Question

A satellite is revolving in a circular equatorial orbit of radius $R=2\times {10}^4$km from east to west. Calculate the interval after which it will appear at the same equatorial town. Given that the radius of the earth $=6400$km and g(acceleration due to gravity) $=10$m $s^{-2}$.

Answer 1

Let $\omega$ be the actual angular velocity of the satellite from east to west and ${\omega }_e$ be the angular speed of the easth(west to east).
Then ${\omega }_{relative}=\omega -(-{\omega }_e)=\omega +{\omega }_e$
$\Rightarrow \omega ={\omega }_{rel}-{\omega }_e$
By the dynamics of circular motion
$\frac{GMm}{R^2}=m{\omega }^{2}R$
or ${\omega }^2=\frac{g{R}^2}{R^3}$ $(\because GM=g{R}_e^3)$
$\Rightarrow \omega =\sqrt{\frac{g{R}_e^3}{R^3}}$
$\therefore {\omega }_{rel}=\sqrt{\frac{g{R}_e^3}{R^3}}+{\omega }_e$
$\Rightarrow {\omega }_{rel}=\sqrt{\frac{10\times {6.4}^2\times {10}^{12}}{2^3\times {10}^{21}}}+7.27\times {10}^{-5}$
$(\because {\omega }_e=\frac{2\pi }{86400}=7.27\times {10}^{-5})$
$\Rightarrow {\omega }_{rel}=22.6\times {10}^{-5}+7.27\times {10}^{-5}=30\times {10}^{-5}rad{s}^{-1}$
$\therefore \Gamma =\frac{2\pi }{{\omega }_{rel}}=\frac{2\pi }{30\times {10}^{-5}}=2.09\times {10}^{4}s=5h48$min