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

Gravitation PYQ — JEE Main 2025 · 4 Apr · Shift 2 · Q59 with Solution

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Correct answer: C

The object is at a height of $3R$ above the earth's surface.

So its distance from the center of the earth is $r = R + 3R = 4R.$

For the object to just escape and not return, it must be projected with the escape speed from distance $r = 4R$ .

Escape speed from a distance $r$ from the center of the earth is $v_e = \sqrt{\frac{2GM}{r}}.$

Substituting $r = 4R$ , $v_e = \sqrt{\frac{2GM}{4R}} = \sqrt{\frac{GM}{2R}}.$

Thus the required minimum speed is $\boxed{\sqrt{\frac{GM}{2R}}}.$

This matches Option C.

Initial total energy at $r=4R$ : $E = \frac{1}{2}mv^2 - \frac{GMm}{4R}.$

For just escaping, final energy at infinity is zero, so minimum condition is $\frac{1}{2}mv^2 - \frac{GMm}{4R} = 0.$

Hence, $\frac{1}{2}mv^2 = \frac{GMm}{4R}$ $v^2 = \frac{GM}{2R}$ $v = \sqrt{\frac{GM}{2R}}.$

So the answer is confirmed.