How much energy is needed to move a body of $10^{3} k g$ mass from Earth's surface to a height of $2 R_{E}$ ? $(R_{E} = r a d i u s o f E a r t h = 6400 k m)$

42 \times 10^{9} J

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36 \times 10^{9} J

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64 \times 10^{3} J

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84 \times 10^{3} J

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Solution

The correct option is A $42 \times 10^{9} J$
$U_{i} = - \frac{G M m}{R}$
$U_{f} = - \frac{G M m}{3 R}$
$Δ U = U_{f} - U_{i} = \frac{G M m}{R} - \frac{G M m}{3 R} = \frac{2 G M m}{3 R} = \frac{2}{3} \frac{G M}{R^{2}} m R$
$= \frac{2}{3} m R g = \frac{2}{3} \times 10^{3} \times 6400 \times 10^{3} \times 10 = 42.67 \times 10^{9} J$

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How much energy is needed to move a body of 103 kg mass from Earth's surface to a height of 2RE?(RE=radius of Earth=6400 km)

The correct option is A 42×109 JUi=−GMmR Uf=−GMm3R ΔU=Uf−Ui=GMmR−GMm3R=2GMm3R=23GMR2mR =23mRg=23×103×6400×103×10=42.67×109 J

How much energy is needed to move a body of $10^{3} k g$ mass from Earth's surface to a height of $2 R_{E}$ ? $(R_{E} = r a d i u s o f E a r t h = 6400 k m)$