Gravitation Ques 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 $v_{\mathrm{sec}}=v \sqrt{N}$, then the value of $N$ is (ignore energy loss due to atmosphere)
(2015 Adv.)
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Answer:
Correct Answer: 13.2
Solution:
Formula:
- At height $h$
$$ g^{\prime}=\frac{g}{\sqrt{1+\left(\frac{h}{R}\right)^{2}}} $$
Given, $g^{\prime}=\frac{g}{4}$
Substituting into equation (i) we get,
$$ h=R $$
Now, from $A$ to $B$,
decrease in kinetic energy $=$ increase in potential energy
$$ \begin{alignedat} & \Rightarrow \quad \frac{1}{2} m v^{2}=\frac{m g h}{1+\frac{h}{R}} \Rightarrow \frac{v^{2}}{2}=\frac{g h}{1+\frac{h}{R}}=\frac{1}{2} g R \quad(h=R) \\ & \Rightarrow \quad v^{2}=g R \quad \text { or } v=\sqrt{g R} \\ & \text { Now, } \quad v_{\mathrm{esc}}=\sqrt{2 g R}=v \sqrt{2} \\ & \Rightarrow \quad N=2 \\ \end{aligned} $$
BETA
