Question

A satellite of mass $1000kg$ is supposed to orbit the earth at a height of $2000km$ above the earth's surface. Find its time period. (Mass of the earth $=6\times {10}^{24}$ kg)

Answer 1

Let the angular velocity be $\omega$. As per Newton's second law,

$m{\omega }^{2}r=\frac{GMm}{r^2}$

$\omega =\sqrt{\frac{GM}{r^3}}$

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{r^3}{GM}}=2\pi \sqrt{\frac{{\left(8400000\right)}^3}{(6.67\times {10}^{-11})(6\times {10}^{24})}}=2.1hrs$

Answer is $2.1hrs$