<br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Energy of Atomic Bonds<br /><br />in Molecules</u><br /><br /><br /><br />In this lecture we will analyze the energy aspect of chemical reactions.<br /><br />Consider the reaction of burning of methane. This gas is used in regular<br /> gas stoves, so the reaction happens every time we cook something.<br /><br />A molecule of methane consists of one atom of carbon <i><b>C</b></i> and four atoms of hydrogen <i><b>H</b></i>, the chemical formula of methane is <i><b>CH<sub>4</sub></b></i>.<br /><br />You can imagine a molecule of methane as a tetrahedron, in its center is<br /> an atom of carbon and on each of its four vertices is an atom of <br />hydrogen.<br /><br />A molecule of oxygen, as we know, consists of two atoms of oxygen and has a chemical formula <i><b>O<sub>2</sub></b></i>.<br /><br /><br /><br />As a result of the reaction of burning of methane, water and carbon dioxide are produced, according to the following equation:<br /><br /><i><b>CH<sub>4</sub> + 2O<sub>2</sub> = 2H<sub>2</sub>O + CO<sub>2</sub></b></i><br /><br />So, during this reaction<br /><br />(a) four atomic bonds between carbon and hydrogen in one molecule of methane are broken,<br /><br />(b) one atomic bond in each molecule of oxygen (out of two) are broken,<br /><br />(c) two atomic bonds between hydrogen and oxygen in each molecule of water (out of two) are created,<br /><br />(d) two atomic bonds between carbon and oxygen in a molecule of carbon dioxide are created.<br /><br /><br /><br />Amounts of potential energy of the different atomic bonds are <br />experimentally determined, which would lead to calculation of the amount<br /> of chemical energy released (for exothermic) or consumed (by <br />endothermic) reaction.<br /><br /><br /><br />To make experiments to determine potential energy of the bonds inside a <br />molecule, we have to make experiments with known amounts of components <br />in chemical reaction. The reaction above includes one molecule of <br />methane and two molecules of oxygen. Obviously, we cannot experiment <br />with one or two molecules. The solution is to experiment with <b>proportional</b> amounts of components, say, 1 million of molecules of methane and 2 million of molecules of oxygen.<br /><br /><br /><br />To explain how to do this, we have to get deeper into atoms. Physics <br />models atoms as consisting of three kinds of elementary particles - <br />protons (electrically positively charged), neutrons (electrically <br />neutral) and electrons (electrically negatively charged). This is a <br />relatively simple model, that corresponds to most of experiments, though<br /> the reality is more complex than this. For our purposes we can view <br />this model of atom as a nucleus, that contains certain number of protons<br /> and neutrons, and a number of electrons circulating the nucleus on <br />different orbits.<br /><br /><br /><br />Electrons are very light relatively to protons and neutrons, so the mass<br /> of an atom is concentrated, mostly, in its nucleus. Protons and neutron<br /> have approximately the same mass, which is called <i>atomic mass unit</i>. So, the mass of an atom in <i>atomic mass units</i><br /> ("atomic weight") is equal to the number of protons and neutrons in its<br /> nucleus. This mass is known for each element of the Periodic Table of <br />Mendeleev, that is for each known atom.<br /><br />For example, it is determined that atom of hydrogen <i><b>H</b></i> has atomic weight of 1 atomic unit, atom of carbon <i><b>C</b></i> has atomic weight of 12, atom of oxygen <i><b>O</b></i> has atomic weight 16.<br /><br /><br /><br />Knowing atomic weights of atoms, we can calculate atomic weight of molecules. Thus, the atomic weight of a molecule of methane <i><b>CH<sub>4</sub></b></i> is 12+4=16. Atomic weight of a molecule of oxygen <i><b>O<sub>2</sub></b></i> is 16+16=32. Atomic weight of water <i><b>H<sub>2</sub>O</b></i> is 2+16=18.<br /><br /><br /><br />Now we can take components of any chemical reaction proportional to the <br />atomic weight of corresponding molecules, which will result in <br />proportional number of molecules. For example, not being able to <br />experiment with one molecule of methane <i><b>CH<sub>4</sub></b></i> and two molecules of oxygen <i><b>O<sub>2</sub></b></i>, we can experiment with <i><b>16 gram</b></i> of methane and <i><b>64 gram</b></i><br /> of oxygen, and the proportionality of the number of molecules will be <br />preserved - for each molecule of methane there will be two molecules of <br />oxygen.<br /><br /><br /><br />As you see, taking amount of any mono-molecular substance in grams <br />equaled to the atomic weight of the molecules of this substance (called a<br /> <i>mole</i>) assures taking the same number of molecules, regardless of the substance. This number is the Avogadro Number and is equal to <i><b>N=6.02214076·10<sup>23</sup></b></i>.<br /><br />Thus, one <i>mole</i> of methane <i><b>CH<sub>4</sub></b></i> (atomic weight of <i><b>C</b></i> is 12, atomic weight of <i><b>H</b></i> is 1) weighs 16g, one <i>mole</i> of silicon <i><b>Si<sub>2</sub></b></i> (atomic weight of <i><b>Si</b></i> is 14) weighs 28g, one <i>mole</i> of copper oxide <i><b>CuO</b></i> (atomic weight of <i><b>Cu</b></i> is 64, atomic weight of <i><b>O</b></i><br /> is 16) weighs 80g etc. And all those amounts of different substances <br />have the same number of molecules - the Avogadro number (approximately, <br />of course).<br /><br /><br /><br />The theory behind the atomic bonds inside a molecule is quite complex <br />and is beyond the scope of this course. Based on this theory and <br />experimental data, for many kinds of atomic bonds there had been <br />obtained an amount of energy needed to break these bonds, that is its <br />inner chemical energy.<br /><br />Thus, chemical energy of atomic bonds inside a mole of methane <i><b>CH<sub>4</sub></b></i><br /> is 1640 kilo-joules (because a molecule of methane has 4 bonds between <br />carbon and each atom of hydrogen, each bond at 410KJ), inside a molecule<br /> of oxygen <i><b>O<sub>2</sub></b></i> - 494 kilo-joules (1 bond between 2 atoms oxygen at 494KJ), inside a molecule of carbon dioxide <i><b>CO<sub>2</sub></b></i> is 1598 kilo-joules (2 bonds between carbon and each atom of oxygen, each 799KJ), inside a molecule of water <i><b>H<sub>2</sub>O</b></i> is 920 kilo-joules (2 bonds between oxygen and each atom of hydrogen, each 460KJ).<br /><br /><br /><br />Let's go back to methane burning:<br /><br /><i><b>CH<sub>4</sub> + 2O<sub>2</sub> = 2H<sub>2</sub>O + CO<sub>2</sub></b></i><br /><br />This chemical reaction converts 1 mole of methane (16g) and 2 moles of <br />oxygen (64g) into 1 mole of carbon dioxide (44g) and 2 moles of water <br />(36g).<br /><br />The energy we have to spend to break the atomic bonds of 1 mole of methane and 2 moles of oxygen, according to above data, is<br /><br /><i><b>E<sub>in</sub> = 1640 + 2·494 = 2628 KJ</b></i><br /><br />The energy we have to spend to break atomic bonds of 2 moles of water and 1 mole of carbon dioxide, according to above data, is<br /><br /><i><b>E<sub>out</sub> = 2·920 + 1598 = 3438 KJ</b></i><br /><br />The net energy is<br /><br /><i><b>E<sub>net</sub> = 2628 − 3438 = −810 KJ</b></i><br /><br />This net energy is the amount of thermal energy released by burning 16g <br />of methane, using 64g of oxygen, obtaining as a result 44g of carbon <br />dioxide and 36g of water.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-52607433522978610102019-07-18T14:41:00.001-07:002019-07-18T14:41:54.070-07:00Unizor - Physics4Teens - Energy - Atoms<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Atoms and Chemical Reaction</u><br /><br /><br /><br />Discussing <i>mechanical energy</i>, we analyzed the movement of objects.<br /><br />When talking about <i>thermal energy (heat)</i>, we had to go deeper inside the objects and analyzed the movement of <i>molecules</i>, the smallest parts of objects that retained the properties of objects themselves.