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JEE Main 2025 (4 Apr, Shift 2) · Escape Velocity

Escape Velocity question from JEE Main 2025 (4 Apr, Shift 2)

Q14JEE Main 2025 · 4 APR · SHIFT 2Medium

Escape Velocity

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 M=M = mass of earth, G=G = Universal gravitational constant)

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Answer: (c)

  1. Height 3R3R means distance r=4Rr = 4R from the centre.
  2. Escape speed there: v=2GM4R=GM2Rv = \sqrt{\frac{2GM}{4R}} = \sqrt{\frac{GM}{2R}}.
  3. So the answer is (c) GM2R\sqrt{\frac{GM}{2R}}.
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