# Motion of Planets and Satellites in Circular Orbits

## Problems from IIT JEE

Problem (IIT JEE 2011): A satellite is moving with a constant speed $V$ in a circular orbit about earth. An object of mass $m$ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its injection, the kinetic energy of the object is,

1. $\frac{1}{2}mV^2$
2. $mV^2$
3. $\frac{3}{2}mV^2$
4. $2mV^2$

Solution: A particle escapes from the gravitational pull if its total energy ($T$) i.e., sum of kinetic energy ($K$) and potential energy ($U$), is greater than or equal to zero. The condition for just escape is $T=K+U=0$ i.e., \begin{align} \label{vaa:eqn:1} K=-U. \end{align} In a circular orbit of radius $r$, gravitational attraction gives centripetal acceleration, ${mV^2}/{r}={GMm}/{r^2}$, which gives, \begin{align} \label{vaa:eqn:2} r=GM/V^2. \end{align} From above equations, the particle kinetic energy at the time of injection is given by, \begin{align} K=-U=-\left(-\frac{GMm}{r}\right)=\frac{GMm}{GM/V^2}=mV^2.\nonumber \end{align}

Problem (IIT JEE 2002): A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a spy satellite orbiting a few hundred kilometers above the earth surface ($R_e={6400}\;\mathrm{km}$) will approximately be,

1. $\frac{1}{2}$ h
2. 1 h
3. 2 h
4. 4 h

Solution: Time period of a geostationary satellite is $T_{36000}={24}\;\mathrm{h}$. Keplar's third law, $T^2\propto a^3$, gives time period of a satellite in an orbit close to the earth surface as, \begin{align} T_{6400}=24\times\left(6400/36000\right)^{3/2}\approx {1.8}\;\mathrm{h}. \end{align} Since time period increases with $a$, time period of a satellite moving few hundred kilometers above the earth surface is $\approx {2}\;\mathrm{h}$.