Question

A satellite revolving in a circular equatorial orbit of radius $R=2.0\times {10}^4$ from west to east appears over a certain point at the equator every $11.6h$. From these data, the mass of the earth is $x\times {10}^{24}kg$ Find $x$. $G=6.67\times {10}^{-11}N{m}^2{Kg}^{-2}$

Answer 1

Relative angular velocity is given by

${\omega }_{rel}=\frac{2\pi }{11.6\left(3600\right)}=1.5X{10}^{-4}/s$

Angular velocity of earth is ${\omega }_e=\frac{2\pi }{24\left(3600\right)}=7.3X{10}^{-5}/s$

Thus, angular velocity of satellite is $\omega =2.23X{10}^{-4}/s=\sqrt{\frac{GM}{R^3}}=\sqrt{\frac{6.67X{10}^{-11}M}{6400000}}$

$M\approx 6X{10}^{24}kg$

Answer is 6.