What is the period of revolution of a satellite of mass m that orbits the earth in a circular path of radius $7880 \; km$ about $1500 \; km$ above the surface of the earth?

Given,

Radius of the orbit $(r) = 7880\;km = 7880\; * \;10^3\;m$

Height $(h) = 1500\;km = 1500\;*\;10^3\;m$

Period of revolution $(T) = \;?$

Radius of earth $(R) = r - h = (7880 - 1500)\; * \; 6380\;*\;10^3\;m$

We have,

$T = 2 \pi \; * \; \sqrt{ \frac{(R\;+\;h)^{3}}{(g\;*\;R)^2}}$

or, $T = 2 \pi \; * \; \sqrt{ \frac{(6380000\;+\;1500000)^3}{9.8\;*\;(6380000)^2}} = 6955.29\;Sec = 1.91\; hrs$

∴ The time period of the revolution = $1.91\;hrs$.

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