<br /><br />Now we go even deeper, inside the molecules, in search of new kinds of energy.<br /><br /><br /><br />The components of <i>molecules</i> are called <i>atoms</i>. Currently there are more than 100 kinds of <i>atoms</i>, classified in the Mendeleev's Periodic Table.<br /><br /><br /><br />Different combinations of these <i>atoms</i> in different quantities make up all kinds of <i>molecules</i>, each with its own properties.<br /><br /><br /><br />In some cases a single <i>atom</i> makes up a <i>molecule</i>. For example, a single atom of iron (denoted by symbol <b>Fe</b>) makes up a molecule of iron.<br /><br />In some other cases a pair of <i>atoms</i> of the same type makes up a <i>molecule</i>. For example, two atoms of oxygen (denoted by symbol <b>O</b>) make up a molecule of oxygen (denoted by symbol <b>O<sub>2</sub></b>).<br /><br />In more complicated cases a few <i>atoms</i> of different types make up a <i>molecule</i>. For example, two atoms of hydrogen (denoted by symbol <b>H</b>) and one atom of oxygen (<b>O</b>) make up a molecule of water (<b>H<sub>2</sub>O</b>).<br /><br /><br /><br />One of the most complicated molecules that contains many elements in <br />different quantities is a molecule of protein that has about half a <br />million of atoms.<br /><br /><br /><br /><b>Chemical energy</b> is a <i>potential energy</i> of bonds between <i>atoms</i> that hold them together in a <i>molecule</i>.<br /><br /><b>Chemical reaction</b> is a process of re-arranging of <i>atoms</i> in a group of <i>molecules</i>, getting, as a result, a group of other <i>molecules</i>.<br /><br />During <b>chemical reactions</b> some bonds between atoms are broken and some are created. Therefore, the <u>energy might be either released or consumed</u> in the process of <b>chemical reaction</b>.<br /> This energy, stored in the molecules as potential energy of atomic <br />bonds and released or consumed during chemical reaction, is classified <br />as <i>chemical energy</i>.<br /><br /><br /><br />Let's consider a few examples of chemical energy.<br /><br /><br /><br />1. <i>Coal burning</i><br /><br />One molecule of carbon, that consists of one carbon atom <i><b>C</b></i>, and one molecule of oxygen, that consists of two oxygen atoms <i><b>O<sub>2</sub></b></i>, when brought together and lit up, will join into one molecule of carbon dioxide <i><b>CO<sub>2</sub></b></i><br /> in the process of burning. After the chemical reaction of burning is <br />initiated, it will maintain itself, as the process of burning produces a<br /> flame that lights up new molecules of carbon, joining them with oxygen.<br /><br />The chemical reaction<br /><br /><i><b>C + O<sub>2</sub> = CO<sub>2</sub></b></i><br /><br />is endothermic (consumes heat energy) in the very beginning, when we <br />have to light up the carbon, but, as soon as the reaction started, it <br />becomes exothermic, that is it produces heat energy, because the <br />potential energy of atoms inside molecules of carbon and oxygen together<br /> is greater than potential energy of atoms inside a molecule of carbon <br />dioxide.<br /><br /><br /><br />2. <i>Making water from hydrogen and oxygen</i><br /><br />Two molecules of hydrogen, each consisting of two hydrogen atom <i><b>H<sub>2</sub></b></i>, and one molecule of oxygen, that consists of two oxygen atoms <i><b>O<sub>2</sub></b></i>, when brought together and lit up, will join into two molecules of water <i><b>H<sub>2</sub>O</b></i><br /> in the process of hydrogen burning. After the chemical reaction of <br />burning is initiated, it will maintain itself, as the process of burning<br /> produces a flame that lights up new molecules of hydrogen, joining them<br /> with oxygen.<br /><br />The chemical reaction<br /><br /><i><b>2H<sub>2</sub> + O<sub>2</sub> = 2H<sub>2</sub>O</b></i><br /><br />is endothermic (consumes heat energy) in the very beginning, when we <br />have to light up the hydrogen, but, as soon as the reaction started, it <br />becomes exothermic, that is it produces heat energy, because the <br />potential energy of atoms inside two molecules of hydrogen and one <br />molecule of oxygen together is greater than potential energy of atoms <br />inside two molecules of water.<br /><br /><br /><br />3. <i>Photosynthesis</i><br /><br />This is a complicated process, during which the light from sun, air <br />components (such as carbon dioxide, nitrogen and oxygen), water and <br />whatever is in the soil are converted by the plants into chemical energy<br /> that maintains their life. This is an endothermic process, and, as its <br />result, plants grow. In most cases they consume carbon dioxide from the <br />air, break it into carbon and oxygen, consume the water from the soil, <br />break it into hydrogen and oxygen (they need sun's radiation energy to <br />break the molecules of <i><b>CO<sub>2</sub></b></i> and <i><b>H<sub>2</sub>O</b></i>),<br /> use the carbon, hydrogen and part of oxygen to produce new organic <br />molecules they consist of and release the unused oxygen back into <br />atmosphere.<br /><br /><br /><br />4. <i>Battery</i><br /><br />Battery consists of three major components: <b>anode</b>, <b>cathode</b> and <b>electrolyte</b><br /> in-between anode and cathode. As a result of a chemical and <br />electro-magnetic reaction between the molecules of anode and <br />electrolyte, some electrons are transferred from anode to electrolyte. <br />Then, as a result of a chemical and electro-magnetic reaction between <br />the molecules of cathode and electrolyte, some electrons are transferred<br /> from electrolyte to cathode. As the result, there are extra electrons <br />on the cathode, which were taken from the anode, thus creating <br />electrical potential.<br /><br /><br /><br />These simple examples explain the general mechanism of chemical energy, <br />released or consumed in the course of chemical reaction, that transforms<br /> molecules by rearranging their atoms' composition. As a result of a <br />chemical reaction and change in the atomic composition of molecules, <br />potential energy of bonds between atoms in molecules is changing. If the<br /> total potential energy of the resulting molecules is greater than the <br />potential energy of the bonds inside original molecules, the process is <br />endothermic, it consumes energy. In an opposite case the process is <br />exothermic, it produces energy.<br /><br /><br /><br />The exothermic process of extracting chemical energy using chemical <br />reaction is the key to getting energy from gasoline in the car engine, <br />producing heat and light in the fireplace by burning wood, it's the <br />source of energy in all living organisms, including humans. We exist <br />because our body knows how to extract chemical energy from the food.<br /><br /><br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-28046097549836138762019-07-08T12:35:00.001-07:002019-07-08T12:35:28.868-07:00Unizor - Physics4Teens - Energy - Heat Transfer Problems<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Heat Transfer - Problems</u><br /><br /><br /><br /><i>Problem 1</i><br /><br /><br /><br />Determine the power of the heat source inside the room required to <br />maintain a certain difference between inside and outside air <br />temperature, given the following:<br /><br />(a) the difference between inside and outside temperature <i><b>δ=T<sub>out</sub>−T<sub>home</sub></b></i><br /><br />(b) the room has only one wall facing the outside air, and the area of this wall is <i><b>A</b></i><br /><br />(c) the thickness of the wall is <i><b>L</b></i><br /><br />(d) the wall is made of solid material with a coefficient of <i>thermal conductivity</i> <i><b>k</b></i>.<br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />Let <i><b>x</b></i> be the distance from a point inside the wall to its surface facing the room.<br /><br />The <i><b>T(x)</b></i> is a temperature at this point as a function from this distance.