Gravitation - Result Question 48
51. The escape velocity on the surface of earth is $11.2 km / s$. What would be the escape velocity on the surface of another planet of the same mass but $1 / 4$ times the radius of the earth?
(a) $22.4 km / s$
(b) $44.8 km / s[2000]$
(c) $5.6 km / s$
(d) $11.2 km / s$
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Answer:
Correct Answer: 51. (a)
Solution:
- (a) $v _{\text{earth }}=\sqrt{\frac{2 G M_e}{R_e}}$
$v _{\text{planet }}=\sqrt{\frac{2 G M_p}{R_p}}=\sqrt{\frac{2 G M_e}{R_e / 4}}=\sqrt{\frac{8 G M_e}{R_e}}$ $\frac{v _{\text{planet }}}{v _{\text{earth }}}=\sqrt{\frac{8 G M_e}{R_e}} \times \sqrt{\frac{R_e}{2 G M_e}}=2$
$\therefore v _{\text{planet }}=2 \times v _{\text{earth }}=2 \times 11.2=22.4 km / s$
