Monday, October 7, 2019

Unizor - Physics4Teens - Energy - Gravitational Potential - Thin Spheric...





Notes to a video lecture on http://www.unizor.com

Gravity Integration 3 -
Thin Spherical Shell


1. Determine the potential of the gravitational field of a uniform infinitesimally thin spherical shell at any point outside it.

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 P, where we have to determine the gravitational potential.

Assume that the sphere's radius is R and the mass is M. Then its surface is 4πR² and the mass density per unit of surface area is ρ=M/(4πR²).
Assume further that X-coordinate of a point P, where we want to calculate the gravitational potential, is H, which is greater than the radius of a spherical shell R.


If, instead of a spherical shell, we had a point mass M concentrated in its center at point O(0,0,0), its gravitational potential at a point P would be
V0 = −G·M/H
(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 P, and the field performs this work for us, so we perform negative work).

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.

To calculate a gravitational potential of an infinitesimally thin spherical shell at point P 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.

The angle φ from X-axis (that is, from OP) to a radius from an origin of coordinates to any point on a ring will be our variable of integration.
Then the radius of a ring will be
r(φ) = R·sin(φ)
The distance from the origin of coordinates to a center of a ring is R·cos(φ).
The area of a ring between angles φ and φ+dφ will be equal to the product of infinitesimal width of a ring dφ and its circumference 2πR·sin(φ)
Therefore, the mass of a ring will be
dm(φ) = ρ·2πR²·sin(φ)·dφ =
= M·2πsin(φ)·
dφ/(4π) =
= M·sin(φ)·
dφ/2


Knowing the mass of a ring dm(φ), its radius r(φ) and the distance from the ring's center to point of interest P, that is equal to H−R·cos(φ), we can use the formula of the ring's potential from a previous lecture
V = −G·M /R²+H²
substituting
dV(φ) instead of V
dm(φ) instead of M
H−R·cos(φ) instead of H
r(φ) instead of R

Therefore,
dV(φ) = −G·dm(φ) /r²(φ)+[H−R·cos(φ)]² =
= −G·M·sin(φ)·
dφ /2√R²+H²−2R·H·cos(φ)


Now all we need is to integrate this by φ in limits from 0 to π.
Substitute
y = √R²+H²−2R·H·cos(φ)
Incidentally, the geometric meaning of this value is the distance from point of interest P to any point on a ring for a particular angle φ.
Then
dy = R·H·sin(φ)·dφ /R²+H²−2R·H·cos(φ)

The limits of integration for φ from 0 to π in terms of y are from |H−R| (which, for our case of point P being outside the sphere, equals to H−R) to H+R.

In terms of y
dV(y) = −G·M·d/(2R·H)
which we have to integrate by y from H−R to H+R.

Simple integration of this function by y on a segment [H−R;H+R] produces −G·M·y/(2R·H) in limits from H−R to H+R:
V = −G·M·(H+R)/(2R·H) +
+ G·M·(H−R)/(2R·H) =
= −G·M/H


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 V0.

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.

2. Determine the potential of the gravitational field of a uniform infinitesimally thin spherical shell at any point inside it.

Using the same notation as in the previous case, this problem requires the distance from a point of interest P to a center of a spherical shell O to be less than the radius R of a spherical shell.
Doing exactly the same manipulation and substitution
y = √R²+H²−2R·H·cos(φ)
we see that the only difference from the previous case is in the limits of integration in terms of y.
The limits of integration for φ from 0 to π in terms of y are from |H−R| (which, in this case of point P being inside the sphere, equals to R−H) to H+R.

Integration by y on a segment [R−H;H+R] produces −G·M·y/(2R·H) in limits from R−H to H+R:
V = −G·M·(H+R)/(2R·H) + G·M·(R−H)/(2R·H) =
= −G·M/R


Remarkably, it's constant and is independent of the position of point P inside a spherical shell.

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:
F(r)=G·M·m /r²=m·dV(r)/dr

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.

An intuitive explanation of this is in the fact that, if you consider any conical surface with a vertex at point P 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 P, while the gravitational forces produced by these pieces of surface are inversely proportional to a square of a distance from point P, thus both forces from opposite ends of a cone are equal in magnitude and opposite in direction, thus nullify each other.



Friday, October 4, 2019

Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Gravit...





Notes to a video lecture on http://www.unizor.com

Gravity Integration 2

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.

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.
Assume that the ring's radius is R and the mass is M, so the density of mass per unit of length is ρ=M/(2πR).
Assume further that the Z-coordinate of a point P, where we want to calculate the gravitational potential, is H.

If, instead of a ring, we had a point mass M concentrated in its center at point (0,0,0), its gravitational potential at a point P would be
V0 = −G·M/H
(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 P, and the field performs this work for us, so we perform negative work).

Since the mass in our case is distributed along the circumference of a ring, and every point on a ring is on a distance r=√R²+H² from point P, which is further from this point than the center of a ring, the gravitational potential of a ring at point P will be smaller.

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.
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 P and integrate all these potentials.

Let's choose an angle from the positive direction of the X-axis to a point on a ring as the main integration variable φ∈[0;2π]. Its increment dφ gives an increment of the circumference of a ring
dl = R·dφ
The mass of this infinitesimal segment of a ring is
dm = ρ·dl = M·R·dφ /(2πR) = M·dφ /(2π)

The distance from this infinitecimal segment of a ring to a point of interest P is independent of variable φ and is equal to constant r=√R²+H².

Therefore, gravitational potential of an infinitecimal segment of a ring is
dV = −G·d/r = −G·M·dφ /(2π√R²+H²)

Integrating this by variable φ on [0;2φ], we obtain the total gravitational potential of a ring at point P:
V = [0;2π]dV = −[0;2π]G·M·dφ /(2π√R²+H²)
Finally,
V = −G·M /R²+H²

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.

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.
Assume that the ring's radius is R and the mass is M, so the density of mass per unit of surface is ρ=M/(πR²).
Assume further that the Z-coordinate of a point P, where we want to calculate the gravitational potential, is H.

Let's split our disc into infinite number of infinitely thin concentric rings of radius from x=0 to x=R of width dx each and use the previous problem to determine the potential of each ring.

The mass of each ring is
dm(x) = ρ·2πx·dx
This gravitational potential of this ring at point P, according to the previous problem, is
dV(x) = −G·dm(x) /x²+H² =
= −G·ρ·2πx·
d/x²+H² =
= −G·M·2πx·
d/(πR²√x²+H²) =
= −G·M·2x·
d/(R²√x²+H²)


To determine gravitational potential of an entire disc, we have to integrate this expression in limits from x=0 to x=R.
V = [0;R]dV(x) = −k·[0;R]2x·d/x²+H²
where k = G·M /

Substituting y=x²+H² and noticing the dy=2x·dx, we get
V = −k·[H²;H²+R²] d/y
The derivative of y is /(2√y) Therefore, the indefinite integral of /y is
2√y + C

Finally,
V = −k·(2√H²+R²−2H) = −2G·M·(√H²+R²−H) /

Tuesday, September 24, 2019

Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Gravit...