<br /><br />To be equal to <i><b>T<sub>home</sub></b></i> at the surface facing the room and to be <i><b>T<sub>out</sub></b></i> at the surface facing the outside and to be linearly changing from one value to another inside the wall of the thickness <i><b>L</b></i>, the temperature inside the wall at distance <i><b>x</b></i> from the surface facing the room should be<br /><br /><i><b>T(x)=T<sub>home</sub>+(T<sub>out</sub>−T<sub>home</sub>)·x/L</b></i><br /><br />From this we can determine the <i>heat flux</i> through the wall at a distance <i><b>x</b></i>, using the Fourier's law of thermal conduction:<br /><br /><i><b>Q(x,A) = −k·A·</b>d<b>T(x)/</b>d<b>x =<br /><br />= −k·A·(T<sub>out</sub>−T<sub>home</sub>)/L =<br /><br />= −k·A·δ/L</b></i><br /><br />That is exactly how much heat we need to maintain the difference between temperature in the room and outside.<br /><br /><br /><br /><i>Problem 2</i><br /><br /><br /><br />Calculate the heating requirement of the room with only one concrete <br />wall facing outside with no windows, assuming the following:<br /><br />(a) the temperature in the room must constantly be <i>T<sub>room</sub>=20°C</i><br /><br />(b) the temperature outside is also constant <i>T<sub>out</sub>=5°C</i><br /><br />(c) the thickness of the concrete wall facing outside is <i>L=0.2m</i><br /><br />(d) the area of the wall facing outside is <i>A=12m²</i><br /><br />(e) <i>heat conductivity</i> of concrete is <i>k=0.6W/(m·°K)</i><br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />Using the above, we can determine the heat flux through the wall at a distance <i><b>x</b></i>:<br /><br /><i><b>Q(x,A) = −k·A·δ/L</b></i><br /><br />(for any distance <i><b>x</b></i> from the surface of the wall that faces the room)<br /><br />Outside surface of the wall is at 5°C, inside is at 20°C.<br /><br />So, <i><b>δ=15°</b></i>.<br /><br />Therefore, the <i>heat flux</i> through the wall (at any distance from the inside surface) will be<br /><br /><i><b>Q = 0.6·12·15/0.2 = 540W</b></i><br /><br />That is exactly how much heat we need to maintain the temperature in the room.<br /><br /><br /><br /><i>Problem 3</i><br /><br /><br /><br />Our task is to determine the law of cooling of a relatively small hot <br />object immersed in the cool infinitely large reservoir with liquid or <br />gaseous substance.<br /><br />We assume the volume of substance this object is immersed in to be <br />"infinite" to ignore its own change of temperature related to heat <br />emitted by our hot object.<br /><br />This law of cooling should be expressed in terms of object's temperature <i><b>T</b></i> as a function of time <i><b>t</b></i>, that is, we have to find the function <i><b>T(t)</b></i>.<br /><br />Assumptions:<br /><br />(a) the initial temperature of the object at time <i><b>t=0</b></i> is assumed to be <i><b>T<sub>0</sub></b></i><br /><br />(b) the object has a shape of a thin flat square (so, its temperature is changing simultaneously in all its volume) of size <i><b>L</b></i>x<i><b>L</b></i> and mass <i><b>m</b></i><br /><br />(c) the <i>specific heat capacity</i> of the object's material is <i><b>C</b></i><br /><br />(d) the temperature of the substance surrounding our object is constant and equals <i><b>T<sub>s</sub></b></i><br /><br />(e) the <i>convective heat transfer coefficient</i> of the substance around our object is <i><b>h</b></i>.<br /><br /><br /><br /><i>Solution</i><br /><br /><br /><br />Consider a small time interval from <i><b>t</b></i> to <i><b>t+</b></i>Δ<i><b>t</b></i>.<br /><br />The temperature of an object is <i><b>T(t)</b></i>, while the temperature of the substance around it is constant <i><b>T<sub>s</sub></b></i>.<br /> Then the amount of heat transferred from our object to the substance <br />around it per unit of time per unit of its surface area is proportional <br />to the difference in temperatures with the <i>convective heat transfer coefficient</i> as a factor:<br /><br /><i><b>q = −h·</b></i>[<i><b>T(t)−T<sub>s</sub></b></i>]<br /><br />Since total surface area of our thin flat square, ignoring its thickness, is <i><b>2L²</b></i> and the time interval we consider is Δ<i><b>t</b></i>, the total amount of heat transferred by our object through its surface during this time period is<br /><br />Δ<i><b>Q(t) = −2L²·h·</b></i>[<i><b>T(t)−T<sub>s</sub></b></i>]<i><b>·</b></i>Δ<i><b>t</b></i><br /><br />This is amount of heat taken away by the substance from our object during a time interval Δ<i><b>t</b></i>.<br /><br />The same amount of heat is lost by the object, taking its temperature from <i><b>T(t)</b></i> to <i><b>T(t+</b></i>Δ<i><b>t)</b></i>.<br /><br />As we know, changes of heat and temperature are proportional and related to the object mass and specific heat capacity:<br /><br />Δ<i><b>Q(t) = C·m·</b></i>Δ<i><b>T</b></i><br /><br />where Δ<i><b>T = T(t+</b></i>Δ<i><b>t)−T(t)</b>.</i><br /><br />Equating the amount of heat lost by our object to the amount of heat <br />carried away by convection of the substance around it, we have come to <br />an equation<br /><br /><i><b>−2L²·h·</b></i>[<i><b>T(t)−T<sub>s</sub></b></i>]<i><b>·</b></i>Δ<i><b>t =<br /><br />= C·m·</b></i>[<i><b>T(t+</b></i>Δ<i><b>t)−T(t)</b></i>]<br /><br />Dividing both parts by Δ<i><b>t</b></i>, diminishing this time interval to zero and using a derivative by time <i><b>t</b></i> to express the limit, we get the following differential equation<br /><br /><i><b>−2L²·h·</b></i>[<i><b>T(t)−T<sub>s</sub></b></i>]<i><b> =<br /><br />= C·m·</b>d<b>T(t)<span style="font-size: medium;">/</span></b>d<b>t</b></i><br /><br />To solve it, let's make two simple substitutions:<br /><br />(a) <i><b>A = 2h·L²/(C·m)</b></i><br /><br />(b) <i><b>X(t) = T(t) − T<sub>s</sub></b></i><br /><br />Then, since <i><b>T<sub>s</sub></b></i> is constant,<br /><br /><i>d<b>T(t)<span style="font-size: medium;">/</span></b>d<b>t = </b>d<b>X(t)<span style="font-size: medium;">/</span></b>d<b>t</b></i><br /><br />and our differential equation looks like<br /><br /><i>d<b>X(t)<span style="font-size: medium;">/</span></b>d<b>t = −A·X(t)</b></i><br /><br />To solve this, we convert it as follows:<br /><br /><i>d<b>X(t)<span style="font-size: medium;">/</span>X(t) = −A·</b>d<b>t</b></i><br /><br /><i>d</i>[<i>ln<b>(X(t))</b></i>]<i><b> = −A·</b>d<b>t</b></i><br /><br />Integrating:<br /><br /><i>ln<b>(X(t))</b></i>]<i><b> = −A·t + B</b></i>,<br /><br />where <i><b>B</b></i> is any constant, defined by initial condition<br /><br /><i><b>X(0)=T(0)−T<sub>s</sub>=T<sub>0</sub>−T<sub>s</sub></b></i>.<br /><br />From the last equation:<br /><br /><i><b>X(t) = e<sup>B</sup>·e<sup>−A·t</sup></b></i><br /><br />Using initial condition mentioned above,<br /><br /><i><b>X(t) = (T<sub>0</sub>−T<sub>s</sub>)·e<sup>−A·t</sup></b></i><br /><br />Since <i><b>X(t)=T(t)−T<sub>s</sub></b></i><br /><br /><i><b>T(t) − T<sub>s</sub> = (T<sub>0</sub>−T<sub>s</sub>)·e<sup>−A·t</sup></b></i><br /><br />or<br /><br /><i><b>T(t) = T<sub>s</sub> + (T<sub>0</sub>−T<sub>s</sub>)·e<sup>−A·t</sup></b></i><br /><br />where <i><b>A = 2h·L²/(C·m)</b></i><br /><br />So, the difference in temperature between a hot object and infinitely <br />large surrounding substance is exponentially decreasing with time.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-21346508164714122752019-06-25T14:41:00.000-07:002019-06-25T14:46:12.921-07:00Unizor - Physics4Teens - Energy - Heat Transfer - Radiation<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/ABTz7UVEuQo" width="480"></iframe><br /><br /><br/><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Heat Transfer - Radiation</u><br /><br /><i>Heat radiation</i> IS NOT the same as <i>radioactivity</i>. Though, under certain circumstances (like an explosion of an atomic bomb) the heat radiation and radioactivity are both present. When we discuss the <i>heat radiation</i> we talk about a process that occurs in any object with a temperature greater than absolute zero, while <i>radioactivity</i> occurs in extreme cases of very high energy output.<br/><br/>Heat transfer through <i>radiation</i> is totally different from <i>conduction</i> and <i>convection</i>. The most important property of heat transfer by <i>radiation</i> is that heat transfer occurs without any visible material conduit that carries the heat, like molecular movement in two other cases.