A bullet is fired vertically upwards with velocity from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is th of its value of the surface of the planet. [2015]

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 $\frac{1}{4}$ th of its value of 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). [2015]

Smart Achievers · Accepted Answer

(2)

Let $\frac{1}{4}$ be the height to which the bullet rises with the height acceleration due to gravity varies as

$g^{1} = g {(1 + \frac{h}{R})}^{- 2} \Rightarrow \frac{g}{4} = g {(1 + \frac{h}{R})}^{- 2} \Rightarrow h = R$

We know escape speed, $v_{e} = \sqrt{\frac{2 G M}{R}} = v \sqrt{N} (given) \dots (i)$

Now applying conservation of energy principle

Loss of kinetic energy = gain in gravitational potential energy

$∴$ $\frac{1}{2} m v^{2} = - \frac{G M m}{2 R} - (- \frac{G M m}{R})$

$∴$ $v = \sqrt{\frac{G M}{R}} \dots (i i)$

$Comparing eq. (i) & (ii) we get N = 2$

Solution

Want Us to Email you About Special Offers & Updates?