**Question:**

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 $12 \mathrm{~km} / \mathrm{s}$. 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 $11.2 \mathrm{~km} / \mathrm{s}$ )? Give your answer to the nearest integer in kilometer/s

**Solution:**

$(16.00)$

Using law of conservation of energy

Total energy at height $10 \mathrm{R}=$ total energy at earth

$-\frac{G M_{E} m}{10 R}+\frac{1}{2} m V_{0}^{2}=-\frac{G M_{E} m}{R}+\frac{1}{2} m V^{2}$

$\left[\because\right.$ Gravitational potential energy $\left.=-\frac{G M m}{r}\right]$

$\Rightarrow \frac{G M_{E}}{R}\left(1-\frac{1}{10}\right)+\frac{V_{0}^{2}}{2}=\frac{V^{2}}{2}$

$\Rightarrow V^{2}=V_{0}^{2}+\frac{9}{5} g R$

$\Rightarrow V=\sqrt{V_{0}^{2}+\frac{9}{5} g R} \approx 16 \mathrm{~km} / \mathrm{s}$

$\left[\because \mathrm{V}_{0}=12 \mathrm{~km} / \mathrm{s}\right.$ given $]$

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