(9am to 6pm)

Ask Questions, Get Answers

Want help in doing your homework? We will solve it for you. Click to know more.
Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Gravitation

A geo stationary satellite orbits around the earth in a circular orbit of radius $36000km$. Then the time period of a spy satellite orbiting a few Kilometers above the earth's surface $(R_{earth}=6400 Km)$ will be approximately

\[(a)\;\frac{1}{2}\;hour \quad (b)\;2\;hour \quad (c)\;1\;hour \quad (d)\;4\;hour \]

1 Answer

Need homework help? Click here.
We know time period of geostationary satellite is $24 hrs$
Also $T^2 \;\alpha\; R^3$
$\large\frac{T_2}{T_1}=\bigg (\large\frac{R_2}{R_1}\bigg)^{\large\frac{3}{2}}$
$\large\frac{T_2}{24}=\bigg (\large\frac{6400}{36000}\bigg)^{\large\frac{3}{2}}$
$T_2=24 \times \bigg (\large\frac{6400}{36000}\bigg)^{\large\frac{3}{2}}$
Hence b is the correct answer.


answered Aug 26, 2013 by meena.p
edited Feb 17, 2014 by meena.p

Related questions