Ask Questions, Get Answers

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

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

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

Can you answer this question?

1 Answer

0 votes
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

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