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An asteroid is moving directly towards the centre of the earth. When at a distance of $10 \mathrm{R}$ ($R$ 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 $km/s$
$20$
$24$
$14$
$16$
Solution
$\mathrm{U}_{1}+\mathrm{K}_{1}=\mathrm{U}_{2}+\mathrm{K}_{2}$
$-\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{10 \mathrm{R}}+\frac{1}{2} \mathrm{mv}_{0}^{2}=-\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}}+\frac{1}{2} \mathrm{mv}^{2}$
$+\frac{9}{10} \times \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}}+\frac{1}{2} \mathrm{mv}_{0}^{2}=\frac{1}{2} \mathrm{mv}^{2}$
$\frac{9}{10} \times \frac{1}{2} \mathrm{M} \times \mathrm{v}_{\mathrm{e}}^{2}+\frac{1}{2} \mathrm{mv}_{0}^{2}=\frac{1}{2} \mathrm{mv}^{2}$
$\mathrm{v}^{2}=\frac{9}{10} \mathrm{v}_{e}^{2}+\mathrm{v}_{0}^{2}$
$=\frac{9}{10} \times(11.2)^{2}+(12)^{2}$
$\mathrm{v}^{2}=112.896+144$
$\mathrm{v}=16.027$
$\mathrm{v}=16 \mathrm{km} / \mathrm{s}$
