QJEEAdv15Phb3f6: JEE Advanced 2015 Physics - Class 11 JEEAdv15Phb3f6 Solution: A bullet is fired vertically upwards with velocity v from the…

Given situation is shown in the figure. Let acceleration due to gravity at the surface of the planet be g. At height h above planet's surface v = 0.

According to question, acceleration due to gravity of the planet at height h above its surface becomes g/4.

${g_h} = {g \over 4} = {g \over {{{\left( {1 + {h \over R}} \right)}^2}}}$

$4 = {\left( {1 + {h \over R}} \right)^2} \Rightarrow 1 + {h \over R} = 2$

${h \over R} = 1 \Rightarrow h = R$ .

So, velocity of the bullet becomes zero at h = R.

Also, ${v_{esc}} = v\sqrt N \Rightarrow \sqrt {{{2GM} \over R}} = v\sqrt N$ ...... (i)

Applying energy conservation principle,

Energy of bullet at surface of earth = Energy of bullet at highest point

${{ - GMm} \over R} + {1 \over 2}m{v^2} = {{ - GMm} \over {2R}}$

${1 \over 2}m{v^2} = {{GMm} \over {2R}}$ $\therefore$ $v = \sqrt {{{GM} \over R}}$

Putting this value in eqn. (i), we get

$\sqrt {{{2GM} \over R}} = \sqrt {{{NGM} \over R}}$

$\therefore$ N = 2