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.