Chapter 7, Problem 86A

To determine

To identify: Whether satellite M or E has a smaller period.

Answer to Problem 86A

Earth’s satellite E will have smaller period.

Explanation of Solution

Given:

Mass of Mars = $\frac{1}{9}$ Mass of Earth

M is the satellite of Mars

E is the satellite of Earth

Radius $r_M=r_E$

Formula used:

According to Newton’s form of Kepler equation period is given as:

  $T=2\pi \sqrt{\frac{r^3}{Gm}}$

Calculation:

Since $r_M=r_E$ and $G$ is constant. So, $T$ will be inversely proportional to square root of mass.

Since, Mass of Mars = $\frac{1}{9}$ Mass of Earth

So, $T$ of Mars will be three times $T$ of Earth.

Hence, Earth’s satellite E will have smaller period.

Conclusion:

Therefore, the satellite having smaller period is E.