JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Work, Power & Energy

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 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 form the earth is

11.2 km / s

)? Given your answer to the nearest integer in kilometer/s ____.

Answer: 16

Solution

Mechanical energy is conserved as the asteroid falls in. Between

10 R

and the surface,

\frac{1}{2} u_{0}^{2} - \frac{GM}{10 R} = \frac{1}{2} v^{2} - \frac{GM}{R}

, so

v^{2} = u_{0}^{2} + \frac{2 GM}{R} (1 - \frac{1}{10}) = u_{0}^{2} + \frac{9}{5} \cdot \frac{GM}{R}

. Writing

\frac{GM}{R} = \frac{v _{e}^{2}}{2}

with escape speed

v_{e} = 11.2

km/s,

v^{2} = 1 2^{2} + \frac{9}{5} \cdot \frac{( 11.2 ) ^{2}}{2} = 144 + 112.9 \approx 256.9

, so

v \approx 16

km/s.

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Integer answer

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JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Work, Power & Energy

Solution

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