\[(a)\;\frac{m}{a} \bigg(\frac{R}{GM}\bigg)\sqrt {n-1} \quad (b)\;\frac{m}{a} \sqrt{\frac{R}{GM}}(\sqrt n -1) \quad (c)\;\frac{m}{a} \sqrt {\frac{RG}{M}}(\sqrt {n+1}) \quad (d)\;\frac{m}{a} \sqrt {\frac{RG}{M}}(\sqrt n-1) \]

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Orbital velocity at distance r from center of earth

$v_0=\sqrt {\large\frac{GM}{r}}$

$KE= \large\frac{1}{2} $$ mv_0^2=\large\frac{GMm}{2r}$

$P.E= \large\frac{-GMm}{r}$

Total energy $E= KE+PE$

$E= \large\frac{-GMm}{2r}$

Now $\large\frac{dE}{dt}$$=F.v$

$\large\frac{GMm}{2r^2}\bigg(\large\frac{dr}{dt}\bigg)$$=-av^3$

$\qquad= -a \bigg(\sqrt {\large\frac{GM}{r}}\bigg)^3$

$\large\frac{GMm}{2r^2}\bigg(\large\frac{dr}{dt}\bigg)$$=-a \large\frac{GM}{r} \sqrt {\frac{GM}{r}}$

$\large\frac{a}{m}$$ \int \limits_0^t dt=-\large\frac{1}{2}\sqrt {\frac{1}{GM}}\int \limits_{r_i}^{r_f} $$r^{-1/2} dr$

$\large\frac{a}{m} $$\int \limits_0^t dt=-\large\frac{1}{2}\sqrt {\frac{1}{GM}}\int \limits_{nR}^{R} $$r^{-1/2} dr$

$t= \large\frac{m}{a} \sqrt {\frac{R}{GM} }$$ (\sqrt n -1)$

Hence b is the correct answer

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