The ratio of the energy required to raise a satellite up to a height R (radius of earth) from the surface of earth to that required to put it into orbit there is

1:1

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8:1

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4:1

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2:3

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Solution

The correct option is D
2:3

$△ E_{1} = - \frac{G M m}{2 R} + \frac{G M m}{R} = \frac{G M m}{2 R}$

$△ E_{2} = - \frac{G M m}{2 (2 R)} + \frac{G M m}{R} = \frac{3 G M m}{4 R}$

$∴$ $\frac{△ E_{1}}{△ E_{2}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{2}{3}$

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The ratio of the energy required to raise a satellite up to a height R (radius of earth) from the surface of earth to that required to put it into orbit there is

The correct option is D 2:3 △E1=−GMm2R+GMmR=GMm2R △E2=−GMm2(2R)+GMmR=3GMm4R ∴ △E1△E2=1234=23

The correct option is D
2:3

$△ E_{1} = - \frac{G M m}{2 R} + \frac{G M m}{R} = \frac{G M m}{2 R}$

$△ E_{2} = - \frac{G M m}{2 (2 R)} + \frac{G M m}{R} = \frac{3 G M m}{4 R}$

$∴$ $\frac{△ E_{1}}{△ E_{2}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{2}{3}$