Notes to a video lecture on http://www.unizor.com

Gravity Integration 1

Determine the potential of the gravitational field of an infinitely thin solid rod at any point outside of it.

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 A(a,0) and another at point B(b,0).
Assume that the rod's length is L=b−a and the mass is M, so the density of mass per unit of length is ρ=M/L.
Assume further that the coordinates of a point P, where we want to calculate the gravitational potential, are (p,q).

If, instead of a rod, we had a point mass M concentrated in the midpoint of a rod at point ((a+b)/2,0), its gravitational potential at a point (p,q) would be
V0=G·M/r
where r is the distance between the midpoint of a rod and a point of measurement of gravitational potential P:
r = {[(p−(a+b)/2]2 + q2}1/2

Since the mass in our case is distributed along the rod, the gravitational potential will be different.

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.
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 P and integrate all these potentials.

Consider a picture below (we recommend to save it locally to see in the bigger format).

As a variable, we will use an X-coordinate of a point on a rod Q and calculate the gravitational potential at point of interest P(p,q) from an infinitesimal segment of a rod of the length dx around point Q(x,0).
Knowing that, we will integrate the result by x on a segment [a;b] to get the gravitational potential of the rod.

The infinitesimal segment of a rod dx, positioned around a point Q(x,0), has an infinitesimal mass dm that can be calculated based on the total mass of a rod M and its length L=b−a as
dm = M·d/L

The gravitational potential of this segment depends on its mass dm and its distance r(x) to a point of interest P(p,q).
dV = G·d/r(x)
Obviously,
r(x) = [(p−x)2+q2]1/2
Combining all this, the full gravitational potential of a rod [a;b] of mass M at point P(p,q) will then be
V(p,q) = abd/r(x) = abG·M·dx/{[(p−x)2+q2]1/2}

We can use the known indefinite integral
d/(t²+c²) = ln|t+√(t²+c²)|

Let's substitute in the integral for gravitational potential t=x−p.
Then
V(p,q) = G·M·d/[t2+q2]1/2
where integration is from t=a−p to t=b−p.
V(p,q) = (G·M/L)·[ln|b−p+√(b−p)²+q²| − ln|a−p+√(a−p)²+q²|]
where L = b−a

Since the difference of logarithms is a logarithm of the result of division,
V(p,q) = G·M·ln(R) /L
where
L = b−a and
R = |b−p+√(b−p)²+q²/ |a−p+√(a−p)²+q²|

Friday, September 20, 2019

Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Energy...





Notes to a video lecture on http://www.unizor.com

Gravitational Energy Conservation

While moving an object from a distance r1 to a distance r2 from the center of gravity, the gravitational field has performed certain work W[r1,r2], spending certain amount of energy. Since energy must be conserved, it should materialize in some other way.

Indeed, the kinetic energy of a probe object at the end of its movement from point {r1,0,0} to {r2,0,0} must be equal to the work performed by the field.

We have positioned our probe object at point {r1,0,0} without any initial speed, that is Vr1=0. Therefore, the kinetic energy Kr1 at this initial point is zero.
At the end of a motion at point {r2,0,0} the speed Vr2 must have such a value that the kinetic energy Kr2 would be equal to work W[r1,r2] performed by the field.
Kr2 = m·V²r2 /2 = W[r1,r2]

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 {r1,0,0}, where it was at rest, to point {r2,0,0}:
m·V²r2 /2 =
(G·M /r2 − G·M /r1
)·m
r2 = 2·G·M·(1/r2 − 1/r1)

In a particular case, when r1=∞ and r2=r, 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 r from it, the formula is simplified:
r = 2·G·M /r

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 r=0 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 r less than the radius of Earth.

Back to energy conservation.
The potential energy 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.

As we know (see the previous lecture), amount of work we need to move a probe object of mass m from an infinite distance to a distance r from a source of gravity equals to
Wr = −G·M·m /r
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.

In this expression, skipping over the universal gravitational constant G, components M (mass of a source of gravitational field) and r (distance from the center of the gravitational field) characterize the gravitational field, while m (mass of a probe object) characterizes the object, whose potential energy we measure.

This energy is transferred to a probe object as its potential energy. 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 potential energy and gaining the kinetic energy because it will move faster and faster.

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.
The Universal Gravitational Constant is
G=6.67408·10−11,
its units are m3·kg−1·sec−2.
The mass of the Moon is M=7.34767309·1022 kg.
The radius of the Moon is r=1.7371·106 m.
Let's assume that an asteroid falling on the Moon is relatively small one, say, m=50 kg.

According to the formula above, the gravitational field of the Moon did the work that equals to
W ≅ 6.67408·10−11·7.34767309·1022·50 /(1.7371·106) ≅ 141,151,800 (joules)
let's check the units to make sure we get joules, the units of work
m3·kg−1·sec−2·kg·kg·m−1 = kg·m2·sec−2 = N·m = J

The final speed V can be calculated by equating this amount of work and kinetic energy of an asteroid:
V2 ≅ 2·6.67408·10−11·7.34767309·1022 /(1.7371·106) ≅ 5646072
let's check the units to make sure we get the square of speed units
m3·kg−1·sec−2·kg·m−1 = m2·sec−2 = (m/sec)2

From this the speed of an asteroid falling from infinity onto Moon's surface is
V ≅ √5646072 ≅ 2376 (m/sec)
or about 2.4 km/sec.

Incidentally, this is the so-called escape speed 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.

Let's do similar calculations for the Earth, using the same assumptions, the same asteroid and the same units of measurement.
M = 5.972·1024 kg
r = 6.371·106 m
W ≅ 6.67408·10−11·5.972·1024·50 /(6.371·106) ≅ 3,128,049,424 (joules)

V2 ≅ 2·6.67408·10−11·5.972·1024 /(6.371·106) ≅ 125121977 (m/sec)2
From this the speed of an asteroid falling from infinity onto Earth's surface is
V≅√125121977≅11186 m/sec
or about 11.2 km/sec.

This is also the escape speed needed to fly away from Earth's gravitational field.

Monday, September 16, 2019

Unizor - Physics4Teens - Energy - Gravitational Field - Problems





Notes to a video lecture on http://www.unizor.com

Problems on Gravity

Problem 1
Gravitational potential of a spherical gravitational field around a point-mass M at a distance r 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
Vr = −G·M /r
Why is this formula independent of trajectory of a probe object or its exact final position relative to the point-mass M, but only on a distance itself from the source of gravity?

