If Earth has mass nine times and radius twice that of a planet P, then (vₑ/3)√x m s⁻¹ will be the minimum velocity required by a rocket to escape from gravitational force of P, where vₑ is escape velocity on Earth. The value of x is:
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Question

If Earth has mass nine times and radius twice that of a planet P, then (vₑ/3)√x m s⁻¹ will be the minimum velocity required by a rocket to escape from gravitational force of P, where vₑ is escape velocity on Earth. The value of x is:

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G1 Physics · Accepted Answer

Escape speed on P is √(2GM(P)/R(P)). Given M(E)=9M(P) and R(E)=2R(P), so M(P)=M(E)/9 and R(P)=R(E)/2. Thus v(P)=√[2G(M(E)/9)/(R(E)/2)] = (√2/3)vₑ = (vₑ/3)√2. Hence x=2.