JEE PYQ: Gravitation Question 32
Question 32 - 2020 (08 Jan Shift 2)
An asteroid is moving directly towards the centre of the earth. When at a distance of 10$R$ ($R$ is the radius of the earth) from the earth’s centre, it has a speed of 12 km/s. Neglecting the effect of earth’s 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 km/s.
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Answer: 16
Solution
By energy conservation: $\frac{1}{2}mv_0^2 - \frac{GMm}{10R} = \frac{1}{2}mv^2 - \frac{GMm}{R}$. $v^2 = v_0^2 + \frac{2GM}{R}\left(1 - \frac{1}{10}\right) = v_0^2 + \frac{9}{10}v_e^2 = 144 + 0.9 \times 125.44 = 144 + 112.9 = 256.9$. $v \approx 16$ km/s.
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