Question

What is the minimum energy required to launch a satellite of mass $m$ from the surface of a planet of mass $M$ and radius $R$ in a circular orbit at an altitude of $2R$.
A) $\dfrac{2GmM}{3R}$
B) $\dfrac{GmM}{2R}$
C) $\dfrac{GmM}{3R}$
D) $\dfrac{5GmM}{6R}$

Answer 1

Hint: To get the minimum energy required, we have to find the difference in the gravitational potential energy in the initial state (on the surface of the planet) and the final state (in orbit). This difference in energy will be the minimum energy that has to be provided to the satellite, so that it can reach the altitude of the desired orbit.

Formula used:
The gravitational potential energy $PE$ of a satellite of mass $m$ orbiting around a planet of mass $M$ and radius $R$ at an altitude $h$ is given by
$PE=-\dfrac{GMm}{\left( R+h \right)}$
Where $G=6.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}$. The negative sign implies that the gravitational force is attractive.

Complete step by step answer:
When a satellite is launched from the surface of the planet into an orbit, its total mechanical energy remains conserved, since the force of gravity is a conservative force. However, the kinetic energy decreases as the satellite reaches the orbit and its potential energy increases. The decrease in the kinetic energy is equal to the increase in the gravitational potential energy of the satellite.
To get the minimum energy that must be imparted to the satellite so that it reaches the orbit, we have to find the increase in the potential energy of the satellite. This will be equal to the required minimum energy that has to be imparted.
The gravitational potential energy $PE$ of a satellite of mass $m$ orbiting around a planet of mass $M$ and radius $R$ at an altitude $h$ is given by
$PE=-\dfrac{GMm}{\left( R+h \right)}$ --(1)
Where $G=6.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}$. The negative sign implies that the gravitational force is attractive.
Now, let us analyze the question.
Let the required minimum energy that has to be given to the satellite be $E$.
The mass of the satellite is $m$.
The mass of the planet is $M$.
The radius of the planet is $R$.
The altitude of the orbit is $2R$.
Let the initial gravitational potential energy of the satellite (on the surface of the planet) be $P{{E}_{i}}$.
Let the final gravitational potential energy of the satellite (in orbit) be $P{{E}_{f}}$.
Using (1), we get,
$P{{E}_{i}}=-\dfrac{GMm}{\left( R+0 \right)}=-\dfrac{GMm}{R}$ $\left( \because \text{at the surface of the planet }h=0 \right)$ --(2)
$P{{E}_{f}}=-\dfrac{GMm}{\left( R+2R \right)}=-\dfrac{GMm}{3R}$ $\left( \because \text{Given the altitude of the orbit is }h=2R \right)$ --(3)
Now as explained above, the energy that has to be given to the satellite will be
$E=P{{E}_{f}}-P{{E}_{i}}$
Hence, using (2) and (3), we get,
$E=-\dfrac{GMm}{3R}-\left( -\dfrac{GMm}{R} \right)=-\dfrac{GMm}{3R}+\dfrac{GMm}{R}=\dfrac{-GMm+3GMm}{3R}=\dfrac{2GMm}{3R}$
Hence, the required minimum energy is $\dfrac{2GMm}{3R}$.

Hence, the correct answer is option A, $\dfrac{2GMm}{3R}$.

Note: Students might be thinking that if the kinetic energy of the satellite becomes zero upon reaching the orbit in this question, what will be the use of sending out the satellite. However, this question involves only the process of sending the satellite up to the required orbit altitude. After reaching this altitude, fuel is burnt to provide thrust and the required tangential velocity is imparted to the satellite according to the specifications and requirements of the orbit. However, if this velocity is not imparted after reaching the orbit altitude, the satellite will fall back to the planet. Hence, it is critical that the thrust is given such that the satellite attains the required tangential velocity exactly at the moment that it reaches the altitude of the orbit.