Gravitation Ques 52
52. The escape velocity from the surface of the earth is $v_e$. The escape velocity from the surface of a planet whose mass and radius are three times those of the earth, will be
[1995]
(a) $v_e$
(b) $3 v_e$
(c) $9 v_e$
(d) $1 / 3 v_e$
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Answer:
Correct Answer: 52.(a)
Solution:
- (a) Escape velocity on surface of earth
$ \begin{aligned} & v_e=\sqrt{\frac{2 G M_e}{R_e}} \propto \sqrt{\frac{M_e}{R_e}} . \\ & \therefore \frac{v_e}{v_P}=\sqrt{\frac{M_e}{R_e}} \times \sqrt{\frac{R_p}{M_p}} \\ & =\sqrt{\frac{M_e}{R_e}} \times \sqrt{\frac{3 R_e}{3 M_e}}=\frac{1}{1}=1 \end{aligned} $
or, $v_P=v_e$.
BETA
