Question

The escape velocity for a rocket from earth is 11.2 km/sec. Its value on a planet where acceleration due to gravity is double that on the earth and diameter of the planet is twice that of earth will be in km/sec

• A. 11.2
• B. 5.6
• C. 22.4
• D. 53.6

Answer 1

Answer: (C)

22.4

Answer 2

$\frac { v _ { p } } { v _ { e } } = \sqrt { \frac { g _ { p } } { g _ { e } } \times \frac { R _ { p } } { R _ { e } } }$= $\sqrt { 2 \times 2 } = 2$

⇒ $v _ { p } = 2 \times v _ { e } = 2 \times 11.2 = 22.4 \mathrm {~km} / \mathrm { s }$