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MHT CET 2020 (16 Oct, Shift 1) · Escape Velocity

Escape Velocity question from MHT CET 2020 (16 Oct, Shift 1)

Q58MHT CET 2020 · 16 OCT · SHIFT 1Medium

Escape Velocity

A body is thrown from the surface of the earth velocity v/sv/\text{s}. The maximum height above the earth's surface upto which it will reach is (R=R = radius of earth, g=g = acceleration due to gravity)

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Answer: (d)

  1. Energy conservation: 12v2=GM(1R−1R+h)=gR2hR(R+h)\frac{1}{2}v^2 = GM\left(\frac{1}{R} - \frac{1}{R+h}\right) = gR^2\frac{h}{R(R+h)}.
  2. v2=2gRhR+h⇒v2R+v2h=2gRhv^2 = \frac{2gRh}{R+h} \Rightarrow v^2R + v^2h = 2gRh.
  3. h=v2R2gR−v2h = \frac{v^2R}{2gR - v^2}.
  4. So the answer is (d) v2R2gR−v2\frac{v^{2}R}{2gR-v^{2}}.
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