Gravitation 2 Question 10
13. 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 at the surface of the planet. If the escape velocity from the planet is $v _{sec}=v \sqrt{N}$, then the value of $N$ is (ignore energy loss due to atmosphere)
(2015 Adv.)
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Answer:
Correct Answer: 13. 3
Solution:
- At height $h$
$$ g^{\prime}=\frac{g}{1+\frac{h}{R}^{2}} $$
Given, $g^{\prime}=\frac{g}{4}$
Substituting in equation (i) we get,
$$ h=R $$
Now, from $A$ to $B$,
decrease in kinetic energy $=$ increase in potential energy
$$ \begin{aligned} & \Rightarrow \quad \frac{1}{2} m v^{2}=\frac{m g h}{1+\frac{h}{R}} \Rightarrow \frac{v^{2}}{2}=\frac{g h}{1+\frac{h}{R}}=\frac{1}{2} g R \quad(h=R) \\ & \Rightarrow \quad v^{2}=g R \quad \text { or } v=\sqrt{g R} \\ & \text { Now, } \quad v _{esc}=\sqrt{2 g R}=v \sqrt{2} \\ & \Rightarrow \quad N=2 \\ & \text { 14. } g=\frac{G M}{R^{2}}=\frac{G \quad \frac{4}{3} \pi R^{3} \rho}{R^{2}} \text { or } g \propto \rho R \quad \text { or } R \propto \frac{g}{\rho} \end{aligned} $$
Now escape velocity, $v _e=\sqrt{2 g R}$
or $\quad v _e \propto \sqrt{g R}$ or $\quad v _e \propto \sqrt{g \times \frac{g}{\rho}} \propto \sqrt{\frac{g^{2}}{\rho}}$ $\therefore \quad\left(v _e\right) _{\text {planet }}=\left(11 km s^{-1}\right) \sqrt{\frac{6}{121} \times \frac{3}{2}}=3 km s^{-1}$
$\therefore$ The correct answer is 3 .
