A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth’s radius $R_{e}$ . By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that it become $\sqrt{\frac{3}{2}}$ times larger. Due to this the farthest distance from the center of the earth that the satellite reaches is R. Value of R is :

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$4 R_{e}$

$2 R_{e}$

$25 R_{e}$

$3 R_{e}$

answer is D.

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Detailed Solution

$\begin{array}{l} E_{i} = \frac{- GMm}{2 R_{e}} \\ A f t e r s u p p l y i n g k i n e t i c e n e r g y, \\ E_{f} = \frac{- GMm}{2 R_{e}} + \frac{1}{2} m {(\sqrt{\frac{3}{2}} \sqrt{\frac{GM}{R_{e}}})}^{2} \\ E_{f} = \frac{- GMm}{2 R_{e}} + \frac{1}{2} m \frac{3}{2} \frac{GM}{R_{e}} \\ = \frac{- GMm}{4 R_{e}} \\ L e t s a y a i s s e m i m a j o r a x i s o f n e w o r b i t . \\ E_{f} = \frac{- GMm}{2 a} \\ \Rightarrow \frac{- GMm}{4 R_{e}} = \frac{- GMm}{2 a} \\ \Rightarrow 2 a = 4 R_{e} \\ W e k n o w t h a t m a j o r a x i s c a n b e g i v e n a s, \\ 2 a = r_{\min} + r_{\max} \\ \Rightarrow 4 R_{e} = R_{e} + r_{\max} \\ \Rightarrow r_{\max} = 3 R_{e} \end{array}$

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