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JEE Main 2020 (8 Jan, Shift 2) · Escape Velocity

Escape Velocity question from JEE Main 2020 (8 Jan, Shift 2)

Q59JEE Main 2020 · 8 JAN · SHIFT 2Difficult

Escape Velocity

An asteroid is moving directly towards the centre of the earth. When at a distance of 10R (R is the radius of the earth) from the earths centre, it has a speed of 12 km/s. Neglecting the effect of earths atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is 11.2 km/s)? Give your answer to the nearest integer in kilometer/s _____.

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

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Answer: 16

  1. Energy conservation: v2=u2+2GM(1R−110R)=u2+0.9 ve2v^2 = u^2 + 2GM\left(\dfrac{1}{R} - \dfrac{1}{10R}\right) = u^2 + 0.9\,v_e^2, since ve2=2GMRv_e^2 = \dfrac{2GM}{R}.
  2. v2=122+0.9×11.22=144+112.9=256.9v^2 = 12^2 + 0.9 \times 11.2^2 = 144 + 112.9 = 256.9
  3. v≈16.03 km/sv \approx 16.03\text{ km/s}
  4. So the speed is 16 km/s16\text{ km/s}.
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