Solution
Any movement can be represented as infinitely many infinitesimal displacements, combined together into a trajectory.
In our three-dimensional world the force and an infinitesimal displacement of a probe object are vectors, so the infinitesimal work dW performed by the force of gravity F during the movement of a probe object, described by the infinitesimal displacement dS, is a scalar product of these two vectors:
dW = F·dS
Note that the vector of gravitational force F is always directed towards the source of gravity.
Since a displacement vector dS can be represented as a sum of radial (towards the source of gravity) dSr and tangential (perpendicular to radius) dSt components, the above expression for a differential of work can be written as
dW = F·(dSr + dSt) =
F·
dSr + F·dSt
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 r from it.

Problem 2
Given two point-masses of mass M each, fixed at a distance 2R from each other.
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.

Solution
Let's draw a diagram of this problem (you can download it to display in a bigger format).

Our two point-masses are at points A and B, the probe object is at point D on a perpendicular bisector of a segment AB going through point C.
The force of gravity towards point A is a segment DE, the force of gravity towards point B is a segment DF.
We will calculate the potential of a combined gravitational field of two point-masses at point D, where the probe object is located.
Let's assume that the segment CD equals to h.
The magnitude of each gravitational force equals to
F = G·M·m /(h2+r2)
Represent each of these forces as a sum of two vectors, one (green on a drawing) going vertically along the bisector CD, another (red) going horizontally parallel to AB.
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
Ftot = 2·G·M·m·sin(φ)/(h2+r2)
Since sin(φ) = CD/AD,
sin(φ) = h /[(h2+r2)1/2]
Ftot = 2·G·M·m·h /(h2+r2)3/2
If the gravitational field pulls a probe object along the perpendicular bisector of a segment AB from infinity to a distance h from the segment, the magnitude of a combined force of gravity, as a function of a distance from the segment x is changing, according to a similar formula:
Ftot(x) = 2·G·M·m·x /(x2+r2)3/2
To calculate work performed by a gravitational field pulling a probe object from infinity to height h above the segment AB, we have to integrate
Wtot = [∞;h]Ftot(x)·dx
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.
Wtot = 2GMm·x·d/(x2+r2)3/2
(within the same limits of integration [∞;h])
This integral can be easily calculated by substituting
y=x2+r2,
2·x·dx = dy,
infinite limit of integration remaining infinite and the x=h limit transforming into y=h2+r2. Now the work expression is
Wtot = G·M·m·y−3/2·dy
with limits from y=∞ to y=h2+r2.
The indefinite integral (anti-derivative) of y−3/2 is −2·y−1/2.
Therefore, the value of integral and the work are
Wtot = −2·G·M·m·(h2+r2)−1/2
For a unit mass m=1 this work is a gravitational potential of a combined gravitational field produced by two point-masses on a distance h from a midpoint between them along a perpendicular bisector
Vtot = −2·G·M·(h2+r2)−1/2
At the same time, the gravitational potential of a field produced by each one of the point-masses, considered separately, equals to
Vsingle = −G·M·(h2+r2)−1/2
As we see, the gravitational potential of two point-masses equals to a sum of gravitational potential of each of them, considered separately.
IMPORTANT NOTE
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 gravitational potential is additive, that is the gravitational potential of any set of objects at any point in space equals to sum of their individual gravitational potentials.

Problem 3
Express mass M of a spherical planet in terms of its radius R and a free fall acceleration g on its surface.

Solution
Let m be a mass of a probe object lying on a planet's surface.
According to the Newton's 2nd Law, its weight is
P = m·g
According to the Universal Law of Gravitation, the force of gravitation between a planet and a probe object is
Fgravity = G·M·m /R2
Since the force of gravitation is the weight Fgravity = P,
m·g = G·M·m /R2
from which
M = g·R2 /G

Problem 4
Express gravitational potential VR of a spherical planet on its surface in terms of its radius R and a free fall acceleration g on its surface.

Solution
From the definition of a gravitational potential on a distance R from a source of gravity
VR = −G·M /R
Using the expression of the planet's mass in terms of its radius R and a free fall acceleration g on its surface (see above),
M = g·R2 /G
Substituting this mass into a formula for potential,
VR = −G·g·R2 /(G·R) = −g·R


Tuesday, September 10, 2019

Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Gravitational Field



Notes to a video lecture on http://www.unizor.com

Gravitational Field

Studying forces, we have paid attention to a force of attraction, that exists between any material objects, the force of gravity.
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.

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.
In Physics this concept of force acting on a distance is described by a term field. Basically, field 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.

Gravitational field 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.
As described in the "Gravity, Weight" chapter of "Mechanics" part of this course, the magnitude of the gravitational force F is proportional to a product of masses of a source object and a probe object, M and m, and it is inversely proportional to a square of a distance r between these objects:
F = G·M·m /
where G - 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.

The direction of the gravitational force acting on a probe object is towards the source object.

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 gravitational field has certain amount of energy at each point that it spends by applying the force onto a probe object.

To quantify this, assume that the source of gravity is a point mass M fixed at the origin of coordinates. Position a probe object of mass m at coordinates {r1,0,0} 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 {r2,0,0}, where r2 is smaller then r1. Let x be a variable X-coordinate (distance to the origin).

According to the Universal Law of Gravitation, the force of attraction of a probe object towards the source of a gravitational field at distance x from the origin equals to
F = −G·M·m /
where minus in front of it signifies that this force is directed opposite to increasing the X-coordinate.
This force causes the motion and, therefore, does some work, moving a probe object from point {r1,0,0} to point {r2,0,0} along the X-axis. To calculate the work done by this variable force, we can integrate dx from x=r1 to x=r2:
W[r1,r2] = [r1,r2]dx =
= −[r1,r2]G·M·m·
dx /x² =
= G·M·m /x
|[r1,r2] =
= G·M·m /r2 − G·M·m /r1 =
= (G·M /r2 − G·M /r1
)·m

The expression
V(r) = −G·M/r
is called gravitational potential.
It's a characteristic of a gravitational field sourced by a point mass M at a distance r from a source.
It equals to work needed by external forces to bring a probe object of mass m=1 to a point at distance r from a source of the field from infinity.
Indeed, set m=1, r1=∞ and r1=r in the above formula for work W[r1,r2] and take into consideration that gravitational field "helps" external forces to move a probe object, so the external forces spend negative amount of energy.

Using this concept of gravitational potential V(r), we can state that, to move a probe object of a unit mass from distance r1 relative to a source of gravitational field to a distance r2 relative to its source in the gravitational field with gravitational potential V(r), we have to spend the amount of energy equal to V(r1)−V(r2).
For a probe object of any mass m this amount should be multiplied by m.
If r2 is greater than r1, 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 r2 is smaller than r1, 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.

Therefore, an expression EP=m·V(r) represents potential energy of a probe object of mass m at a distance r from a source of a gravitational field with gravitational potential V(r).

