JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Gravitation

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An object of mass 1 kg is taken to a height from the surface of earth which is equal to three times the radius of earth. The gain in potential energy of the object will be
[If,

g = 10 ms^{- 2}

and radius of earth

= 6400 km

]

(A)48 MJ✓ Correct
(B)24 MJ
(C)36 MJ
(D)12 MJ

Answer: (A)

Solution

Initial position: at Earth's surface (distance R from center)

U_{i} = - \frac{GM m}{R}

Final position: at height 3R from surface (distance 4R from center)

U_{f} = - \frac{GM m}{4 R}

Change in potential energy:

Δ U = U_{f} - U_{i} = - \frac{GM m}{4 R} - (- \frac{GM m}{R}) = \frac{3 GM m}{4 R}

Since

g = \frac{GM}{R ^{2}}

, we have

GM = g R^{2}

Δ U = \frac{3 g R ^{2} m}{4 R} = \frac{3 m g R}{4}

Δ U = \frac{3 \times 1 \times 10 \times 6400 \times 1 0 ^{3}}{4} = \frac{3 \times 64 \times 1 0 ^{6}}{4} = 48 \times 1 0^{6}

J = 48 MJ

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Gravitation

Solution

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