Tuesday, October 15, 2019
Unizor - Physics4Teens - Energy - Gravitational Potential
Notes to a video lecture on http://www.unizor.com
Gravity Integration 4 -
1. Determine the potential of the gravitational field of a uniform solid sphere at any point outside it.
Let's establish a system of coordinates with a sphere's center at the origin of coordinates and X-axis going through a point of interest P, where we have to determine the gravitational potential.
Assume that the sphere's radius is R and the mass is M. Then its volume is 4πR³/3; and the mass density per unit of volume is ρ=3M/(4πR³).
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 sphere R.
If, instead of a sphere, 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(H) = −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 a solid sphere at point P on the X-axis, let's divide it into infinite number of infinitesimally thin concentric spherical shells, all centered at the origin of coordinates and use the results presented in the previous lecture about spherical shell.
As the variable of integration we will chose a radius of a spherical shell r that varies from 0 to R. Its outside surface area is 4πr², its thickness is dr and, therefore, its volume is 4πr²·dr.
This allows us to calculate the mass of this spherical shell using the volume and mass density calculated above.
dm = ρ·4πr²·dr =
= 3M·4πr²·dr/(4πR³) =
The formula for gravitational potential of a spherical shell, derived in the previous lecture was V=−G·M/H, where G is a gravitational constant, M is a mass of a spherical shell and H is a distance from a center of a shell to a point of interest.
In case of a solid sphere divided into infinite number of infinitesimally thin concentric spherical shells the distance H remains the same. So, all we have to do is to substitute the mass in the formula for a shell with the variable mass of a shells we divided our solid sphere and to integrate by variable radius r:
V(H) = −(G/H)∫[0;R]dm =
= −(G/H)∫[0;R]3M·r²·dr/R³ =
The indefinite integral of r² is r³/3, which gives the value of the integral
∫[0;R]r²·dr = R³/3 − 0 = R³/3
V(H) = −G·M/H
Remarkably, the formula is exactly the same as if the whole mass was concentrated in the center of a sphere, the same as in case of a spherical shell.
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.
In theory, this result was easily predictable. The gravitational potential of each spherical shell is the same as if its mass is concentrated at its center. All shells are concentric, therefore the masses of all of them are concentrated in the origin of coordinates and can be added together, since the gravitational potential is additive. Thus, we come to the same value of gravitational potential of a sphere, as if its total mass is concentrated in one point - its center.
Let's analyze the force of gravity, acting on a probe object of a mass m at a point of interest on the distance H from a center of a solid sphere. This is a mass of a prove object multiplied by a derivative of the gravitational potential:
F(H) = m·dV(H)/dH =
which is a well known Newton's Law of Gravitation.
2. Determine the potential of the gravitational field of a uniform solid sphere of radius R at a point inside it at distance r from a center.
For this problem, as in the Problem 1 above, we will need a mass density per unit of volume ρ=3M/(4πR³).
Assume that a probe object is a distance r from a center of sphere, which is less than the radius of a sphere R. Let's calculate the gravitational potential V(r) of the combination of two separate sources - the solid sphere of radius r with a probe object on its surface and a thick empty spherical object between a surface of a sphere of the radius r and surface of a sphere of the radius R.
The gravitational potential of a uniform solid sphere of radius r on its surface is discussed above as a Problem 1. To use the results of this problem, we need a mass M1(r) of a source of gravity and the distance of a point of interest from a center H.
The mass is
M1(r) = ρ·4πr³/3 = M·r³/R³
The distance form a center is
H = r
The gravitational potential on the surface of this solid sphere of the radius r equals to
V1(r) = −G·M·r²/R³
Consider now the second source of gravity - a thick empty sphere between the radiuses r and R.
As in Problem 1 above, we will divide a thick empty sphere into an infinite number of concentric infinitesimally thin spherical shells of a variable radius x and thickness dx.
The mass of each shell is
dm(x) = ρ·4πx²·dx =
= 3M·4πx²·dx/(4πR³) =
The potential inside such an infinitesimally thin spherical shell of radius x is, as we know from a previous lecture, constant and equals to
dV(x) = −G·dm(x)/x =
To get a full potential inside such a thick empty sphere we have to perform integration of this expression from x=r to x=R.
V2(r) = (−3G·M/R³)∫[r;R]x·dx =
The total potential at distance r from a center equal to sum of two potentials calculated above
V(r) = V1(r) + V2(r) =
= −G·M·r²/R³ −
On the outer surface of this sphere, when r=R, the above formula converts into the one derived in Problem 1:
V(R) = −G·M·/R
In the center of a solid sphere, when r=0, the potential is
V(R) = −(3/2)·G·M·/R
Let's analyze the force of gravity, acting on a probe object of a mass m at a point of interest on the distance r from a center of a solid sphere. This is a mass of a prove object multiplied by a derivative of the gravitational potential:
Case inside a solid sphere:
F(r) = m·dV(r)/dr =
So, as we move from a center of a solid sphere (r=0) towards its outer surface (r=R), the force is linearly growing from zero at the center to G·m·M/R² at the end on the surface.
Case outside a solid sphere (using the results of Problem 1 above for V(H)=−G·M/H, where H=r is greater than R):
F(r) = m·dV(r)/dr =
So, as we move from a surface of a solid sphere (r=R) outwards to infinity, increasing r, the force is decreasing inversely to a square of a distance from the center from G·m·M/R² to zero at infinity.
It's quite interesting to graph the force of gravitation as a function of a distance of a probe object from a center of a solid sphere. We have two different functions that represent this force, one inside and one outside the surface of a sphere.
The graph looks like this: