The ratio of the energy required to raise a satellite upto 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 $2 : 3$
$: Δ E_{1} = \frac{- G M m}{2 R} + \frac{G M m}{R} = \frac{G M m}{2 R}$
$Δ E_{2} = \frac{- G M m}{2 (2 R)} + \frac{G M m}{R} = \frac{3}{4} \frac{G M m}{R}$
$⟹ \frac{Δ E_{1}}{Δ E_{2}} = \frac{2}{3}$

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The ratio of the energy required to raise a satellite upto 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:ΔE1=−GMm2R+GMmR=GMm2RΔE2=−GMm2(2R)+GMmR=34GMmR⟹ΔE1ΔE2=23

The correct option is D $2 : 3$
$: Δ E_{1} = \frac{- G M m}{2 R} + \frac{G M m}{R} = \frac{G M m}{2 R}$
$Δ E_{2} = \frac{- G M m}{2 (2 R)} + \frac{G M m}{R} = \frac{3}{4} \frac{G M m}{R}$
$⟹ \frac{Δ E_{1}}{Δ E_{2}} = \frac{2}{3}$