A satellite revolving around earth in a circular orbit of radius 4R. Due to ejection of a rocket the speed of satellite is reduced to half as a result of which satellite starts falling toward earth. Find the angle at which it will hit the earth surface.

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$θ = 33^{o}$

$θ = 32^{o}$

$θ = 37^{o}$

$θ = 40^{o}$

answer is A.

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Detailed Solution

with respect to earth, we conserve energy and angular momentum

$m (\frac{v}{2}) 4 R = m (v_{1} \cos θ) R$

$2 \sqrt{\frac{GM}{4 R}} = v_{1} \cos θ$ ……..(1) [by conservation of angular momentum ]

$\frac{1}{2} m (v / 2)^{2} - \frac{{GM}_{e} m}{4 R} = \frac{1}{2} {mv}_{1}^{2} - \frac{{GM}_{1} m}{R}$ ………(2)

$v_{1} = \frac{5}{4} \sqrt{\frac{G M}{R}}$

from (i) $\sqrt{\frac{GM}{R}} = \frac{5}{4} \sqrt{\frac{G M}{R}} \cos θ$

$\cos θ = 4 / 5 \Rightarrow θ = 37^{\circ}$

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