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Work Energy Theorem

The additiona...

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 $2 R$ to another orbit of radius $3 R$ is:

\frac{G m M}{12 R}

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\frac{G m M}{6 R}

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\frac{G m M}{3 R}

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\frac{G m M}{R}

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Solution

The correct option is A $\frac{G m M}{12 R}$
When the satellite is revolving around the planet in an orbit of radius $r$ , its total energy is given by:

$T E = \frac{- G M_{p} m_{p}}{2 r}$

Initially, when orbital radius is $2 R$ ,

$T E_{i} = \frac{- G m M}{2 (2 R)} = \frac{- G m M}{4 R}$

Finally, when the orbital radius is $3 R$ ,

$T E_{f} = \frac{- G m M}{2 (3 R)} = \frac{- G m M}{6 R}$

Kinetic energy imparted to the satelite is given by,

$K E = Δ T E = T E_{f} - T E_{i}$

$K E = \frac{- G m M}{6 R} - (\frac{- G m M}{4 R})$

$K E = \frac{G m M}{R} (\frac{1}{4} - \frac{1}{6})$

$K E = \frac{G m M}{12 R}$

Hence, option $(A)$ is the correct answer.

Why this Question?
Note: The additional kinetic energy required to change the orbital radius from $R_{1}$ and $R_{2}$ is given by,

$K E = \frac{G m M}{2} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

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