*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

*and the mass is*

**R***, so the density of mass per unit of length is*

**M***.*

**ρ=M/(2πR)**Assume further that the Z-coordinate of a point

*, where we want to calculate the gravitational potential, is*

**P***.*

**H**If, instead of a ring, we had a point mass

*concentrated in its center at point*

**M***, its gravitational potential at a point*

**(0,0,0)***would be*

**P**

**V**_{0}= −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

*, and the field performs this work for us, so we perform negative work).*

**P**Since the mass in our case is distributed along the circumference of a ring, and every point on a ring is on a distance

*from point*

**r=√R²+H²***, which is further from this point than the center of a ring, the gravitational potential of a ring at point*

**P***will be smaller.*

**P**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

*and integrate all these potentials.*

**P**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***]. Its increment*

**0;2π***d*gives an increment of the circumference of a ring

**φ***d*

**l = R·**d**φ**The mass of this infinitesimal segment of a ring is

*d*

**m = ρ·**d**l = M·R·**d**φ /(2πR) = M·**d**φ /(2π)**The distance from this infinitecimal segment of a ring to a point of interest

*is independent of variable*

**P***and is equal to constant*

**φ***.*

**r=√R²+H²**Therefore, gravitational potential of an infinitecimal segment of a ring is

*d*

**V = −G·**d**m /r = −G·M·**d**φ /(2π√R²+H²)**Integrating this by variable

*on [*

**φ***], we obtain the total gravitational potential of a ring at point*

**0;2φ***:*

**P**

**V = ∫**d_{[0;2π]}**V = −∫**d_{[0;2π]}G·M·**φ /(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

*and the mass is*

**R***, so the density of mass per unit of surface is*

**M***.*

**ρ=M/(πR²)**Assume further that the Z-coordinate of a point

*, where we want to calculate the gravitational potential, is*

**P***.*

**H**Let's split our disc into infinite number of infinitely thin concentric rings of radius from

*to*

**x=0***of width*

**x=R***d*each and use the previous problem to determine the potential of each ring.

**x**The mass of each ring is

*d*

**m(x) = ρ·2πx·**d**x**This gravitational potential of this ring at point

*, according to the previous problem, is*

**P***d*

**V(x) = −G·**d**m(x) /√x²+H² =**

= −G·ρ·2πx·d= −G·ρ·2πx·

**x /√x²+H² =**

= −G·M·2πx·d= −G·M·2πx·

**x /(πR²√x²+H²) =**

= −G·M·2x·d= −G·M·2x·

**x /(R²√x²+H²)**To determine gravitational potential of an entire disc, we have to integrate this expression in limits from

*to*

**x=0***.*

**x=R**

**V = ∫**d_{[0;R]}**V(x) = −k·∫**d_{[0;R]}2x·**x /√x²+H²**where

**k = G·M /R²**Substituting

*and noticing the*

**y=x²+H²***d*, we get

**y=2x·**d**x**

**V = −k·∫**d_{[H²;H²+R²]}**y /√y**The derivative of

*is*

**√y***Therefore, the indefinite integral of*

**1 /(2√y)***is*

**1 /√y**

**2√y + C**Finally,

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