<br/><br/>Let's start from an example.<br/>The brightest example is our Sun, as a source of heat energy. Between Sun and Earth there is no visible material conduit, yet the heat comes to Earth and is a source of life on our planet.<br/><br/>The fundamental concept that lies in the foundation of a process of <i>heat radiation</i> is a concept of a <i><b>field</b></i>.<br/>The <i>field</i> is a region of space, where certain forces act on certain objects without visible material medium.<br/>As an example of the <i>field</i>, consider <i>gravity</i>. The Sun keeps planets on their orbits, the Earth keeps the Moon circling around, people are walking on the ground without flying away to stars etc. We did not know much about WHY the <i>gravitational field</i> exist, yet we did study its behavior, the forces involved and the laws of motion in this field.<br/><br/>There are other fields.<br/><i>Magnetic field</i> around our planet, acting on a compass, forces the arrow to point North.<br/><i>Electric field</i> exists around electrically charged objects, so other electrically charged objects are attracted to or repulsed from it.<br/><br/>In Physics we successfully study these fields, but complete understanding of WHY they have the properties that we observed is not completely clear. So, we will concentrate on properties, answering the question HOW?, not on a more fundamental question WHY?.<br/><br/>Let's start with a particular field called <i>electro-magnetic</i>. Very simplified description of this field is as follows.<br/><br/>Any electron creates an <i>electric field</i> around itself. Moving electrons, which we call <i>electric current</i> or <i><b>electric field that changes in time</b></i>, also create a <i>magnetic field</i> around them. So, <i><b>changing in time electric field creates magnetic field that changes in space</b></i>. It's an experimental fact, and we have the whole theory about properties of these fields.<br/><br/>Consider an experiment, when you move a metal rod or any other electrical conductor in a <i>magnetic field</i> or change a <i>magnetic field</i> around any electrical conductor, thus creating a <i><b>magnetic field that changes in time</b></i>. You will observe that there is an electric current in the conductor, thus creating an <i><b>electric field that changes in space</b></i>. It's an experimental fact, and we also have the whole theory about properties of this process.<br/><br/>So, <i>electric field</i> creates <i>magnetic field</i>, which, in turn, creates <i>electric field</i> etc. This is a loop of energy conversion that propagates extremely fast, with a speed of light, about 3·10<sup>8</sup>m/sec.<br/><br/>The combination of electric and magnetic forces form <i>electro-magnetic field</i> that propagates much faster than its physical medium - electrons. So, the propagation of the <i>electro-magnetic field</i> seems to be a self-sufficient process, occurring without the medium. This is a very brief and unsatisfactory explanation of the nature of the <i>electro-magnetic field</i>. We will not go much further in this explanation, but rather concentrate on the properties of the <i>electro-magnetic field</i>.<br/><br/>Assuming that we accept the existence of the <i>electro-magnetic field</i> and, however uncomfortably we feel about it, but accept that there is no need for a medium to propagate this field, we can talk about frequency of electro-magnetic transformations, that can be considered similar to oscillation of molecules in a solid. Inasmuch as the oscillations of molecules in a metal are propagated, thus transferring heat energy from hot area to cold one, oscillations of the electric and magnetic components of the <i>electro-magnetic field</i> transfers energy.<br/>This energy transfer by <i>electro-magnetic field</i> is called <i><b>radiation</b></i>.<br/><br/>As in a case of oscillating molecules in a solid, carrying more heat energy when oscillation is more intense (higher frequency), the <i>electro-magnetic field</i> oscillation carries energy with higher frequencies being more "energetic" than lower.<br/><br/>Interestingly, receptors in our skin feel the temperature of a solid object, that is, we feel the intensity of oscillation of its molecules. Similarly, we feel the warm rays of Sun on our skin, that is, we feel the intensity of electro-magnetic oscillation of the electro-magnetic field.<br/>What's more remarkable, we see the light. Apparently, electro-magnetic oscillations in certain frequency range act upon censors in our eyes, thus we see the light. Moreover, in this visible range of frequencies different frequencies of electro-magnetic oscillation produce effect of different colors in our eyes.<br/><br/>As you see, the light and heat of radiation have the same source - the oscillation of electro-magnetic field, the only difference is the frequency. In other words, the <i>light</i> and <i>heat radiation</i> are manifestations of the same process of transferring energy by the oscillations of the components of the <i>electro-magnetic field</i>.<br/><br/>An object does not have to have a temperature of the Sun to emit <i>heat radiation</i>. All objects that have temperature higher than absolute zero emit thermal radiation of some frequencies. Usually, the whole spectrum of frequencies of electro-magnetic oscillations is emitted by objects. Lower frequencies (usually called infrared) are felt by skin receptors, higher frequencies are visible by an eye. Frequencies higher than those visible by a human eye are called ultraviolet. Even higher frequencies are called X-rays, which can be produced by special equipment and, depending on intensity and time of exposure, can represent a health hazard. Even higher intensity and high frequencies are called gamma rays, and they are produced in extreme cases like nuclear explosion or nuclear reactor meltdown, and they are extremely dangerous and are usually meant, when the term <i>radioactivity</i> is used. All frequencies can be observed using some scientific instruments.<br/><br/>Any object, placed in the outer space will emit its heat energy through radiation until its temperature will reach absolute zero. Our Sun emits huge amounts of energy in all spectrum of frequencies in all directions and, eventually, run out of heat energy and go dark.<br/><br/>The intensity of radiation, that is amount of heat radiated per unit of time per unit of area of an object depends, as in other cases of heat transfer, on the temperature of an object and temperature of surrounding environment.<br/>In the complete vacuum with no other source of energy the radiation intensity of an object is proportional to the fourth degree of its absolute temperature in °K:<br/><i><b>q = </b>σ<b>·T<sup>4</sup></b></i> where<br/><i>σ = 5.67·10<sup>−8</sup> W/(m<sup>2</sup>·°K<sup>4</sup>)</i><br/>is the Stefan-Boltzmann constant.<br/>This is the Law of Stefan-Boltzmann. Its derivation is complex and is outside of the scope of this course.<br/><br/>Radiation is not only emitted by objects with temperatures above absolute zero, but also can be absorbed by them and even reflected. While ability to absorb the heat is common for other heat transfer types (conduction and convection), reflection is a specific property of <i>heat radiation</i>. More precisely, it's a specific property of oscillations of the electro-magnetic field.<br/>Obvious application of this property is the usage of mirrors that reflect the oscillations of the electro-magnetic field in a very broad spectrum of frequencies, including the visible light.<br/>An example of absorbed radiation is a slice of bread toasted in the electric toaster. It absorbs the thermal radiation emitted by electric coils, that changes the bread's structure.