A useful consequence from a concept of a gravitational potential is that the force of gravity can be expressed as the derivative of the gravitational potential.
F = G·M·m /r² = m·dV(r)/dr
which emphasizes the statement that the gravitational potential is a characteristic of a field itself, not its source.
We, therefore, can discuss gravitational field as an abstract concept defined only by the function called gravitational potential.

Sunday, August 18, 2019

Unizor - Physics4Teens - Energy - Energy of Nucleus - Fusion



Notes to a video lecture on http://www.unizor.com



Nucleus Fusion



Fusion is a nuclear reaction, when light nuclei are brought together and combined into a heavier ones.

The reason for this reaction to release the energy is the difference
between amount of energy needed to overcome the repulsion between nuclei
because they have the same positive electric charge (this energy is
consumed by fusion) and the potential energy released by strong forces, when the formation of a combined nucleus occurs (this energy is released by fusion).

The former is less than the latter.



When the light nuclei are fused into a heavier one, the excess of potential energy of strong forces, released in the process of fusion,
over the energy needed to squeeze together protons against their
repulsion is converted into thermal and electromagnetic field energy.



Analogy to this process can be two magnets separated by a spring.


The magnets represent two separate protons, the magnetic force of attraction between them represents the strong force
that is supposed to hold the nucleus together, when these particles are
close to each other, the spring represents the electrical repulsive
force between them, acting on a larger distance, as both are positively
charged.

It's known that magnetic force is inversely proportional to a square of a
distance between objects, while the resistance of a spring against
contraction obeys the Hooke's Law and is proportional to the length of
contraction.

On the picture magnets are separated. To bring them together, we have to
spend certain amount of energy to move against a spring that resists
contraction. But the magnetic attraction grows faster then the
resistance of the spring, so, at some moment this attraction will be
greater than the resistance of a spring. At this moment nothing would
prevent magnets to fuse.



As is in the above analogy, if we want to fuse two protons, we have to bring them together sufficiently close for strong forces to overtake the repulsion of their positive charges.



Consider the following nuclear reaction of fusion.

One nucleus of hydrogen isotope deuterium 1H2 with atomic mass 2 contains one proton and one neutron.

One nucleus of hydrogen isotope tritium 1H3 with atomic mass 3 contains one proton and two neutrons.

If we force these two nuclei to fuse, they will form a nucleus of helium 2He4 and releasing certain amount of energy:

1H2 + 1H3 = 2He4 + 0n1



It's not easy to overcome the repulsion of protons. High temperature and
pressure, like in the core of our Sun, are conditions where it happens.
On Earth these conditions are created in the nuclear bomb, using the
atomic bomd to achieve proper amount of heat and pressure, thus creating
an uncontrlled fusion.

Controlled nuclear reaction of fusion is what scientists are working on right now. So far, it's still in the experimental stage.

Monday, August 12, 2019

Unizor - Physics4Teens - Energy - Energy of a Nucleus - Fission



Notes to a video lecture on http://www.unizor.com

Nucleus Fission

Fission, first of all, is a nuclear reaction, when heavier nuclei are split into lighter ones.
The reason for this reaction to release the energy is the difference between amount of energy needed to break strong forces that hold the nucleus together (this energy is consumed by fission) and amount of potential energy in positively charged and repelling protons inside nucleus (this energy is released by fission).
The former is less than the latter.

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.

Analogy of this is a spring squeezed tightly and held in this position by a thread. A thread plays the role of strong forces, 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.

Electrically positively charged protons repel each other and, at the same time, are bonded together by strong forces inside a nucleus. At the same time neutrons are also bonded by strong forces among themselves and with protons without any repulsion.
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.

Uranium-238 with 92 protons and 146 neutrons (92U238) naturally occurs on Earth and is relatively stable.
Uranium-235 with the same 92 protons and 143 neutrons (92U235) has less "bonding material" (less neutrons) and is more susceptible to fission.

All it takes to break the nucleus of 92U235 is a little "push" from outside, which can be accomplished by bombarding it with neutrons. In the process of fission, caused by hitting a nucleus of 92U235 with a neutron, it can transforms into Barium-141 with 56 protons and 85 neutrons 56Ba141, Krypton-92 with 36 protons and 56 neutrons 36Kr92 and 3 free neutrons.
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 92U235 total 144, output: 85 in a nucleus 56Ba141, 56 in a nucleus of 36Kr92 and 3 new free neutrons total 144).

Let's express this reaction in a formula (letter n denotes a neutron):
0n1 + 92U235 =
56Ba141 + 36Kr92 + 3·0n1


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 chain reaction and, potentially, an explosion (atomic bomb). However, if we absorb extra neutrons, it will allow to slowly release of nuclear energy (nuclear power stations).

Monday, August 5, 2019

Unizor - Physics4Teens - Energy - Energy of a Nucleus



Notes to a video lecture on http://www.unizor.com



Energy of Nucleus



In this lecture we will analyze the energy aspect of nucleus - the central part of an atom.



By now we have built a pyramid of energy types, related to the depth of our view inside the matter.



First, we analyzed the mechanical energy - the energy of moving macro-objects.



Our next view deep into the world of macro-objects uncovered the molecules - the smallest parts of macro-objects that retain their characteristics. The movement of these molecules was the source of thermal energy, which we often call the heat.



Next step inside the molecules uncovered atoms, as the molecules'
components. There are about 100 types of atoms and their composition
inside the molecules creates all the thousands of different molecules. Chemical reactions
change the composition of atoms in molecules, thereby creating new
molecules from the atoms of old molecules. This process broke some
inter-atomic bonds and created the new ones and is the source of chemical energy.



Now we look deep inside the atoms and find there 3 major elementary particles - electrically positively charged protons and electrically neutral neutrons inside a small but heavy nucleus and electrically negatively charged electrons,
circulating around nucleus on different orbits. For electrically
neutral atoms the numbers of protons and electrons are equal. Nuclear energy is hidden inside the nucleus and is the subject of this lecture.



The first question we would like to answer is "What holds nucleus, its
protons and neutrons, together, considering protons, as electrically
positively charged particles must repel each other?"



The answer is simple. There are other forces in the Universe, not only
electrostatic ones, that act in this case. These intra-nucleus forces
that hold the nucleus together are called strong forces. They are strong
because they are the source of attraction between the protons that is
stronger than electrostatic repelling. However, these strong forces act
only on a very small distance, comparable to the size of a nucleus
inside an atom. For example, at a distance 10−15m the strong force is more than 100 times stronger than electrostatic one.



If, by regrouping protons and neutrons, we will be able to create different atoms (inasmuch as regrouping atoms in chemical reaction we create new molecules), a new source of energy, based on strong forces, the nuclear energy, can be uncovered in the course of nuclear reaction.



There is another form of nuclear reaction related to
transformation of elementary particles. Under certain circumstance a
neutron inside a nucleus can transform into proton and, to keep the
total electrical charge in balance, it emits an electron. This reaction
is called beta-decay and it also produces energy in the form of electromagnetic waves of very high frequency (gamma-rays).



