Given:
Radius of earth, [tex]R_{e} = 6.38\times 10^{6} m[/tex]
Radius of of orbit, [tex]r_{o} = 4.43R_{e} = 4.43\times 6.38\times 10^{6} = 2.826\times 10^{7} m[/tex]
mass of earth, [tex]M_{e} = 5.98\times 10^{24}[/tex]
[tex]G = 6.67\times 10^{-11} m^{3}kg{-1}s^{-2}[/tex]
Formula used:
By Kepler's third law,
[tex]T^{2} = \frac{4\pi^{2} r^{3}}{GM_{e}}[/tex]
Solution:
Using the given values in above mentioned formula:
[tex]T^{2} = \frac{4\pi^{2} (4.43R)^{3}}{GM_{e}}[/tex]
[tex]T = {\sqrt \frac{4\pi^{2} (2.826\times 10^{7})^{3}}{6.67\times 10^{-11}\times 5.98\times 10^{24}}}[/tex]
[tex]T= 4.73\times 10^{4} s[/tex]
Therefore, time period of satellite is [tex]T= 4.73\times 10^{4} s[/tex]
