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MHT CET 2024 (11 May, Shift 2) · Escape Velocity

Escape Velocity question from MHT CET 2024 (11 May, Shift 2)

Q22MHT CET 2024 · 11 MAY · SHIFT 2Medium

Escape Velocity

A body is projected in vertically upward direction from the surface of the earth of radius ' RR ' into space with velocity ' nVenV_e ' (n<1)(n < 1). The maximum height from the surface of earth to which a body can reach is

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Answer: (a)

  1. Energy conservation: 12mn22GMR=GMm(1R−1R+h)\frac{1}{2}m n^2\frac{2GM}{R} = GMm\left(\frac{1}{R} - \frac{1}{R+h}\right).
  2. 1R+h=1−n2R⇒R+h=R1−n2\frac{1}{R+h} = \frac{1-n^2}{R} \Rightarrow R+h = \frac{R}{1-n^2}.
  3. h=n2R1−n2h = \frac{n^2R}{1-n^2}.
  4. So the answer is (a) n2R(1−n2)\frac{n^{2}R}{(1-n^{2})}.
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