Nuclear reactions are a very powerful source of nuclear energy, which is
so much more powerful than other types of energy, that, if misused, it
might represent a danger for life on our planet.



There is a clear analogy between nuclear and chemical reactions.

What happens with atoms in the chemical reaction, happens with protons
and neutrons in nuclear reaction. Some atomic bonds break in a chemical
reaction, some are created. Some nuclear bonds between protons and
neutrons break in a nuclear reaction, some are created.



Sometimes the chemical reaction happens by itself, as long as
participating substances are close together, but sometimes we have to
initiate it, like lighting methane gas with a spark or a flame of a
match to initiate continuous burning.

Similar approach is valid for nuclear reaction. Sometimes it happens by
itself, but sometimes it should be started, like bombarding the nucleus
with neutrons, after which it continues by itself.



Here is an interesting fact.

Physicists have measured the masses of protons, neutrons and many
different nuclei that contain these protons and neutrons and have
discovered that the sum of masses of individual protons and neutrons is
greater than the mass of a nucleus that contain these exact particles.

For example,

mass of proton is 1.0072766 atomic mass units or 1.6726·10-27kg,

mass of neutron is 1.0086654 atomic mass units or 1.6749·10-27kg.

At the same time, mass of deuterium nucleus, that contains 1 proton and 1
neutron is 2.0135532 atomic mass units, which is smaller than the sum
of masses of proton and neutron (1.0072766 + 1.0086654 = 2.015942).

This so-called "mass defect" is directly related to nuclear energy - the energy of strong forces that hold the nucleus together.



A simplified explanation of this effect is based on the law of energy
conservation. Consider the force of gravity between a planet and an
object above its surface. The object has certain potential energy and,
if dropped to the ground, this potential energy transforms into other
forms, like kinetic, thermal etc.



Similarly, if we consider two independent neutrons (or neutron and
proton, or two protons) on a very small distance from each other, but
not forming a nucleus, there is a potential energy of the strong forces
acting between them. If we let these two particles to form a nucleus,
analogously to an object falling towards the surface of a planet, this
potential energy should be transformed into other forms, like thermal.



Now the Theory of Relativity comes to play, that has established the equivalence of mass and energy by a famous formula E=m·c².
According to this equivalence, if some energy is released during the
formation of a nucleus from individual protons and neutrons, there must
be certain amount of mass released associated with this energy. That is
the explanation of "mass defect".



It should be noted that to form a nucleus of deuterium from 1 proton and
1 neutron is easier than to form a nucleus that contains more than one
proton, because electrostatic repulsion between positively charged
protons prevents their bonding. So, to bring protons sufficiently close
to each other for strong forces to overcome the electrostatic
repulsion, we have to spend some energy. The net energy released by
forming a nucleus from protons and neutrons is the difference between
the energy released from strong forces taking hold of these particles inside a nucleus and the energy consumed to overcome repulsion of protons.



Actually, as we attempt to form bigger nuclei, the energy we have to
spend to overcome electrostatic repulsion forces become greater than
amount of energy released by forming a nucleus. This border line is
approximately around the nucleus of iron Fe. Forming iron
and heavier elements from protons and electrons is a process that
consumes more energy than releases. These heavier nuclei will produce
energy, if we reverse the procedure, breaking them into individual
protons and neutrons.



The mechanisms described above are used in nuclear reactors and atomic
bomb, where heavier elements are broken into lighter ones (fission),
releasing energy, and in hydrogen bomb, where lighter elements are
bonded together to release the energy (fusion).

Monday, July 29, 2019

Unizor - Physics4Teens - Energy - Atoms and Chemical Reactions - Interat...





Notes to a video lecture on http://www.unizor.com



Interatomic Bonds



Atoms in a molecule are bonded together to form a stable chemical substance or compound.

The mechanism of bonding is quite complex and different for different
molecules. In fact, the complexity of these bonds is outside of the
scope of this course. However, certain basic knowledge about molecular
bonding and molecular structure is necessary to understand the following
lecture, where we will make certain calculations related to energy
produced or consumed in chemical reactions.



The key to a mechanism of bonding atoms into molecules lies in an internal structure of atoms.

For our purposes we can consider the orbital model of atom as consisting
of electrically positive nucleus and electrically negative electrons
circulating on different orbits around a nucleus. This is only a model,
not an exact representation of what's really happening inside the atom,
but this model gives relatively good results that correspond to some
simple experiments.



Two different particles can be found in a nucleus - positively charged
protons and electrically neutral neutrons. The number of protons inside a
nucleus and electrons circulating on different orbits around a nucleus
should be the same for electrically neutral atoms in their most common
state.



For reasons not well understood by many physicists, each orbit can have
certain maximum number of electrons that can circulate on it without
"bumping" into each other. The higher the orbit - the more electrons it
can hold. The lowest orbit can hold no more than 2 electrons, the next -
no more than 8, the next - no more than 14 etc.



Consider a few examples.



1. Let's consider the structure of a simplest molecule - the molecule of
hydrogen, formed by two atoms of hydrogen. Each hydrogen atom has one
electron on the lowest orbit around a nucleus. The maximum number of
electrons on this orbit is two, in which case the compound becomes much
more stable. So, two atoms of hydrogen grab each other and the two
electrons, each from its own atom, are shared by a couple of atoms, thus
creating a stable molecule of hydrogen with symbol H2. The bond between two atoms of hydrogen is formed by one pair of shared electrons, so structurally the molecule of hydrogen H2 can be pictured as

H−H.



2. Atom of oxygen has 8 electrons - 2 on the lowest orbit and 6 on the
next higher one. The next higher orbit is stable when it has 8
electrons. So, two atoms of oxygen are grabbing each other and share 2
out of 6 electrons on the outer orbit with another atom. So, each atom
has 4 "personal" electrons, 2 electrons that it shares with another atom
and 2 electrons that the other atom shares with it. Thus, the orbit
becomes full, all 8 spots are filled. The bond between two atoms of
oxygen is formed by two pairs of shared electrons, so structurally the
molecule of oxygen O2 can be pictured as

O=O

(notice double link between the atoms).



3. Our next example is gas methane. Its molecule consists of one atom of
carbon (6 electrons, 2 of them on the lowest orbit, 4 - on the next
one) and 4 atoms of hydrogen (1 electron on the lowest orbit of each
atom). Obviously, having only 4 electrons on the second orbit, carbon is
actively looking for electrons to fill the orbit. It needs 4 of them to
complete an orbit of 8 electrons. Exactly this it finds in 4 atoms of
hydrogen that need to complete their own lowest orbit. Sharing
electrons, one atom of carbon and 4 atoms of hydrogen fill their
corresponding orbits, thus creating a molecule of methane CH4 with can be pictured as

     H

      |

H−C−H

      |

     H




4. Carbon dioxide molecule contains 1 atom of carbon, that needs 4
electrons to complete its orbit, and 2 atoms of oxygen, each needs 2
electrons to complete its orbit: CO2. By
sharing 2 electrons from each atom of oxygen with 4 electrons from atom
of carbon they all fill up their outer orbit of electrons and become a
stable molecule, pictured as

O=C=O

(notice double link between the atoms).



