A bullet is fired vertically upwards with velocity $v$ from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is $1/4^{\text{th}}$ of its value at the surface of the planet. If the escape velocity from the planet is $v_{\text{esc}}=v \sqrt{N}$, then the value of $N$ is (ignore energy loss due to atmosphere)

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Gravitation

A bullet is fired vertically upwards with velocity

v

from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is

1/ 4^{th}

of its value at the surface of the planet. If the escape velocity from the planet is

v_{esc} = v N

, then the value of

N

is (ignore energy loss due to atmosphere)

Answer: 2

Solution

At the surface,

g = \frac{GM}{R ^{2}}

. At maximum height

r

g^{'} = \frac{GM}{r ^{2}} = \frac{g}{4}

, giving

r = 2 R

. By conservation of energy:

- \frac{GM m}{R} + \frac{1}{2} m v^{2} = - \frac{GM m}{2 R}

, so

\frac{1}{2} m v^{2} = \frac{GM m}{2 R}

. The escape velocity is

v_{esc} = \frac{2 GM}{R} = 2 \cdot v

. Therefore

N = 2

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Integer answer

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