# 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,

- $\frac{1}{2}mV^2$
- $mV^2$
- $\frac{3}{2}mV^2$
- $2mV^2$

**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,

- $\frac{1}{2}$ h
- 1 h
- 2 h
- 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 $\approx {2}\;\mathrm{h}$.