Chapter 7, Problem 95A

To determine

The orbital period of satellite in units of the Moon’s period.

Answer to Problem 95A

  $\text{Orbital Period of Satellite}=\frac{1}{\sqrt{8}}\left(\text{Orbital period of the Moon}\right)$

Explanation of Solution

Given:

Satellite is orbiting around the Earth just like the Moon.

The orbital radius of the satellite is half of that of the Moon’s.

Formula used:

Period of a satellite orbiting the Earth can be obtained by:

  $T=2\pi \sqrt{\frac{r^3}{Gm}}$

Where, $G$ is the gravitational constant, $r$ is the orbital radius, and $m_E$ is the mass of the Earth.

Calculation:

The value of gravitational constant is:

  $G=6.67\times {10}^{-11}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Now, let the time period of the Moon be $T_m$ , then

  $T_m=2\pi \sqrt{\frac{r_M^3}{G{m}_E}}$

Similarly, the time period of a satellite having the orbital radius half of the Moon’s orbital radius will be,

  $\begin{array}{c}{T}_S=2\pi \sqrt{\frac{r_S^3}{G{m}_E}}\\ T_S=2\pi \sqrt{\frac{{\left(\frac{r_{Moon}}{2}\right)}^3}{G{m}_E}}\\ T_S=\frac{1}{\sqrt{8}}{T}_m\end{array}$

Conclusion:

Thus, the orbital period of the satellite will be $\frac{1}{\sqrt{8}}$ times the orbital period of the Moon.