An earth satellite moves in a circular orbit with a speed of $6.2\;km/s$. Find the time of one revolution and its centripetal acceleration.

Given,

Orbital velocity $v_o = 6.2\;km/s = 6200\;m/s$

Radius of earth $(R) = 6400000\;m$

Time Period $(T) = \; ?$

Centripetal Acceleration $(a) = \;?$

We have,

$ v_{0} = R \sqrt { \frac{g}{R+h}} $

or, $6200 = 6400000\;*\; \sqrt{\frac{10}{6400000 \; + \; h}}$

or, $9.38 \; *\; 10^{-7} = \sqrt{\frac{10}{6400000 + h}}$

or, $h = 10655301 - 6400000$

h = $4255301\;m$

Now, 

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

or, $2\pi \; \sqrt{\frac{(4255301 + 6400000)^3}{10 * 6400000}}$

or, $2 \; \pi \;*\;1718.57 = 10792.65\;Sec = 2.99\;hrs$

∴ Time of revolution = $2.99\;hrs$

Again,

Centripetal Acceleration $a = \frac{v_0^2}{h + R} = \frac{(6200)^2}{4255301 + 6400000} = 3.6\;m/s^2$

∴ Centripetal Acceleration = $3.6\;m/s^2$

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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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