<br/><br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-9125641408687227262019-06-21T14:21:00.001-07:002019-06-21T14:21:58.347-07:00Unizor - Physics4Teens - Energy - Heat Transfer - Convection<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/WfhCtIcRJTU" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Heat Transfer - Convection</u><br /><br />As in a case of conduction, we start with a statement: <b>heat</b> is a form of <b>internal energy</b> that is related to molecular movement.<br />However, while heat transfer during the process of <i>conduction</i> occurs between molecules oscillating around their relatively fixed positions and transferring their internal energy by "shaking" the neighboring molecules, <i>convection</i> occurs when molecules are free to travel in different directions and carry their internal energy with them.<br /><br />In other words, <i>conduction</i> is a pure transfer of energy on a micro level from one oscillating molecule in a relatively fixed position to another such molecule, while <i>convection</i>occurs when molecules freely fly away from their positions, carrying their internal energy with themselves, thus transferring energy on a macro level.<br /><br />It should be noted that, when dealing with solid objects, <i>conduction</i> is a prevailing way of heat transfer, while in liquids and gases the main way of heat transfer is <i>convection</i>. It does not mean that <i>conduction</i> does not occur in liquids or gases, it does, but it does not constitute the major way of heat transfer. Much more heat is transferred through the mechanism of <i>convection</i><br /><br />Here are a few examples of heat transfer through <i>convection</i>:<br />(a) heating up water in a pot; heat is carried from hot bottom of a pot up by hot (fast moving with high kinetic energy) molecules;<br />(b) circulation of air in the atmosphere from hot places to cold;<br />(c) circulation of water in oceans from hot places to cold.<br /><br />Describing <i>convection</i>mathematically is not a simple task.<br />While in case of <i>conduction</i> we can use a relatively simple Fourier's Law of Thermal Conduction<br /><i><b>q(x) = −k·</b>d<b>T(x)<span style="font-size: medium;">/</span></b>d<b>x</b></i><br />that describes the heat flow as a function of how fast the temperature between the layers of conducting material changes (<i>d<b>T(x)<span style="font-size: medium;">/</span></b>d<b>x</b></i>) and properties of the material itself (<i>conductivity coefficient <b>k</b></i>), the process of <i>convection</i> is significantly more complex, described by convection-diffusion differential equations that are beyond the scope of this course.<br /><br />However, for practical purposes we can use a similar formula that puts the amount of heat transferred by <i>convection</i>process in a liquid or gas during a unit of time through a unit of area as proportional to a difference of temperatures between the layers of liquid or gas and a <i>convective heat transfer coefficient <b>h</b></i> that depends on the physical properties of this liquid or gas:<br /><i><b>q = −h·(T<sub>2</sub>−T<sub>1</sub>)</b></i><br /><br />This formula puts amount of heat <i><b>q</b></i> going through a layer of a unit area of liquid or gas during a unit of time as proportional to a difference of temperatures between bounding surfaces of this layer <i><b>T<sub>2</sub>−T<sub>1</sub></b></i> and some physical properties of liquid or gas expressed in <i>convective heat transfer coefficient <b>h</b></i> that, in turn, depend on such properties as <i>viscosity</i>, <i>density</i>, the type of flow (turbulent or laminar) etc.<br /><br />Consider an example.<br />A round steam pipe of temperature 100°C goes through a room with air temperature 25°C. We have to calculate the amount of heat from the pipe to select an air conditioner required to neutralize the heat from a pipe and keep the room temperature at that level.<br />Assume that the pipe's length is 4m, diameter 0.2m and the <i>convective heat transfer coefficient</i> of air is 40J/(sec·m²·°C). As we know, J/sec is a unit called "watt", so we will use W instead of J/sec.<br /><br />The heat transfer per unit of time through a unit of area of a pipe is, therefore,<br /><i><b>q = 40·(100−25) = 3000(W/m²)</b></i><br />The pipe's area is<br /><i><b>A = π·0.2·4 = 2.512(m²)</b></i><br />Therefore, the pipe is producing the following amount of heat:<br /><i><b>Q = 3000·2.512 = 7536(W)</b></i><br /><br />So, we need an air conditioner that can extract 7536W of heat from the room to maintain stable temperature of 25°C.<br />Usually, the power of air conditioners is measured in BTU/hr (1 watt = 3.41 BTU/hr). So we need an air conditioner of approximately 2200 BTU/hr - a relatively small one.<br /><br />Another example.<br />Outside temperature is 40°C, inside a room we want temperature 25°C. The glass wall between a room and outside air has an area of 20m². What kind of air conditioner is needed to maintain the room temperature at 25°C, assuming the <i>convective heat transfer coefficient</i> of air is 40W/(m²·°C)?<br /><br /><i><b>Q = 40·(40−25)·20 = 12000(W)</b></i><br />This is equivalent to about 3500 BTU/hr.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-2963730674516771812019-06-20T12:36:00.001-07:002019-06-20T12:36:23.797-07:00Unizor - Physics4Teens - Energy - Heat Transfer - Conduction<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/0UaoqluO4OE" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Heat Transfer - Conduction</u><br /><br />As we know, heat is a form of internal energy that is related to molecular movement.<br /><br />For solids the molecular movement is usually restricted to molecules' oscillation around some neutral positions.<br /><br />For liquids the freedom of molecular motion is greater, but still restricted by external forces, like gravity, and surface tension. The average distance between molecules of liquids is relatively constant.<br /><br />Gas molecules are usually taking all the space available for them. Such forces as gravity also restrict their movement (otherwise, the air molecules would fly away from our planet), but still allow substantial freedom. The average distance between molecules of gas mostly depends on a reservoir the gas is in, the larger the reservoir - the larger average distance between molecules.<br /><br />Transfer of heat is transfer of molecular movement from one object or part of an object to another object or part of an object, from an object or part of an object with more intense molecular movement (relatively warmer) to an object or part of an object with less intense movement (relatively cooler).<br /><br />There are three major ways to transfer heat from a hot object to a cold one:<br /><i>Conduction</i>,<br /><i>Convection</i>,<br /><i>Radiation</i>.<br />This lecture explains a concept of <i>conduction</i>.<br /><br /><i><b>Conduction</b></i><br /><br /><i>Conduction</i> of heat energy is a transfer of molecular movement mostly applicable to solid objects - between two solid objects that touch each other, having an area of a contact, or within one object, one part of it having different temperature than another.<br />The <i>conductivity</i> is present in heat transfer in liquids and gases, but there it's usually combined with another form of heat transfer - <i>convection</i>, while in solids it's not the case, and we can study <i>conductivity</i> by itself.<br /><br />An example is a building wall, one side of which towards the outside having temperature of the air outside, while inner surface of the wall having room temperature. The heat energy constantly flows from a warmer surface of the wall to the opposite cooler one with some <i>rate of flow</i> that depends on the <i>thermal conductivity</i> of the wall material. The wall material with higher level of <i>thermal conductivity</i> will transfer more heat energy to a cooler side within unit of time per unit of area, which is usually not a desirable property of the building walls.<br /><br />The mechanism of heat transfer through <i>conductivity</i> can be explained as follows.