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 1/4th of its value at the surface of the planet. If the escape velocity from the planet is vesc = , then the value of N is (ignore energy loss due to atmosphere)

Engineering

Physics

Gravitational Acceleration

Escape and Orbital Velocity

Question

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 = $v \sqrt{N}$ , then the value of N is (ignore energy loss due to atmosphere)

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$g = g_{0} \frac{R^{2}}{{(R + h)}^{2}} = \frac{g_{0}}{4}$

⇒ 4R² = (R + 4)²

2R = R + h ⇒ h = R

$So, - \frac{GMm}{R} + \frac{1}{2} {mV}^{2} = \frac{- GMm}{2 R} + 0$

$\frac{1}{2} {mV}^{2} = \frac{GMm}{2 R} \Rightarrow V = \sqrt{\frac{GM}{R}} = \frac{V_{e}}{\sqrt{2}}$

$as Ve = \sqrt{\frac{2 GM}{R}}$

So, Ve = $\sqrt{2}$ V

N = 2