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 centre, it has a speed of $12 k m / s$ . Neglecting the effect of the earths atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is $11.2 k m / s$ )? Give your answer to the nearest integer in kilometer/s _____

Question

Verified by Toppr

Answer 1

Conserving energy between (1) and (2) of Earth and asteroid system

$\frac{1}{2} m v_{1}^{2} + (- \frac{G m M _{e}}{1 0 R}) = \frac{1}{2} m v_{2}^{2} + (- \frac{G m M _{e}}{R})$

Given $v_{1} = 12 K m / s$

$v_{1} = 12000 m / s$

$M_{e} =$ Mass of earth

From above equation

$\frac{1}{2} m (v_{2}^{2} - v_{1}^{2}) = (\frac{G M e}{R} - \frac{G m e}{1 0 R}) . m$

$[v_{1}^{2} - (12000)^{2}] = \frac{9}{5} \frac{G M e}{R} = \frac{9}{5} \frac{6 M e}{R ^{2}} . R$

$= \frac{9}{5} . g . R = \frac{9}{5} \times 9.8 \times 6400000$

$v_{1}^{2} - 1200 0^{2} = 112896000$

$v_{2} = 16031.57 m / s$

$v_{2} = 16 k m / s$

Answer 2

Conserving energy between (1) and (2) of Earth and asteroid system

\frac{1}{2} m v_{1}^{2} + (- \frac{G m M _{e}}{1 0 R}) = \frac{1}{2} m v_{2}^{2} + (- \frac{G m M _{e}}{R})

Given

v_{1} = 12 K m / s

v_{1} = 12000 m / s

M_{e} =

Mass of earth

From above equation

\frac{1}{2} m (v_{2}^{2} - v_{1}^{2}) = (\frac{G M e}{R} - \frac{G m e}{1 0 R}) . m

[v_{1}^{2} - (12000)^{2}] = \frac{9}{5} \frac{G M e}{R} = \frac{9}{5} \frac{6 M e}{R ^{2}} . R

= \frac{9}{5} . g . R = \frac{9}{5} \times 9.8 \times 6400000

v_{1}^{2} - 1200 0^{2} = 112896000

v_{2} = 16031.57 m / s

v_{2} = 16 k m / s