Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A satellite is seen after each $8$ hours over equator at a place on the earth when its sense of rotation is opposite to the earth. The time interval after which it can be seen at the same place when the sense of rotation of earth and satellite is same will be :

Gravitation

Solution:

Given $ 8 = \frac{2\pi}{\omega_1 + \omega_2}$
$ = \frac{2\pi}{\frac{2\pi}{T_1} + \frac{2\pi}{T_2}}, T_1 = 24$ hours for earth.
$\Rightarrow T_2 = 12$ hours ($T_2$ being the time period of satellite, it will remain same as the distance from the centre of the earth remains constant).
$\Rightarrow T = \frac{2\pi}{\omega_2 - \omega_1}$
$= \frac{2 \pi}{\frac{2\pi}{T_2} - \frac{2\pi}{T_1}} = 24$ hours