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