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MHT CET 2024 (4 May, Shift 2) · Escape Velocity

Escape Velocity question from MHT CET 2024 (4 May, Shift 2)

Q16MHT CET 2024 · 4 MAY · SHIFT 2Easy

Escape Velocity

The escape velocity from earth surface is $11\text{ km/s}$ . The escape velocity from a planet having twice the radius and same mean density as earth is

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Answer: (a)

$v_e \propto R\sqrt{\rho}$
Same density, twice radius: $v_e' = 2 \times 11 = 22$ km/s
So the answer is (a) 22 km/s.

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