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Standard XII

Physics

Gravitational Potential Energy

The change in...

Question

The change in the P.E., when a body of mass $m$ is displaced from Earth's surface to a vertical height equal to radius of earth (g $=$ acceleration due to gravity on earth surface) is:

\frac{m g R}{2}

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\frac{2 m g R}{3}

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\frac{3 m g R}{4}

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\frac{m g R}{3}

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Solution

The correct option is A $\frac{m g R}{2}$
$Δ P E = \frac{G M m}{R} {1 - \frac{1}{2}} = \frac{G M m}{2 R}$
$g R = \frac{G M}{R}$
$Δ P E = \frac{m g R}{2}$

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The change in the P.E., when a body of mass m is displaced from Earth's surface to a vertical height equal to radius of earth (g = acceleration due to gravity on earth surface) is:

The correct option is A mgR2ΔPE=GMmR{1−12}=GMm2RgR=GMRΔPE=mgR2

The change in the P.E., when a body of mass $m$ is displaced from Earth's surface to a vertical height equal to radius of earth (g $=$ acceleration due to gravity on earth surface) is:

The correct option is A $\frac{m g R}{2}$
$Δ P E = \frac{G M m}{R} {1 - \frac{1}{2}} = \frac{G M m}{2 R}$
$g R = \frac{G M}{R}$
$Δ P E = \frac{m g R}{2}$