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MHT CET 2026 (16 Apr, Shift 2) · Gravitational Potential & Energy

Gravitational Potential & Energy question from MHT CET 2026 (16 Apr, Shift 2)

Q1MHT CET 2026 · 16 APR · SHIFT 2Easy

Gravitational Potential & Energy

A body of mass m is taken from earth surface to a height h equal to twice the radius of earth, the increase in potential energy will be ( $g$ = acceleration due to gravity on earth's surface) (R - radius of earth)

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Answer: (a)

$\Delta U = \frac{mgh}{1 + \frac{h}{R}}$
With $h = 2R$ : $\Delta U = \frac{mg(2R)}{1+2} = \frac{2}{3}mgR$
So the answer is (a) $\frac{2}{3}mgR$ .

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