5. Ethanol molecule contains 2 atoms of carbon, 1 atom of oxygen and 6 atoms of hydrogen connected as follows

     H   H

      |     |

H−C−C−O−H

      |     |

     H   H


(notice single bond between atoms of carbon and oxygen in ethanol, while
the bond between them in carbon dioxide has double link)



6. Hydrogen peroxide molecule contains 2 atoms of hydrogen and 2 atoms of oxygen connected as follows

H−O−O−H

(notice single bond between atoms of oxygen, not like in a molecule of oxygen)



Numerous examples above illustrate that bonds between atoms can be
different, even between the same atoms in different molecules. That's
why it is important to understand the structure of molecules, how
exactly the atoms are linked and what kind of links exist between them.
This is the basis for calculation of the amount of energy produced or
consumed by chemical reactions that rearrange the atoms from one set of
molecules to another.



Obviously, bonds O−O and O=O are different.
The first one is facilitated by one shared electron, the second one - by
two. The amounts of energy, needed to break these bonds, are different
too. Therefore, when calculating the energy of chemical reaction, it's
important to understand the kind of bond between atoms in each separate
case.

Tuesday, July 23, 2019

Unizor - Physics4Teens - Energy - Chemical Energy of Atomic Bonds







Notes to a video lecture on http://www.unizor.com



Energy of Atomic Bonds

in Molecules




In this lecture we will analyze the energy aspect of chemical reactions.

Consider the reaction of burning of methane. This gas is used in regular
gas stoves, so the reaction happens every time we cook something.

A molecule of methane consists of one atom of carbon C and four atoms of hydrogen H, the chemical formula of methane is CH4.

You can imagine a molecule of methane as a tetrahedron, in its center is
an atom of carbon and on each of its four vertices is an atom of
hydrogen.

A molecule of oxygen, as we know, consists of two atoms of oxygen and has a chemical formula O2.



As a result of the reaction of burning of methane, water and carbon dioxide are produced, according to the following equation:

CH4 + 2O2 = 2H2O + CO2

So, during this reaction

(a) four atomic bonds between carbon and hydrogen in one molecule of methane are broken,

(b) one atomic bond in each molecule of oxygen (out of two) are broken,

(c) two atomic bonds between hydrogen and oxygen in each molecule of water (out of two) are created,

(d) two atomic bonds between carbon and oxygen in a molecule of carbon dioxide are created.



Amounts of potential energy of the different atomic bonds are
experimentally determined, which would lead to calculation of the amount
of chemical energy released (for exothermic) or consumed (by
endothermic) reaction.



To make experiments to determine potential energy of the bonds inside a
molecule, we have to make experiments with known amounts of components
in chemical reaction. The reaction above includes one molecule of
methane and two molecules of oxygen. Obviously, we cannot experiment
with one or two molecules. The solution is to experiment with proportional amounts of components, say, 1 million of molecules of methane and 2 million of molecules of oxygen.



To explain how to do this, we have to get deeper into atoms. Physics
models atoms as consisting of three kinds of elementary particles -
protons (electrically positively charged), neutrons (electrically
neutral) and electrons (electrically negatively charged). This is a
relatively simple model, that corresponds to most of experiments, though
the reality is more complex than this. For our purposes we can view
this model of atom as a nucleus, that contains certain number of protons
and neutrons, and a number of electrons circulating the nucleus on
different orbits.



Electrons are very light relatively to protons and neutrons, so the mass
of an atom is concentrated, mostly, in its nucleus. Protons and neutron
have approximately the same mass, which is called atomic mass unit. So, the mass of an atom in atomic mass units
("atomic weight") is equal to the number of protons and neutrons in its
nucleus. This mass is known for each element of the Periodic Table of
Mendeleev, that is for each known atom.

For example, it is determined that atom of hydrogen H has atomic weight of 1 atomic unit, atom of carbon C has atomic weight of 12, atom of oxygen O has atomic weight 16.



Knowing atomic weights of atoms, we can calculate atomic weight of molecules. Thus, the atomic weight of a molecule of methane CH4 is 12+4=16. Atomic weight of a molecule of oxygen O2 is 16+16=32. Atomic weight of water H2O is 2+16=18.



Now we can take components of any chemical reaction proportional to the
atomic weight of corresponding molecules, which will result in
proportional number of molecules. For example, not being able to
experiment with one molecule of methane CH4 and two molecules of oxygen O2, we can experiment with 16 gram of methane and 64 gram
of oxygen, and the proportionality of the number of molecules will be
preserved - for each molecule of methane there will be two molecules of
oxygen.



As you see, taking amount of any mono-molecular substance in grams
equaled to the atomic weight of the molecules of this substance (called a
mole) assures taking the same number of molecules, regardless of the substance. This number is the Avogadro Number and is equal to N=6.02214076·1023.

Thus, one mole of methane CH4 (atomic weight of C is 12, atomic weight of H is 1) weighs 16g, one mole of silicon Si2 (atomic weight of Si is 14) weighs 28g, one mole of copper oxide CuO (atomic weight of Cu is 64, atomic weight of O
is 16) weighs 80g etc. And all those amounts of different substances
have the same number of molecules - the Avogadro number (approximately,
of course).



The theory behind the atomic bonds inside a molecule is quite complex
and is beyond the scope of this course. Based on this theory and
experimental data, for many kinds of atomic bonds there had been
obtained an amount of energy needed to break these bonds, that is its
inner chemical energy.

Thus, chemical energy of atomic bonds inside a mole of methane CH4
is 1640 kilo-joules (because a molecule of methane has 4 bonds between
carbon and each atom of hydrogen, each bond at 410KJ), inside a molecule
of oxygen O2 - 494 kilo-joules (1 bond between 2 atoms oxygen at 494KJ), inside a molecule of carbon dioxide CO2 is 1598 kilo-joules (2 bonds between carbon and each atom of oxygen, each 799KJ), inside a molecule of water H2O is 920 kilo-joules (2 bonds between oxygen and each atom of hydrogen, each 460KJ).



Let's go back to methane burning:

CH4 + 2O2 = 2H2O + CO2

This chemical reaction converts 1 mole of methane (16g) and 2 moles of
oxygen (64g) into 1 mole of carbon dioxide (44g) and 2 moles of water
(36g).

