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CGP EDU Academic Team
Published on: September 12, 2026
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
Text Solution
Verified by ExpertsThe correct answer is:
C
$\frac{v_{p}}{v_{e}} = \sqrt{\frac{g_{p}}{g_{e}} \times \frac{R_{p}}{R_{e}}}$ = $\sqrt{2 \times 2} = 2$
⇒ ⇒ $v_{p} = 2 \times v_{e} = 2 \times 11.2 = 22.4 \, km/s$
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