Question

The satellite orbits at a distance from the center of the moon. Which of the following is a correct expression for the time it takes the satellite to make one complete revolution around the moon?

A) T = 2π
`\sqrt (R^3)/(Gm)`
B) T = 2π
`\sqrt(R^3)/(GM)`
C) T = 2π
`\sqrt(Gm)/(R^3)`
D) T = 2π
`\sqrt(GM)/(R^3)`

Answer 1

`\displaystyle 2\, \pi\, \sqrt{(R^(3))/(G\m M)}`, where
`R` is the orbital radius,
`M` is the mass of the Moon, and
`G` is the gravitational constant.

Step-by-step explanation:

Let
`m` denote the mass of the satellite. Let
`R` denote the orbital radius, let
`M` denote the mass of the Moon, and let
`G` denote the gravitational constant.

The Moon would exert the following gravitational attraction on the satellite:

`\displaystyle (G\, M\, m)/(R^(2))`.

Let
`\omega` denote the angular velocity of the satellite. For the satellite to stay in this orbit of radius
`R`, the net force on the satellite needs to be:

`m\, \omega^(2)\, R`.

Since the gravitational force is the only force on this satellite, the net force on the satellite would be equal to the gravitational force:

`\displaystyle m\, \omega^(2)\, R = (G\, M\, m)/(R^(2))`.

Rearrange this equation to find the angular velocity:

`\displaystyle \omega^(2) = (G\, M)/(R^(3))`.

`\displaystyle \omega = \sqrt{(G\, M)/(R^(3))}`.

Note that with the Moon as the center, a full revolution around the Moon would take an angular distance of
`2\, \pi`. Divide the angular distance by the angular velocity to find the time required for this revolution:

`\begin{aligned}T &= (2\, \pi )/(\omega) && \genfrac{}{}{0}{}{\text{angular displacement}}{\text{angular velocity}} \\ &= 2\, \pi \, \sqrt{(R^(3))/(G\, M)}\end{aligned}`.