Question

Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23,500 km
from the center.
What is the ratio of the orbital period of Deimos to that of Phobos?

Answer 1

Answer:

The ratio is   $\frac{T_1}{T_2} = 3.965$

Explanation:

From the question we are told that

   The  radius of Phobos orbit is  R_2 =  9380 km

    The radius  of Deimos orbit is  $R_1 = 23500 \ km$

Generally from Kepler's third law

    $T^2 = \frac{ 4 * \pi^2 * R^3}{G * M }$

Here M is the mass of Mars which is constant

        G is the gravitational  constant

So we see that $\frac{ 4 * \pi^2 }{G * M } = constant$

   

    $T^2 = R^3 * constant$      

=>  $[\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3$

Here $T_1$ is the period of Deimos

and  $T_1$ is the period of  Phobos

So

      $[\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}$

=>    $\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}]$

=>    $\frac{T_1}{T_2} = 3.965$