<br />Imagine two objects (or two parts of the same object), a hot one with higher intensity of molecular movement and a cold one with lower intensity level of molecular movement, that touch each other along some surface, while completely insulated from heat around them. For example, you put a cold metal spoon into a styrofoam cup with hot tea.<br /><br />Molecules of a hot object are hitting the molecules of a cold one, thus forcing the molecules of a cold object to move faster. These faster molecules of a cold object, in turn, hit their neighbors, forcing them to move faster. This process of transferring heat energy through contacting surfaces continues until the intensity of molecular movement gradually equalizes on average. A hot object will lose some energy of molecular movement, while a cold one will gain it. As a result, the temperatures of both will equalize.<br /><br />Since the heat energy is a kinetic energy of molecular movement, that is a sort of mechanical energy, we expect that the total amount of energy for an isolated system will remain constant, whatever a hot object loses in its kinetic energy of molecular movement will be gained by a cold object. The total amount of heat energy will remain the same.<br /><br />If you put a silver spoon into a cup of hot tea, it will heat up faster than a spoon made of steel, which, in turn, will heat up much faster then a spoon made of wood. The reason for this is that the <i>thermal conductivity</i> of different materials is different.<br /><br />We can experimentally measure the thermal conductivity of different solids by having a standard rod of any solid material at certain starting temperature and heating its one end by bringing it to contact with some hot object. Measuring the temperature on the other end after different time intervals will give us a picture of growing temperature.<br /><br />Some materials with higher thermal conductivity will have the temperature at the opposite end of a rod growing faster than in case of other materials.<br />Metals have much higher heat conductivity then plastic or wood, for example. That's why the handle of a tea kettle is usually made of plastic or wood. Diamonds have one of the highest thermal conductivity, even higher than silver.<br /><br />More precise definition of <i>thermal conductivity</i> is related to a concept of <i>heat flux</i>(sometimes, called <i>heat flow density</i> or <i>thermal flux</i>, or <i>thermal flow density</i>). <b>Heat flux is an amount of heat energy flowing through a unit of area during a unit of time</b>.<br /><br />Let's examine how heat flows through a building wall made of some uniform material from a warm room to cold air outside the building.<br />Assume, the room temperature is <i><b>T<sub>room</sub></b></i> and the cold air outside the building has temperature <i><b>T<sub>air </sub></b></i>. If the thickness of a wall is <i><b>L</b></i>, the temperature inside the wall <i><b>T(x)</b></i>, as a function of the distance <i><b>x</b></i> from the surface facing outside, gradually changes from <i><b>T(0)=T<sub>air</sub></b></i> to <i><b>T(L)=T<sub>room</sub></b></i>.<br /><br />It is intuitively understandable and experimentally confirmed that amount of heat energy flowing through a unit of area of such a wall during a unit of time (<i>heat conductivity</i> of a wall) is proportional to a difference between temperatures <i><b>T<sub>room</sub></b></i>and <i><b>T<sub>air</sub></b></i> and inversely proportional to a thickness of a wall <i><b>L</b></i>:<br /><i><b>q = −k·(T<sub>room</sub> − T<sub>air </sub>) <span style="font-size: medium;">/</span> L</b></i><br />(negative sign is used because the flow of heat is opposite to a direction of temperature growth).<br /><br />The situation with <i>heat flux</i>might be compared with a water flow down a river between two points A and B. The difference in levels above the sea level of these points is similar to a difference in temperature between the inside and outside walls of a building. The distance between points A and B is similar to a thickness of a wall. It's reasonable to assume that amount of water flowing through a unit of area in a unit of time will be proportional to a difference between the levels of points A and B above the sea level and inversely proportional to a distance between these two points.<br /><br />To make this definition of the <i>heat flux</i> more precise and independent of the way how the heat flows inside the wall, let's consider a thin slice of wall parallel to both sides from a point at distance <i><b>x</b></i> from the outdoor cold side to a point at distance <i><b>x+</b></i>Δ<i><b>x</b></i>.<br />The heat flow through this thin slice of a wall, as a function of distance <i><b>x</b></i>, can be expressed similarly to the above:<br /><i><b>q(x)=−k·</b></i>[<i><b>T(x+</b></i>Δ<i><b>x)−T(x)</b></i>]<i><b> <span style="font-size: medium;">/</span></b></i>Δ<i><b>x</b></i><br /><br />Next step is, obviously, to reduce the thickness of the slice by making Δ<i><b>x</b></i> infinitesimal, that is Δ<i><b>x→0</b></i>, which leads to the following definition of the <i>heat flux</i>:<br /><i><b>q(x) = −k·</b>d<b>T(x)<span style="font-size: medium;">/</span></b>d<b>x</b></i><br />This definition was formulated by Fourier in 1822 and is called <i>Fourier's law of thermal conduction</i>.<br />The coefficient <i><b>k</b></i> is called <i>thermal conductivity</i>.<br /><br />To find the amount of heat <i><b>Q(x,A)</b></i> going through an area <i><b>A</b></i>at distance <i><b>x</b></i> from the outside wall during a unit of time, we have to multiply the <i>heat flux</i> by an area:<br /><i><b>Q(x,A) = −k·A·</b>d<b>T(x)<span style="font-size: medium;">/</span></b>d<b>x</b></i><br />As in a case of water flow along the river, if the temperature is linearly dependent on the distance from the outside wall, the derivative is constant and the flow of heat is constant. But, if the wall material is uneven, like in case of a river bed not being a straight line down, the heat flow rate will change.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-23446821878253532262019-06-12T19:55:00.001-07:002019-06-12T20:09:50.697-07:00Unizor - Physics4Teens - Energy - Measuring Heat - Problems<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/d6iBVozDJ5w" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Measuring Heat - Problems</u><br /><br /><i>Problem 1</i><br/><br/>How much heat energy is required to raise the temperature of <b><i>1 kg</i></b> of water from <i><b>20°C</b></i> to a boiling point of <i><b>100°C</b></i>?<br/>Assume the specific heat capacity of water is<br/><i><b>C</b><sub>w</sub><b> = 4184 J/(kg·°C)</b></i>.<br/><br/><i>Answer</i><br/><b><i>Q = C·m·(T</b><sub>end</sub><b>−T</b><sub>beg</sub><b>) =<br/>= 4184·1·(100−20) =<br/>= 334,720 J</i></b><br/><br/><i>Problem 2</i><br/><br/>A piece of unknown metal of mass <i><b>M<sub>m</sub></b></i> and temperature <i><b>T<sub>m</sub></b></i> was put into an isolated reservoir filled with <i><b>M<sub>w</sub></b></i> mass of water at temperature <i><b>T<sub>w</sub></b></i>. After the system of water and metal came to thermal equilibrium, its temperature became <i><b>T</b></i>.<br/>Assume that the metal is not too hot (so, water will not vaporize) and not too cold (so, the water will not freeze).<br/>Assuming that the water's specific heat capacity is known and equals to <i><b>C<sub>w</sub></b></i>, what is the specific heat capacity <i><b>C<sub>m</sub></b></i> of the unknown metal?<br/><br/><i>Answer</i><br/><br/><i><b>C<sub>w</sub>·M<sub>w</sub>·(T−T<sub>w</sub>) =<br/>= C<sub>m</sub>·M<sub>m</sub>·(T<sub>m</sub>−T)</b></i><br/>from which <i><b>C<sub>m</sub></b></i> equals to<br/><i><b>C<sub>w</sub>·M<sub>w</sub>·(T−T<sub>w</sub>)<font size=4>/</font></b></i>[<i><b>M<sub>m</sub>·(T<sub>m</sub>−T)</b></i>]<i><b></b></i><br/><br/><i>Problem 3</i><br/><br/>A burger has about <i><b>300 kcal</b></i> of energy in it.<br/><b><i>1 kcal = 4184 J</i></b>.<br/>A person, who ate it, wants to spend this energy by climbing up the stairs. A person's mass is <i><b>75 kg</b></i>, the height between the floor is <i><b>3 m</b></i>.<br/>Assume that only 25% of energy in the food can be used for climbing, while the other 75% is needed to maintain our body's internal functions.