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

Ein = 1640 + 2·494 = 2628 KJ

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

Eout = 2·920 + 1598 = 3438 KJ

The net energy is

Enet = 2628 − 3438 = −810 KJ

This net energy is the amount of thermal energy released by burning 16g
of methane, using 64g of oxygen, obtaining as a result 44g of carbon
dioxide and 36g of water.

Thursday, July 18, 2019

Unizor - Physics4Teens - Energy - Atoms















Notes to a video lecture on http://www.unizor.com



Atoms and Chemical Reaction



Discussing mechanical energy, we analyzed the movement of objects.

When talking about thermal energy (heat), we had to go deeper inside the objects and analyzed the movement of molecules, the smallest parts of objects that retained the properties of objects themselves.

Now we go even deeper, inside the molecules, in search of new kinds of energy.



The components of molecules are called atoms. Currently there are more than 100 kinds of atoms, classified in the Mendeleev's Periodic Table.



Different combinations of these atoms in different quantities make up all kinds of molecules, each with its own properties.



In some cases a single atom makes up a molecule. For example, a single atom of iron (denoted by symbol Fe) makes up a molecule of iron.

In some other cases a pair of atoms of the same type makes up a molecule. For example, two atoms of oxygen (denoted by symbol O) make up a molecule of oxygen (denoted by symbol O2).

In more complicated cases a few atoms of different types make up a molecule. For example, two atoms of hydrogen (denoted by symbol H) and one atom of oxygen (O) make up a molecule of water (H2O).



One of the most complicated molecules that contains many elements in
different quantities is a molecule of protein that has about half a
million of atoms.



Chemical energy is a potential energy of bonds between atoms that hold them together in a molecule.

Chemical reaction is a process of re-arranging of atoms in a group of molecules, getting, as a result, a group of other molecules.

During chemical reactions some bonds between atoms are broken and some are created. Therefore, the energy might be either released or consumed in the process of chemical reaction.
This energy, stored in the molecules as potential energy of atomic
bonds and released or consumed during chemical reaction, is classified
as chemical energy.



Let's consider a few examples of chemical energy.



1. Coal burning

One molecule of carbon, that consists of one carbon atom C, and one molecule of oxygen, that consists of two oxygen atoms O2, when brought together and lit up, will join into one molecule of carbon dioxide CO2
in the process of burning. After the chemical reaction of burning is
initiated, it will maintain itself, as the process of burning produces a
flame that lights up new molecules of carbon, joining them with oxygen.

The chemical reaction

C + O2 = CO2

is endothermic (consumes heat energy) in the very beginning, when we
have to light up the carbon, but, as soon as the reaction started, it
becomes exothermic, that is it produces heat energy, because the
potential energy of atoms inside molecules of carbon and oxygen together
is greater than potential energy of atoms inside a molecule of carbon
dioxide.



2. Making water from hydrogen and oxygen

Two molecules of hydrogen, each consisting of two hydrogen atom H2, and one molecule of oxygen, that consists of two oxygen atoms O2, when brought together and lit up, will join into two molecules of water H2O
in the process of hydrogen burning. After the chemical reaction of
burning is initiated, it will maintain itself, as the process of burning
produces a flame that lights up new molecules of hydrogen, joining them
with oxygen.

The chemical reaction

2H2 + O2 = 2H2O

is endothermic (consumes heat energy) in the very beginning, when we
have to light up the hydrogen, but, as soon as the reaction started, it
becomes exothermic, that is it produces heat energy, because the
potential energy of atoms inside two molecules of hydrogen and one
molecule of oxygen together is greater than potential energy of atoms
inside two molecules of water.



3. Photosynthesis

This is a complicated process, during which the light from sun, air
components (such as carbon dioxide, nitrogen and oxygen), water and
whatever is in the soil are converted by the plants into chemical energy
that maintains their life. This is an endothermic process, and, as its
result, plants grow. In most cases they consume carbon dioxide from the
air, break it into carbon and oxygen, consume the water from the soil,
break it into hydrogen and oxygen (they need sun's radiation energy to
break the molecules of CO2 and H2O),
use the carbon, hydrogen and part of oxygen to produce new organic
molecules they consist of and release the unused oxygen back into
atmosphere.



4. Battery

Battery consists of three major components: anode, cathode and electrolyte
in-between anode and cathode. As a result of a chemical and
electro-magnetic reaction between the molecules of anode and
electrolyte, some electrons are transferred from anode to electrolyte.
Then, as a result of a chemical and electro-magnetic reaction between
the molecules of cathode and electrolyte, some electrons are transferred
from electrolyte to cathode. As the result, there are extra electrons
on the cathode, which were taken from the anode, thus creating
electrical potential.



These simple examples explain the general mechanism of chemical energy,
released or consumed in the course of chemical reaction, that transforms
molecules by rearranging their atoms' composition. As a result of a
chemical reaction and change in the atomic composition of molecules,
potential energy of bonds between atoms in molecules is changing. If the
total potential energy of the resulting molecules is greater than the
potential energy of the bonds inside original molecules, the process is
endothermic, it consumes energy. In an opposite case the process is
exothermic, it produces energy.



The exothermic process of extracting chemical energy using chemical
reaction is the key to getting energy from gasoline in the car engine,
producing heat and light in the fireplace by burning wood, it's the
source of energy in all living organisms, including humans. We exist
because our body knows how to extract chemical energy from the food.



Monday, July 8, 2019

Unizor - Physics4Teens - Energy - Heat Transfer Problems











Notes to a video lecture on http://www.unizor.com



Heat Transfer - Problems



Problem 1



Determine the power of the heat source inside the room required to
maintain a certain difference between inside and outside air
temperature, given the following:

(a) the difference between inside and outside temperature δ=Tout−Thome

(b) the room has only one wall facing the outside air, and the area of this wall is A

(c) the thickness of the wall is L

(d) the wall is made of solid material with a coefficient of thermal conductivity k.



Solution



Let x be the distance from a point inside the wall to its surface facing the room.

The T(x) is a temperature at this point as a function from this distance.

To be equal to Thome at the surface facing the room and to be Tout at the surface facing the outside and to be linearly changing from one value to another inside the wall of the thickness L, the temperature inside the wall at distance x from the surface facing the room should be

T(x)=Thome+(Tout−Thome)·x/L

From this we can determine the heat flux through the wall at a distance x, using the Fourier's law of thermal conduction:

Q(x,A) = −k·A·dT(x)/dx =

= −k·A·(Tout−Thome)/L =

= −k·A·δ/L


That is exactly how much heat we need to maintain the difference between temperature in the room and outside.



Problem 2



Calculate the heating requirement of the room with only one concrete
wall facing outside with no windows, assuming the following:

(a) the temperature in the room must constantly be Troom=20°C

(b) the temperature outside is also constant Tout=5°C

(c) the thickness of the concrete wall facing outside is L=0.2m

(d) the area of the wall facing outside is A=12m²

(e) heat conductivity of concrete is k=0.6W/(m·°K)



Solution



Using the above, we can determine the heat flux through the wall at a distance x:

Q(x,A) = −k·A·δ/L

(for any distance x from the surface of the wall that faces the room)

Outside surface of the wall is at 5°C, inside is at 20°C.

