Question

The moon Phobos orbits Mars

Answer 1

$27.9816 \times 10^{3} s$ is the period of orbit.

Explanation:

The equation that is useful in describing satellites motion is Newton form after Kepler's Third Law. The period of the satellite (T) and the average distance to the central body (R) are related as the following equation:

                  $\frac{T^{2}}{R^{3}}=\frac{4 \times \pi^{2}}{G \times M_{c e n t r a l}}$

Where,

T is the period of the orbit

R is the average radius of orbit

G is gravitational constant  $6.673 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}$

Here, given data

$M=6.23 \times 10^{23} \mathrm{kg}$

$R=9.38 \times 10^{6} \mathrm{m}$

Substitute the given values, we get T as

      $\frac{T^{2}}{\left(9.38 \times 10^{6}\right)^{3}}=\frac{4 \times(3.14)^{2}}{\left(6.673 \times 10^{-11}\right) \times 6.23 \times 10^{23}}$

      $T^{2}=\frac{4 \times 9.8596 \times 825.29 \times 10^{18}}{41.57 \times 10^{12}}$

      $T^{2}=\frac{32548.12 \times 10^{18-12}}{41.57}=782.97 \times 10^{6}$

Taking square root, we get

       $T=27.9816 \times 10^{3} s$