Question

A satellite is orbiting the Earth of mass M in equatorial plane in a circular orbit having radius 2 R and same sense of rotation as that of the Earth. find duration of time for which a man standing on the equator will be able to see the satellite continuously. Assume that the man can see the satellite when it is above horizon. Take Earth's angular velocity $=\omega$.

Answer 1

Let the angular velocity of satellite be $\omega$.

Hence, from Newton's Second law,

$m{\omega }^2\left(2R\right)=\frac{GMm}{{\left(2R\right)}^2}$

$\omega =\sqrt{\frac{GM}{8R^3}}$

Thus, relative angular velocity will be

${\omega }_r=\sqrt{\frac{GM}{8R^3}}-{\omega }_e$

Now, since the length of radius of orbit is 2R, the angle subtended at horizon is $\theta =2{cos}^{-1}\left(0.5\right)=\frac{2\pi }{3}$

Hence, required time is

$t=\frac{2\pi }{3{\omega }_r}=\frac{2\pi }{3(\sqrt{\frac{GM}{8R^3}}-{\omega }_e)}$

Answer is $\frac{2\pi }{3{\omega }_r}=\frac{2\pi }{3(\sqrt{\frac{GM}{8R^3}}-{\omega }_e)}$