Question

Assume the radius of the earth to be $6.4\times {10}^6$m.
a. Calculate the time period T of a satellite on equatorial orbit at $1.4\times {10}^3$ above the surface of the earth.
b. What is the speed of the satellite in this orbit?
c. If the satellite is travelling in the same direction as the rotation of the earth, i.e., west to east, what is the interval between two successive times at which it will appear vertically overhead to an observer at a fixed point on the equator?

Answer 1

a. $r=R_{earth}+h=6.4\times {10}^6+1.4\times {10}^6=7.8\times {10}^6$m
$\Rightarrow r=7.8\times {10}^{6}m\Rightarrow T=\left(\frac{4{\pi }^2{r}^3}{G{M}_{earth}}\right)=6831$s
b. Speed of satellite,
$V=\sqrt{\frac{G{M}_{earth}}{r}}=7174m{s}^{-1}$
c. Relative angular velocity $={\omega }_{satellite}-{\omega }_{earth}$
Thus $T=\frac{2\pi }{({\omega }_s-{\omega }_e)}=\frac{2\pi }{[\frac{2\pi }{T_s}-\frac{2\pi }{T_e}]}$
$\Rightarrow T=\frac{T_s{T}_e}{T_e-T_s}=7417$s.