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An asteroid is moving directly towards the centre of the earth. When at a distance of $$10R$$ ($$R$$ is the radius of the earth) from the earth centre, it has a speed of $$12km/s$$. 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 $$11.2km/s$$)? Give your answer to the nearest integer in kilometer/s _____
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Correct option is A. 16
Conserving energy between (1) and (2) of Earth and asteroid system
$$\dfrac{1}{2} mv_1^2 + \left(-\dfrac{GmM_e}{10R}\right) = \dfrac{1}{2}mv_2^2 + \left(-\dfrac{GmM_e}{R}\right)$$
Given $$v_1 = 12Km/s$$
$$v_1 = 12000m/s$$
$$M_e =$$ Mass of earth
From above equation
$$\dfrac{1}{2}m(v_2^2 - v_1^2) = \left( \dfrac{GMe}{R} - \dfrac{Gme}{10R}\right).m$$
$$[v_1^2 - (12000)^2] = \dfrac{9}{5} \dfrac{GMe}{R} = \dfrac{9}{5} \dfrac{6Me}{R^2} . R$$
$$= \dfrac{9}{5} . g. R = \dfrac{9}{5} \times 9.8 \times 6400000$$
$$v_1^2 - 12000^2 = 112896000$$
$$v_2= 16031.57m/s$$
$$v_2 = 16 km/s$$
$$\dfrac{1}{2} mv_1^2 + \left(-\dfrac{GmM_e}{10R}\right) = \dfrac{1}{2}mv_2^2 + \left(-\dfrac{GmM_e}{R}\right)$$
Given $$v_1 = 12Km/s$$
$$v_1 = 12000m/s$$
$$M_e =$$ Mass of earth
From above equation
$$\dfrac{1}{2}m(v_2^2 - v_1^2) = \left( \dfrac{GMe}{R} - \dfrac{Gme}{10R}\right).m$$
$$[v_1^2 - (12000)^2] = \dfrac{9}{5} \dfrac{GMe}{R} = \dfrac{9}{5} \dfrac{6Me}{R^2} . R$$
$$= \dfrac{9}{5} . g. R = \dfrac{9}{5} \times 9.8 \times 6400000$$
$$v_1^2 - 12000^2 = 112896000$$
$$v_2= 16031.57m/s$$
$$v_2 = 16 km/s$$
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