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(B) Let $R$ be the radius of the planet and $M$ be its mass. The acceleration due to gravity at the surface is $g = GM/R^2$.At a height $h$ above the

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$2$

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(B) Let $R$ be the radius of the planet and $M$ be its mass. The acceleration due to gravity at the surface is $g = GM/R^2$.At a height $h$ above the surface,the acceleration due to gravity is $g' = GM/(R+h)^2$.Given $g' = g/4$,we have $GM/(R+h)^2 = (1/4) \cdot (GM/R^2)$,which implies $(R+h)^2 = 4R^2$,so $R+h = 2R$,or $h = R$.Using the conservation of energy at the surface and at the maximum height $h=R$:$-(GMm/R) + (1/2)mv^2 = -(GMm/(R+R))$$-(GMm/R) + (1/2)mv^2 = -(GMm/2R)$$(1/2)mv^2 = GMm/2R \implies v^2 = GM/R$.The escape velocity is defined as $v_{esc} = \sqrt{2GM/R}$.Substituting $GM/R = v^2$,we get $v_{esc} = \sqrt{2v^2} = v\sqrt{2}$.Comparing this with $v_{esc} = v\sqrt{N}$,we find $N = 2$.

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