Question

The additional kinetic energy to be provided to a satellite of mass $m$ revolving around a planet of mass $M$ to transfer it from a circular orbit of radius $R_1$ to another of radius $R_2$ $(R_2 > R_1)$ is

• A. $GmM\left(\frac{1}{R^{2}_{1}}-\frac{1}{R^{2}_{2}}\right)$
• B. $GmM\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
• C. $2GmM\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
• D. $\frac{1}{2}GmM\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

Answer 1

The Correct Option is D

To find the additional kinetic energy required to transfer a satellite from a circular orbit of radius $R_1$ to another orbit of radius $R_2$, we proceed as follows.

1. Start by recalling the formula for the total energy of a satellite in a circular orbit. The total energy \(E\) is given by: $$E = -\frac{G m M}{2R}$$ where $G$ is the gravitational constant, $m$ is the mass of the satellite, $M$ is the mass of the planet, and $R$ is the radius of the orbit.
2. The kinetic energy $K$ of the satellite is one half of the magnitude of the total energy: $$K = -\frac{E}{2} = \frac{G m M}{2R}$$
3. The problem asks for the additional kinetic energy required for moving the satellite from radius $R_1$ to $R_2$. We calculate the difference in kinetic energy: $$\Delta K = K_2 - K_1 = \frac{G m M}{2R_2} - \frac{G m M}{2R_1}$$
4. Simplifying the expression for $\Delta K$: $$\Delta K = \frac{G m M}{2} \left(\frac{1}{R_2} - \frac{1}{R_1}\right)$$ or equivalently: $$\Delta K = \frac{1}{2} G m M \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

This expression matches the option:

$\frac{1}{2} G m M \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

Hence, the correct answer is:

$\frac{1}{2} G m M \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

This is the amount of additional kinetic energy that must be provided to the satellite to move from an orbit of radius $R_1$ to $R_2$.