logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Gravitation

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 ?

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

1 Answer

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

Download clay6 mobile appDownload clay6 mobile app
...
X