Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
0 votes

The mean radius of earth is R, its angular velocity on its own axis is 'w' and acceleration due to gravity is g. What will be the radius of orbit of a geostationary satellite ?

\[(a)\;\bigg(\frac{R^2}{w^2}\bigg)^{1/3} \quad (b)\;\bigg(\frac{Rg}{w^2}\bigg)^{1/3} \quad (c)\;\bigg(\frac{R^2w^2}{g}\bigg)^{1/3} \quad (d)\;\bigg(\frac{R^2g}{w}\bigg)^{1/3} \]

Can you answer this question?

1 Answer

0 votes
Satellite orbitary with a orbital velocity $v_0$, radius of orbit r
$v_0=\sqrt {\large\frac{GM}{r}}$
$\qquad= \sqrt {\large\frac{gR^2}{r}}$
Time period $=\large\frac{2 \pi r}{v_0}$
$\qquad= \large\frac{2 \pi r}{\bigg(\Large\frac{gR^2}{r}\bigg)^{1/2}}$
$\qquad= \large\frac{2 \pi r^{3/2}}{\sqrt {gR^2}}$
$T=\large\frac{2 \pi}{w}$
$\therefore \large\frac{2 \pi}{w}=\frac{2 \pi r^{3/2}}{\sqrt {gR^2}}$
$r^{3/2} =\large\frac{\sqrt {gR^2}}{w}$
$r^3= \large\frac{gR^2}{w^2}$
$r= \bigg( \large\frac{gR^2}{w}\bigg)^{1/3}$
Hence d is the correct answer.
answered Aug 24, 2013 by meena.p
edited Jun 30, 2014 by lmohan717

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App