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MCQMediumJEE 2023Newton's Law of Gravitation

JEE Physics 2023 Question with Solution

An object is allowed to fall from a height RR above the earth, where RR is the radius of the earth. Its velocity when it strikes the earth’s surface, ignoring air resistance, will be:

  • A

    2gR2\sqrt{gR}

  • B

    gR\sqrt{gR}

  • C

    gR2\frac{\sqrt{gR}}{2}

  • D

    2gR\sqrt{2gR}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: An object falls from a height RR above the earth, so its initial distance from the earth's centre is 2R2R and final distance is RR. Air resistance is ignored.

Find: The velocity with which it strikes the earth's surface.

Use conservation of mechanical energy. The loss in gravitational potential energy is equal to the gain in kinetic energy.

(−GMm2R)−(−GMmR)=12mv2\left(-\frac{GMm}{2R}\right)-\left(-\frac{GMm}{R}\right)=\frac{1}{2}mv^2

This gives

v2=GMR=gRv^2=\frac{GM}{R}=gR

Hence,

v=gRv=\sqrt{gR}

Therefore, the correct value of the speed is gR\sqrt{gR}. The solution states "Option A" but this conflicts with the listed options; among the given options, gR\sqrt{gR} corresponds to option B.

Energy Interpretation

The gravitational potential energy at distance rr from the earth's centre is

U=−GMmrU=-\frac{GMm}{r}

At the starting point, r=2Rr=2R, and at the earth's surface, r=Rr=R.

So,

Ui=−GMm2R,Uf=−GMmRU_i=-\frac{GMm}{2R},\qquad U_f=-\frac{GMm}{R}

The decrease in potential energy appears as kinetic energy:

Ui−Uf=12mv2U_i-U_f=\frac{1}{2}mv^2
−GMm2R−(−GMmR)=12mv2-\frac{GMm}{2R}-\left(-\frac{GMm}{R}\right)=\frac{1}{2}mv^2
GMm2R=12mv2\frac{GMm}{2R}=\frac{1}{2}mv^2
v2=GMRv^2=\frac{GM}{R}

Using g=GMR2g=\frac{GM}{R^2}, we get

GMR=gR\frac{GM}{R}=gR

Thus,

v=gRv=\sqrt{gR}

So the correct option from the listed choices is B.

Common mistakes

  • Using the constant-acceleration formula v2=u2+2ghv^2=u^2+2gh with h=Rh=R directly is incorrect because gravitational acceleration is not constant over such a large height. Use gravitational potential energy instead.

  • Taking the initial potential energy as 00 and final as −mgR-mgR is wrong for planetary motion over large distances. The correct potential energy is U=−GMmrU=-\frac{GMm}{r} measured from infinity.

  • Confusing the starting position as distance RR from the centre instead of height RR above the surface leads to the wrong radius. The object starts at distance 2R2R from the earth's centre.

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