Gravitation - Result Question 32
33. The Earth is assumed to be a sphere of radius $R$. A platform is arranged at a height $R$ from the surface of the Earth. The velocity of a body from this platform is $f v$, where $v$ is its velocity from the surface of the Earth. The value of $f$ is
[2006]
(a) $\frac{1}{\sqrt{2}}$
(b) $\frac{1}{3}$
(c) $\frac{1}{2}$
(d) $\sqrt{2}$
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Answer:
Correct Answer: 33. (a)
Solution:
- (a) Potential energy at height $R=-\frac{G M m}{2 R}$
If $m$ be the mass of a body which is thrown with velocity $v_e$ so that it goes out of gravitational field from distance $R$, then
$\frac{1}{2} m v_e^{2}=\frac{G M}{2 R} \cdot m \Rightarrow v_e=\sqrt{\frac{G M}{R}}$
or, $v_e=\sqrt{g R}$
Now, $v=\sqrt{2 g R}$, So, $v=\sqrt{2} v_e$
or, $v_e=\frac{v}{\sqrt{2}}$
Comparing it with given equation, $f=\frac{1}{\sqrt{2}}$.
