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JEE Main 2021 (25 Feb, Shift 2) · Escape Velocity

Escape Velocity question from JEE Main 2021 (25 Feb, Shift 2)

Q49JEE Main 2021 · 25 FEB · SHIFT 2Medium

Escape Velocity

The initial velocity vi required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity ve such that viv_i == xy×\sqrt{\frac{x}{y}} \times vev_e. The value of x will be ____________.

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

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Answer: 10

  1. To reach h=10Rh = 10R (distance 11R11R): 12vi2=GM(1R−111R)=GMR⋅1011\frac{1}{2}v_i^2 = GM\left(\frac{1}{R} - \frac{1}{11R}\right) = \frac{GM}{R}\cdot\frac{10}{11}.
  2. vi2=2GMR⋅1011=1011ve2v_i^2 = \frac{2GM}{R}\cdot\frac{10}{11} = \frac{10}{11}v_e^2.
  3. vi=1011 vev_i = \sqrt{\frac{10}{11}}\,v_e, so x=10x = 10.
  4. Answer: 1010.
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