Chapter 5, Problem 48P

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.

To determine

The period of a satellite orbiting the Moon.

Answer to Problem 48P

Solution:

  $7052\text{s}$

Explanation of Solution

Given:

The height of the satellite from the Moon, $h=95\text{km}$

The radius of the Moon, $R=184\text{0}\text{km}$

Formula Used:

From Kepler’s third law of planetary motion:

  $\frac{T^2}{r^3}=\frac{4{\pi }^2}{GM}$ and $r=R+h$

Where, $G$ is the gravitational constant.

  $M$ is the mass of the moon

  $R$ is the radius of the moon

Calculation:

According to the Kepler’s law of planetary motion, the relation between the time periodTand the mean radius of the orbit of the satellite is given by

  $\frac{T^2}{r^3}=\frac{4{\pi }^2}{GM}$

  $\frac{T^2}{{\left(R+h\right)}^3}=\frac{4{\pi }^2}{GM}$

  $T^2=\frac{4{\pi }^2}{GM}{\left(R+h\right)}^3$

  $T=\sqrt{\frac{4{\pi }^2}{GM}{\left(R+h\right)}^3}$

Substituting all the values, we get

  $T=\sqrt{\frac{4{\pi }^2}{6.67\times {10}^{-11}\times 7.347\times {10}^{22}}{\left(1740\times {10}^3+95\times {10}^3\right)}^3}$

  $T\approx 7052\text{s}$

Conclusion:

Therefore, the period of a satellite orbiting the Moon is $7052\text{s}$ .