So, δ=15°.

Therefore, the heat flux through the wall (at any distance from the inside surface) will be

Q = 0.6·12·15/0.2 = 540W

That is exactly how much heat we need to maintain the temperature in the room.



Problem 3



Our task is to determine the law of cooling of a relatively small hot
object immersed in the cool infinitely large reservoir with liquid or
gaseous substance.

We assume the volume of substance this object is immersed in to be
"infinite" to ignore its own change of temperature related to heat
emitted by our hot object.

This law of cooling should be expressed in terms of object's temperature T as a function of time t, that is, we have to find the function T(t).

Assumptions:

(a) the initial temperature of the object at time t=0 is assumed to be T0

(b) the object has a shape of a thin flat square (so, its temperature is changing simultaneously in all its volume) of size LxL and mass m

(c) the specific heat capacity of the object's material is C

(d) the temperature of the substance surrounding our object is constant and equals Ts

(e) the convective heat transfer coefficient of the substance around our object is h.



Solution



Consider a small time interval from t to t+Δt.

The temperature of an object is T(t), while the temperature of the substance around it is constant Ts.
Then the amount of heat transferred from our object to the substance
around it per unit of time per unit of its surface area is proportional
to the difference in temperatures with the convective heat transfer coefficient as a factor:

q = −h·[T(t)−Ts]

Since total surface area of our thin flat square, ignoring its thickness, is 2L² and the time interval we consider is Δt, the total amount of heat transferred by our object through its surface during this time period is

ΔQ(t) = −2L²·h·[T(t)−Ts]·Δt

This is amount of heat taken away by the substance from our object during a time interval Δt.

The same amount of heat is lost by the object, taking its temperature from T(t) to T(t+Δt).

As we know, changes of heat and temperature are proportional and related to the object mass and specific heat capacity:

ΔQ(t) = C·m·ΔT

where ΔT = T(t+Δt)−T(t).

Equating the amount of heat lost by our object to the amount of heat
carried away by convection of the substance around it, we have come to
an equation

−2L²·h·[T(t)−Ts]·Δt =

= C·m·
[T(t+Δt)−T(t)]

Dividing both parts by Δt, diminishing this time interval to zero and using a derivative by time t to express the limit, we get the following differential equation

−2L²·h·[T(t)−Ts] =

= C·m·
dT(t)/dt


To solve it, let's make two simple substitutions:

(a) A = 2h·L²/(C·m)

(b) X(t) = T(t) − Ts

Then, since Ts is constant,

dT(t)/dt = dX(t)/dt

and our differential equation looks like

dX(t)/dt = −A·X(t)

To solve this, we convert it as follows:

dX(t)/X(t) = −A·dt

d[ln(X(t))] = −A·dt

Integrating:

ln(X(t))] = −A·t + B,

where B is any constant, defined by initial condition

X(0)=T(0)−Ts=T0−Ts.

From the last equation:

X(t) = eB·e−A·t

Using initial condition mentioned above,

X(t) = (T0−Ts)·e−A·t

Since X(t)=T(t)−Ts

T(t) − Ts = (T0−Ts)·e−A·t

or

T(t) = Ts + (T0−Ts)·e−A·t

where A = 2h·L²/(C·m)

So, the difference in temperature between a hot object and infinitely
large surrounding substance is exponentially decreasing with time.

Tuesday, June 25, 2019

Unizor - Physics4Teens - Energy - Heat Transfer - Radiation




Notes to a video lecture on http://www.unizor.com

Heat Transfer - Radiation

Heat radiation IS NOT the same as radioactivity. Though, under certain circumstances (like an explosion of an atomic bomb) the heat radiation and radioactivity are both present. When we discuss the heat radiation we talk about a process that occurs in any object with a temperature greater than absolute zero, while radioactivity occurs in extreme cases of very high energy output.

Heat transfer through radiation is totally different from conduction and convection. The most important property of heat transfer by radiation is that heat transfer occurs without any visible material conduit that carries the heat, like molecular movement in two other cases.

Let's start from an example.
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.

The fundamental concept that lies in the foundation of a process of heat radiation is a concept of a field.
The field is a region of space, where certain forces act on certain objects without visible material medium.
As an example of the field, consider gravity. 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 gravitational field exist, yet we did study its behavior, the forces involved and the laws of motion in this field.

There are other fields.
Magnetic field around our planet, acting on a compass, forces the arrow to point North.
Electric field exists around electrically charged objects, so other electrically charged objects are attracted to or repulsed from it.

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?.

Let's start with a particular field called electro-magnetic. Very simplified description of this field is as follows.

Any electron creates an electric field around itself. Moving electrons, which we call electric current or electric field that changes in time, also create a magnetic field around them. So, changing in time electric field creates magnetic field that changes in space. It's an experimental fact, and we have the whole theory about properties of these fields.

Consider an experiment, when you move a metal rod or any other electrical conductor in a magnetic field or change a magnetic field around any electrical conductor, thus creating a magnetic field that changes in time. You will observe that there is an electric current in the conductor, thus creating an electric field that changes in space. It's an experimental fact, and we also have the whole theory about properties of this process.

So, electric field creates magnetic field, which, in turn, creates electric field etc. This is a loop of energy conversion that propagates extremely fast, with a speed of light, about 3·108m/sec.

The combination of electric and magnetic forces form electro-magnetic field that propagates much faster than its physical medium - electrons. So, the propagation of the electro-magnetic field seems to be a self-sufficient process, occurring without the medium. This is a very brief and unsatisfactory explanation of the nature of the electro-magnetic field. We will not go much further in this explanation, but rather concentrate on the properties of the electro-magnetic field.

Assuming that we accept the existence of the electro-magnetic field 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 electro-magnetic field transfers energy.
This energy transfer by electro-magnetic field is called radiation.

As in a case of oscillating molecules in a solid, carrying more heat energy when oscillation is more intense (higher frequency), the electro-magnetic field oscillation carries energy with higher frequencies being more "energetic" than lower.

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.
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.

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 light and heat radiation are manifestations of the same process of transferring energy by the oscillations of the components of the electro-magnetic field.

An object does not have to have a temperature of the Sun to emit heat radiation. 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 radioactivity is used. All frequencies can be observed using some scientific instruments.

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.

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.
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:
q = σ·T4 where
σ = 5.67·10−8 W/(m2·°K4)
is the Stefan-Boltzmann constant.
This is the Law of Stefan-Boltzmann. Its derivation is complex and is outside of the scope of this course.

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 heat radiation. More precisely, it's a specific property of oscillations of the electro-magnetic field.
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.
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.