Gravitation - Result Question 52

55. 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

(a) $v_e$

(b) $3 v_e$

(c) $9 v_e$

(d) $1 / 3 v_e$

[1995]

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Answer:

Correct Answer: 55. (a)

Solution:

  1. (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, $v_P=v_e$.



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