## 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) /