Gravitation - Result Question 50
53. The escape velocity of a body on the surface of the earth is $11.2 km / s$. If the earth’s mass increases to twice its present value and the radius of the earth becomes half, the escape velocity would become
[1997]
(a) $44.8 km / s$
(b) $22.4 km / s$
(c) $11.2 km / s$ (remains unchanged)
(d) $5.6 km / s$
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Answer:
Correct Answer: 53. (b)
Solution:
- (b) Escape velcocity
$v_e=\sqrt{\frac{2 G M_e}{R_e}}, v_e^{\prime}=\sqrt{\frac{2 G M_e^{\prime}}{R_e^{\prime}}}$
$\therefore \frac{v_e^{\prime}}{v_e}=\sqrt{\frac{M_e^{\prime}}{M_e} \times \frac{R_e}{R_e^{\prime}}}$
Given $M_e^{\prime}=2 M_e$ and $R_e^{\prime}=\frac{R_e}{2}$
$\therefore \frac{v_e^{\prime}}{v_e}=\sqrt{\frac{2 M_e}{M_e} \times \frac{R_e}{R_e / 2}}=\sqrt{4}=2$
$v_e^{\prime}=2 v_e=2 \times 11.2=22.4 km / s$