<br/>Counting from the ground floor as floor #0, to what floor can a person climb using that energy from a burger?<br/><br/><i>Answer</i><br/><b><i>E</b><sub>bur</sub><b> = 300</b>kcal<b> · 4184</b>J/kcal<b> =<br/>= 1,255,200</b>J<b></i></b><br/><b><i>E</b><sub>climb</sub><b> = 0.25·E</b><sub>bur</sub><b> =<br/>= 313,800</b>J<b></i></b><br/><b><i>E</b><sub>floor</sub><b> = 75</b>kg<b> · 9.8</b>m/sec²<b> · 3</b>m<b> =<br/>= 2205</b>J<b></i></b><br/><b><i>N = E</b><sub>climb</sub><b> / E</b><sub>floor</sub><b> ≅ 142</b> floors</i><br/><br/><i>Problem 4</i><br/><br/>An ice of mass <i><b>0.1</b>kg</i> has temperature <i><b>−10</b>°C</i>.<br/>What's the minimum amount of water <i><b>M</b></i> at temperature <i><b>20</b>°C</i> needed to melt it?<br/>Assume, specific heat capacity of water is <i><b>4183</b>J/(kg·°C)</i> and that of ice is <i><b>2090</b>J/(kg·°C)</i>. Assume also that the amount needed to melt ice at <i><b>0</b>°C</i> is <i><b>333,000</b>J/kg</i>.<br/><br/><i>Answer</i><br/><i><b>E</b><sub>warm</sub><b> = 2090·0.1·10 = 2090</b>J</i><br/><i><b>E</b><sub>melt</sub><b> = 333000·0.1 = 33300</b>J</i><br/><i><b>E</b><sub>need</sub><b> = 2090 + 33300 = 35390</b>J</i><br/><i><b>E</b><sub>water</sub><b> = 4183·M·20 = 83660·M</b></i><br/><i><b>E</b><sub>water</sub><b> = E</b><sub>need</sub></i><br/><i><b>35390 = 83660·M</b></i><br/><i><b>M = 0.423</b>kg</i><br/><br/> Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-62844118903892419332019-06-10T16:28:00.001-07:002019-06-10T16:28:10.421-07:00Unizor - Physics4Teens - Energy - Heat - Heat & Temperature<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/QQOqz5qTRXE" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Heat and Temperature</u><br /><br /><i>Specific Heat Capacity</i><br /><br />Let's discuss the relationship between <i>heat</i> and <i>temperature</i>.<br />From the unscientific standpoint these two concepts are almost identical. Heating an object results in increase of its temperature. Increasing a temperature of an object constitutes its heating.<br />Yet, from the strictly scientific viewpoint these two concepts are different.<br /><br /><i>Heat</i> is energy of some specific type, that can be transferred from one object to another, while <i>temperature</i> is an average kinetic energy of molecules of an object.<br />The same amount of <i>heat</i>, transferred into different objects, will result in different growth of their <i>temperatures</i>, depending on many factors, like mass, chemical composition, state etc. of these objects. Analogy of this is that same amount of fuel in different cars results in achieving very different speeds and, therefore, different kinetic energies in different cars, even if the gas pedal is pushed all the way down for all cars. It's just because cars are different and their internal structure converts fuel into movement differently.<br /><br />It has been experimentally observed that to increase a temperature of an isolated object of a unit mass by a unit of temperature is independent (within reasonable level of precision) of the initial temperature of an object, but depends only on the type of object's material, composition, state etc. In other words, amount of heat needed to increase a temperature of 1 kg of water from 20°C to 21°C is the same as from 50°C to 51°C. If, instead of water, we take copper, the amount of heat, needed to increase its temperature from 20°C to 21°C, will be the same as to increase it from 50°C to 51°C, but different than that for water.<br /><br />The above experimental fact allowed to establish a concept of <i><b>specific heat capacity</b></i> for each material as an <b>amount of heat required to increase the temperature of a unit of mass of this material (1 kg in SI) by a unit of temperature (1°C or 1°K in SI)</b>.<br /><br />Thus, <i>specific heat capacity</i> of water is, as we know, one kilocalorie per kilogram per degree - <i>1 kcal/(kg·°K</i>), that is about 4184 joules per kilogram per degree - <i>4183 J/(kg·°K</i>).<br />For copper the specific heat capacity is <i>385 J/(kg·°K</i>).<br />Gold has the specific heat capacity of <i>129 J/(kg·°K</i>).<br />Uranium's specific heat capacity is <i>116 J/(kg·°K</i>).<br />Cotton's specific heat capacity is <i>1400 J/(kg·°K</i>).<br />Hydrogen's specific heat capacity is <i>14304 J/(kg·°K</i>).<br />Generally speaking, but not always, more dense, more solid materials have less specific heat capacity than less dense or liquids, which have, in turn, less specific heat capacity than gases.<br /><br />Knowing <i>specific heat capacity</i><b><i>C</i></b> of material of an object and its mass <b><i>m</i></b>, we can easily determine amount of energy Δ<b><i>Q</i></b>needed to heat it up by Δ<b><i>T</i></b>degrees:<br />Δ<b><i>Q = C·m·</i></b>Δ<b><i>T</i></b><br />Inversely, knowing the amount of heat supplied, we can determine an increase in temperature:<br />Δ<b><i>T = </i></b>Δ<b><i>Q<span style="font-size: medium;">/</span>(C·m)</i></b><br />Notice, that increment of temperature Δ<b><i>T</i></b> and increment of heat energy Δ<b><i>Q</i></b> can be both positive or both negative, which means that an object, that has increased its temperature, has increased (gained, consumed) energy, and the object that decreased its temperature, has decreased (lost, released) energy.<br /><br /><i>Change of State</i><br /><br />Consider specific heat capacity of ice and water:<br />Ice: <i>2090 J/(kg·°K)</i><br />Water: <i>4183 J/(kg·°K)</i><br />Both these substances exist at temperature about 0°C=273°K. That means that we can heat <nobr>1 kg</nobr> of ice up to 0°C spending 2090 joules per each degree of temperature, but to increase the temperaature of the water around 0°C we have to spend 4183 joules per each degree. But, while ice is melting into water, which takes some time, both states, solid and liquid, exist side by side and the temperature of the water will not rise while the ice is not completely melted, we have to spend heat energy just on melting without actually changing the temperature of a substance, that remains around 0°C during the melting process.<br /><br />This experimental observation leads us to believe that, while graduate change of temperature for any specific state of matter linearly depends on the amount of heat supplied, change of state (like <b>melting</b> or <b>freezing</b>, or <b>evaporating</b> etc.) brings an element of non-linearity to this dependency.<br />More precisely, the temperature, as a function of amount of heat supply, in case of an object going through transformation of state from solid to liquid, looks like this (for better view, right click on the picture and open it in another tab of a browser):<br /><img src="http://www.unizor.com/Pictures/HeatTemperature.png" style="height: 150px; width: 200px;" /><br />As seen on this graph, supplying heat to ice will increase its temperature proportionally to amount of heat supplied.<br />Then, when the temperature reaches 0°C, ice will start melting and new heat will not change the temperature of ice/water mix, but will be used to change the state of matter from solid (ice) to liquid (water).<br /><br />This process of melting consumes heat without increasing the temperature until all ice is melted. Such a transformation of state that requires supply of heat energy is called <i><b>endothermic</b></i>.<br />Then, when the process of melting is complete, and all ice is transformed into water, the temperature of water will start increasing, as the new heat is supplied, but with a different coefficient of linearity relatively to the heat than in case of ice, since water's <i>specific heat capacity</i> is different than that of ice.<br />The graph above is characteristic to any heating process, where the transformation of the state of matter is involved. It is relatively the same for transforming liquid to gas (evaporation) or solid to gas (sublimation).