Consider two satellites S1 and S2 with periods of revolution 1 hr. and 8 hr. respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S1 to the angular velocity of satellite S2 is :
Solution
Given, period of revolution of first satellite,<br/><br/>T<sub>1</sub> = 1h<br/><br/>Period of revolution of second satellite,<br/><br/>T<sub>2</sub> = 8h<br/><br/>$\therefore$ ${{{T_1}} \over {{T_2}}} = {1 \over 8}$<br/><br/>We know that, $\omega = {{2\pi } \over T}$<br/><br/>$\Rightarrow \omega \propto {1 \over T}$<br/><br/>$\because$ $${{{\omega _1}} \over {{\omega _2}}} = {{{T_2}} \over {{T_1}}} \Rightarrow {{{\omega _1}} \over {{\omega _2}}} = {8 \over 1}$$<br/><br/>or ${\omega _1}:{\omega _2} = 8:1$
About this question
Subject: Physics · Chapter: Gravitation · Topic: Satellites and Orbital Motion
This question is part of PrepWiser's free JEE Main question bank. 129 more solved questions on Gravitation are available — start with the harder ones if your accuracy is >70%.