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JEE Main 2021 (17 Mar, Shift 1) · Escape Velocity

Escape Velocity question from JEE Main 2021 (17 Mar, Shift 1)

Q47JEE Main 2021 · 17 MAR · SHIFT 1Medium

Escape Velocity

The radius in kilometer to which the present radius of earth (R = 6400 km) to be compressed so that the escape velocity is increased 10 times is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 64

  1. For the same mass, ve=2GMR∝1Rv_e = \sqrt{\frac{2GM}{R}} \propto \frac{1}{\sqrt{R}}.
  2. ve′=10ve⇒R′=R100v_e' = 10v_e \Rightarrow R' = \frac{R}{100}.
  3. R′=6400100=64R' = \frac{6400}{100} = 64 km.
  4. Answer: 6464 km.
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