Question

A satellite circles the earth in an orbit whose radius is 4.43 times the earth's radius. The earth's mass is 5.98 x 10^24 kg, and its radius is 6.38 x 10^6 m. What is the period of the satellite?

Answer 1

Given:

Radius of earth, $R_{e} = 6.38\times 10^{6} m$

Radius of of orbit,  $r_{o} = 4.43R_{e} = 4.43\times 6.38\times 10^{6} = 2.826\times 10^{7} m$

mass of earth, $M_{e} = 5.98\times 10^{24}$

$G = 6.67\times 10^{-11} m^{3}kg{-1}s^{-2}$

Formula used:

By Kepler's third law,

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

Solution:

Using the given values in above mentioned formula:

$T^{2} = \frac{4\pi^{2} (4.43R)^{3}}{GM_{e}}$

$T = {\sqrt \frac{4\pi^{2} (2.826\times 10^{7})^{3}}{6.67\times 10^{-11}\times 5.98\times 10^{24}}}$

$T= 4.73\times 10^{4} s$

Therefore, time period of satellite is $T= 4.73\times 10^{4} s$