<br /><br />In case of a reversed transformation from liquid to solid (freezing) or gas to liquid (condensation) or gas to solid (deposition) the heat energy must be taken away from a substance. It is the same amount in absolute value as was needed to supply to solid to melt it into liquid or to liquid to vaporise it into gas, or to solid to vaporize it into gas. So, in case of freezing, condensation or deposition we deal with the process of decreasing heat energy of a substance. Such a process is called <i><b>exothermic</b></i>.<br /><br />Amount of heat energy needed to transform a substance from one state to another is also experimentally determined and, obviously, depends on a substance, its mass and a kind of transformation it undergoes.<br />For example, melting of 1 kg of ice requires 333,000 joules of heat energy to be supplied. The same amount of heat energy should be extracted from the water at 0°C to freeze it into ice.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-89101864694300283532019-06-10T14:19:00.001-07:002019-06-10T14:19:39.219-07:00Unizor - Physics4Teens - Energy - Heat - Measuring<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/5ekDwNd0PCo" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Measuring Heat - Calorie</u><br /><br />Heat and temperature were known and researched by scientists long before the molecular movement was determined as their essence. As a result, attempts of measuring the amount of heat were unrelated to kinetic energy of the molecules.<br /><br />The unit of amount of heat was defined as amount of heat needed to increase the temperature of one gram of water by one degree Celsius (or Kelvin), and this unit of amount of heat was called a <i>calorie (cal)</i>.<br /><br />Obviously, this definition has its flaws. For example, the amount of heat needed to heat the same amount of water by the same temperature depends where exactly on Earth we are. The higher we are above the sea level - the less heat is needed. It also depends on the chemical purity of water. Also it's not obvious that the amount of heat needed to heat the water from 1°C to 2°C is the same as amount of heat needed to heat it from 88°C to 89°C, though within reasonable level of precision we do assume that this is true.<br /><br />With the development of molecular theory of heat and establishing relationship between heat and kinetic energy of molecules there was a need to put into correspondence existing units of heat (<i>calories</i>) and units of mechanical energy (<i>joules</i>). A simple experiment allowed to do just that.<br /><br />Imagine a standing on the ground large reservoir with known amount of water <i><b>M</b></i> of depth <i><b>H</b></i> from the surface to the bottom, kept at certain known temperature <i><b>T</b></i>°, and a relatively small stone of known mass <i><b>m</b></i>, having the same temperature <i><b>T</b></i>°, kept on the level of the surface of the water in a reservoir. This stone has certain known amount of potential energy <i><b>E=m·g·H</b></i>relatively to the bottom level of a reservoir.<br />Now we let the stone go down the reservoir to its bottom. Its potential energy relatively to the bottom of a reservoir decreases to zero. Where did the potential energy go? It's used to stir the water, thereby increasing the kinetic energy of its molecules.<br /><br />Because of this more intense movement of molecules, the water will increase its temperature. Potential energy of a stone will turn into kinetic energy of the molecular movement of water. If the temperature of the water has risen by Δ<i><b>T</b></i>°, potential energy of a stone <i><b>m·g·H</b></i> (in <i>joules</i>) equals to <i><b>M</b></i>·Δ<i><b>T</b></i> (in <i>calories</i>).<br /><br />Precise experiments like the above allowed to determine the correspondence between historical measure of the amount of heat in <i>calories</i> under different conditions and contemporary one in units of energy in SI - <i>joules</i>. This correspondence had been established approximately as<br /><i><b>1 calorie = 4.184 joules</b></i><br />but under different conditions (initial temperature, air pressure etc.) it might be equal to a slightly different value.<br /><br />Besides <i>calorie</i>, which is a relatively small amount of heat, the unit <i>kilocalorie (kcal)</i> had been introduced.<br />As is obvious from its name, <nobr><i><b>1 kilocalorie = 1000 calories</b></i>.</nobr><br /><br />The amount of energy contained in food and in some other practical cases very often is measured in <i>kilocalories</i>, which sometimes are called <i>large calories</i>, while a <i>calorie</i> is sometimes called <i>small calorie</i>. Unfortunately, the word "large" in many cases is omitted, which might cause misunderstanding.<br />From the experimental viewpoint, <i><b>1 kcal</b></i> is amount of heat needed to heat <i><b>1 kg</b></i> of water by one degree Celsius (or Kelvin).<br /><br />To avoid problems with the definition of <i>calorie</i> as amount of heat needed to warm up one gram of water by one degree Celsius, the contemporary <b>scientific definition</b> of <i>thermocalorie</i> is<br /><i><b>1 thermocalorie = 4.1833 joules</b></i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-79690377419782082032019-06-03T14:31:00.003-07:002019-06-03T14:32:23.557-07:00Unizor - Physics4Teens - Energy - Ideal Gas Kinetics - Problems <a href="https://youtu.be/MYEgnMaBJ0w"></a> <i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Problem 1</u><br /><br />How much kinetic energy have all the molecules in the room?<br/>How fast should an average size car move to have this amount of kinetic energy?<br/><br/>Assume the following:<br/>(a) the room dimensions are <i>4</i>x<i>4</i>x<i>3 meters</i> (that is, <i><b>V=48m³</b></i>);<br/>(b) normal atmospheric pressure is <i>100,000 Pascals</i> (that is, <i><b>p=100,000N/m²</b></i>);<br/>(c) a mass of an average size car is <i>2,000 kg</i> (that is, <i><b>M=2,000kg</b></i>).<br/><br/><i>Solution</i><br/><br/>From the lecture on <i>Kinetics of Ideal Gas</i> we know the relationship between the pressure on the walls of a reservoir, volume of a reservoir and total kinetic energy of gas inside this reservoir:<br/><i><b>p = (2/3)E</b><sub>tot </sub><b><font size=5>/</font>V</b></i><br/><br/>From this we derive a formula for total kinetic energy:<br/><i><b>E</b><sub>tot</sub><b> = (3/2)·p·V</b></i><br/><br/>Substituting the values for pressure and volume, we obtain<br/><i><b>E</b><sub>tot</sub><b> = (3/2)·100,000·48 =<br/>= 7,200,000</b>(joules)</i><br/><br/>The kinetic energy of a car is <br/><i><b>E = M·v²<font size=5>/</font>2</b></i><br/>Therefore, given the kinetic energy and mass, we can determine the car's speed:<br/><i><b>v = √<span style='text-decoration:overline'>2·E/M</span></b></i><br/>Substituting calculated above <i><b>E</b><sub>tot</sub><b>=7,200,000</b>(J)</i> for <i><b>E</b></i> and the value for mass <i><b>M=2,000</b>(kg)</i>, we obtain<br/><i><b>v = √<span style='text-decoration:overline'>2·7,200,000/2,000</span> ≅<br/>≅ 85</b>(m/sec)<b> ≅<br/>≅ 306</b>(km/hour)<b> ≅<br/>≅ 190</b>(miles/hour)</i><br/><br/><br/><u>Problem 2</u><br /><br />Given the temperature, pressure and volume of the air in a room, determine the number of gas molecules in it.<br/><br/>Assume the following:<br/>(a) the room dimensions are <i>4</i>x<i>4</i>x<i>3 meters</i> (that is, <i><b>V=48m³</b></i>);<br/>(b) normal atmospheric pressure is <i>100,000 Pascals</i> (that is, <i><b>p=100,000=10<sup>5</sup>N/m²</b></i>);<br/>(c) temperature is <i>20°C</i> (that is, <i><b>T=20+273=293°K</b></i>).<br/><br/><i>Solution</i><br/>Recall the combined law of ideal gas<br/><i><b>p·V<font size=5>/</font>T = k</b><sub>B</sub><b>·N = const</b></i><br/>where<br/><i><b>k</b><sub>B</sub><b> = 1.381·10<sup>−23</sup> (J/°K)</b></i> is Boltzmann's constant and<br/><i><b>N</b></i> is the number of gas molecules in a reservoir.<br/>From this we derive the number of molecules<br/><i><b>N = p·V<font size=5>/</font>(k</b><sub>B</sub><b>·T)</b></i><br/>Substituting the values,<br/><i><b>N = 10<sup>5</sup>·48<font size=5>/</font>(1.381·10<sup>−23</sup>·293) = 0.12·10<sup>28</sup></b></i><br/>It's a lot!<br/><br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0