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MCQEasyJEE 2025Escape Velocity

JEE Physics 2025 Question with Solution

An object is kept at rest at a distance of 3R3R above the earth’s surface where RR is earth’s radius. The minimum speed with which it must be projected so that it does not return to earth is: (Assume MM = mass of earth, GG = Universal gravitational constant)

  • A

    GM2R\sqrt{\frac{GM}{2R}}

  • B

    GMR\sqrt{\frac{GM}{R}}

  • C

    3GMR\sqrt{\frac{3GM}{R}}

  • D

    2GMR\sqrt{\frac{2GM}{R}}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The object is at rest at a distance 3R3R above the earth’s surface, so its distance from the earth’s center is 4R4R. Find: The minimum projection speed so that it does not return to earth.

For escape, the total mechanical energy at the starting point must be zero.

12mv2−GMm4R=0\frac{1}{2}mv^2 - \frac{GMm}{4R} = 0

Rearranging,

12mv2=GMm4R\frac{1}{2}mv^2 = \frac{GMm}{4R}

Cancelling mm,

12v2=GM4R\frac{1}{2}v^2 = \frac{GM}{4R}

So,

v2=2GM4R=GM2Rv^2 = \frac{2GM}{4R} = \frac{GM}{2R}

Hence,

v=GM2Rv = \sqrt{\frac{GM}{2R}}

Therefore, the minimum speed required is GM2R\sqrt{\frac{GM}{2R}}. The correct option is A.

Escape Velocity at Distance $$4R$$

Given: The object is located 3R3R above the earth’s surface, so the distance from the center is 4R4R. Find: The minimum speed to escape.

Use the escape speed formula at distance rr from the earth’s center:

vescape=2GMrv_{\text{escape}} = \sqrt{\frac{2GM}{r}}

Here r=4Rr = 4R, so

vescape=2GM4R=GM2Rv_{\text{escape}} = \sqrt{\frac{2GM}{4R}} = \sqrt{\frac{GM}{2R}}

Therefore, the required minimum speed is GM2R\sqrt{\frac{GM}{2R}}, so the correct option is A.

Common mistakes

  • Using RR instead of 4R4R in the escape velocity formula is incorrect because the object is 3R3R above the surface, not on the surface. Always convert the height above the surface into distance from the earth’s center first.

  • Applying the surface escape speed 2GMR\sqrt{\frac{2GM}{R}} directly is wrong because that formula is only for launch from the earth’s surface. Here the starting point is farther away, so the required speed is smaller.

  • Confusing gravitational potential energy with −GMm3R-\frac{GMm}{3R} is incorrect because gravitational potential depends on distance from the center of the earth, not height above the surface. The correct distance is 4R4R.

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