Question

A satellite P is revolving around the earth at a height h $=$ radius of earth (R) above equator. Another satellite Q is at a height 2h revolving in opposite direction. At an instant the two are at same vertical line passing through centre of sphere.
The least time after which again they are in this situation is given as $\frac{2\pi R^{3/2}\left(x\sqrt{6}\right)}{\sqrt{GM}(2\sqrt{2}+3\sqrt{3})}$. Find $x$.

Answer 1

${\omega }_p=({\frac{GM}{R+h}}^3{)}^{0.5}=(\frac{GM}{(2R{)}^3}{)}^{0.5}$

Similarly, ${\omega }_q=(\frac{GM}{(3R{)}^3}{)}^{0.5}$

Relative $\omega ={\omega }_p+{\omega }_q=(\frac{GM}{8R^3}{)}^{0.5}+(\frac{GM}{27R^3}{)}^{0.5}=(\frac{GM}{R^3}{)}^{0.5}\times \frac{8^{0.5}+{27}^{0.5}}{{216}^{0.5}}$

Least time taken $=\frac{2\pi }{(\frac{GM}{R^3}{)}^{0.5}\times \frac{8^{0.5}+{27}^{0.5}}{{216}^{0.5}}}=\frac{2\pi R^{1.5}6(6{)}^{0.5}}{\left(GM{)}^{0.5}\right(8^{0.5}+{27}^{0.5})}$
So, $x=6$