Chapter 7, Problem 53A

To determine

The period of Earth’s Moon if the orbital radius were doubled.

Answer to Problem 53A

79 days

Explanation of Solution

Given:

The orbital radius of Earth’s Moon is $3.9\times {10}^8\text{m}$ .

Formula used:

Period of a Planet orbiting the Sun is given by,

  $T=2\pi \sqrt{\frac{r^3}{G{m}_S}}$

Where $G$ is the gravitational constant, $r$ is the orbital radius, and $m_S$ is the mass of the Sun given as $1.99\times {10}^{30}\text{kg}$ .

Calculation:

The value of gravitational constant is,

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

Also, the mass of the Earth is $m_E=5.98\times {10}^{24}\text{}\text{kg}$ .

Then, considering the motion of the Moon around the Earth as the motion of a planet around the Sun, the orbital period of the Moon if its orbital radius is doubled will be,

  $\begin{array}{c}T=2\pi \sqrt{\frac{{\left(2r\right)}^3}{G{m}_E}}\\ =2\pi \sqrt{\frac{{\left(2\times 3.9\times {10}^8\right)}^3}{6.67\times {10}^{-11}\times 5.98\times {10}^{24}}}\\ =2\pi \left(1.09\times {10}^6\right)\\ =6.84\times {10}^6\text{s}\times \frac{1\text{}\text{hr}}{3600\text{s}}\times \frac{1\text{}\text{day}}{24\text{hr}}\\ \approx 79\text{days}\end{array}$

Conclusion:

Thus, the orbital period of the Moon around the Earth would be 79 days if the Moon’s orbital radius were doubled.