Question

Calculate the period of a satellite orbiting the Moon, 95 km above the Moon’s surface. Ignore effects of the Earth. The radius of the Moon is 1740 km.

Short answer

The period of a satellite is $7.05\times {10}^3\mathrm{s}$.

Step-by-step solution

Step 1. Given Data

The distance above the Moon’s surface is $d=95\mathrm{km}$.

The radius of the Moon is $r=1740\mathrm{km}$.

Step 2. Understanding the relation of speed of the satellite

In this problem, the time period of the satellite will be calculated by using the relation of the speed of the satellite in a circular orbit having radius r and mass m.

Step 3. Estimating the time period of the satellite

The relation of speed of a satelliteis given by,

$\begin{array}{c}v=\sqrt{\frac{Gm}{r+d}}\\ \frac{2\pi \left(r+d\right)}{T}=\sqrt{\frac{Gm}{r+d}}\\ T=\frac{2\pi \left(r+d\right)}{\sqrt{\frac{Gm}{\left(r+d\right)}}}\end{array}$

Here, m is the mass of the Moon, G is the gravitational constant and T is the time period.

On plugging the values in the above relation.

$\begin{array}{c}T=\frac{2\pi \left(1740\mathrm{km}\times \frac{1000\mathrm{m}}{1\mathrm{km}}+95\mathrm{km}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\right)}{\sqrt{\frac{\left(6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(7.35\times {10}^{22}\mathrm{kg}\right)}{\left(1740\mathrm{km}\times \frac{1000\mathrm{m}}{1\mathrm{km}}+95\mathrm{km}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\right)}}}\\ T=7.05\times {10}^3\mathrm{s}\end{array}$

Thus, $T=7.05\times {10}^3\mathrm{s}$is the required period of a satellite.