Motion of Planets and Satellites in Circular Orbits
By The orbital velocity of a satellite moving in a circular orbit of radius $r$ is given by \begin{align} v_o=\sqrt{\frac{GM}{r}}=\sqrt{\frac{GM}{R+h}} \end{align}
Kepler's three laws are:
- The planets moves in elliptical orbit with sun at one of the focus.
- Areal velocity is constant. ($\because {\mathrm{d}\vec{L}}/{\mathrm{d}t}=0$).
- The square of the time period is proportional to the cube of the semi-major axis i.e., $T^2\propto a^3$. In circular orbit \begin{align} T^2=\frac{4\pi^2}{GM} a^3. \end{align}
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} 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} 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 $\approx {2}\;\mathrm{h}$.
