A bullet is fired vertically upwards with velocity $\mathrm{\upsilon }$ from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is $\frac{1}{4^{\mathrm{t}\mathrm{h}}}$ of its value at the surface of the planet. If the escape velocity from the planet is ${\mathrm{\upsilon }}_{\mathrm{e}\mathrm{s}\mathrm{c}}=\mathrm{\upsilon }\sqrt{\mathrm{N}}$ , then the value of N is (ignore energy loss due to atmosphere)

Solution

${\mathrm{g}}^{\mathrm{\prime }}=\frac{\mathrm{g}}{{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)}^2}=\frac{\mathrm{g}}{4}$
$1+\frac{\mathrm{h}}{\mathrm{R}}=2$
$\mathrm{h}=\mathrm{R}$ ...(i)
So velocity of particle becomes zero at $\mathrm{h}=\mathrm{R}$
Given ${\mathrm{v}}_{\mathrm{e}\mathrm{s}\mathrm{c}}=\mathrm{v}\sqrt{\mathrm{N}}$
So $\sqrt{\frac{2\mathrm{G}\mathrm{M}}{\mathrm{R}}}=\mathrm{v}\sqrt{\mathrm{N}}$ ...(ii)
Applying conservation of energy
$\frac{-\mathrm{G}\mathrm{M}\mathrm{m}}{\mathrm{R}}+\frac{1}{2}\mathrm{m}{\mathrm{v}}^2=-\frac{\mathrm{G}\mathrm{M}\mathrm{m}}{2\mathrm{R}}+0$
On solving
${\mathrm{v}}^2=\frac{\mathrm{G}\mathrm{M}}{\mathrm{R}}$
So $\mathrm{v}=\sqrt{\frac{\mathrm{G}\mathrm{M}}{\mathrm{R}}}$ putting in equation (ii)
$\sqrt{\frac{2\mathrm{G}\mathrm{M}}{\mathrm{R}}}=\mathrm{v}\sqrt{\mathrm{N}}$
Comparing $\mathrm{N}=2$

Hence, the correct answer is 2.