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A skylab of mass m kg is first launched from the surface of the earth in a circular orbit of radius 2R (from the centre of the earth) and then it is shifted from this circular orbit to another circular orbit of radius 3R. The minimum energy required to place the lab in the first orbit and to shift the lab from first orbit to the second orbit are

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$2 mgR, mgR$

$mgR, mgR$

$\frac{3}{4} mgR, \frac{mgR}{12}$

$\frac{3}{4} mgR, \frac{mgR}{6}$

answer is B.

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Detailed Solution

Energy of the skylab in the first orbit is

$- \frac{GMm}{2 (2 R)} = - \frac{GMm}{4 R}$

Total energy required to place the skylab into the orbit of radius 2R from the surface of earth is

$\begin{array}{l} - \frac{GMm}{4 R} - (- \frac{GMm}{R}) = \frac{3 GMm}{4 R} \\ = \frac{3 {gR}^{2} m}{4 R} = \frac{3}{4} mgR \end{array}$

Energy of the skylab in the second orbit $= - (GMm) / 6 R$
Energy needed to shift the skylab from the first orbit to the
second orbit is

$- \frac{GMm}{4 R} - \frac{GMm}{6 R} = \frac{GMm}{R} \times \frac{2}{24} = \frac{mgR}{12}$

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