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(C) The velocity $v$ required to jump to a height $h$ is given by $v = \sqrt{2gh}$.To escape the gravitational pull,the boy's jump velocity must be eq

Vedclass · Accepted Answer

$[\frac{3}{4\pi} \frac{gh}{Gd}]^{1/2}$

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(C) The velocity $v$ required to jump to a height $h$ is given by $v = \sqrt{2gh}$.To escape the gravitational pull,the boy's jump velocity must be equal to the escape velocity $v_e$ of the planet.The escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.Substituting the mass $M = \frac{4}{3}\pi R^3 d$,we get $v_e = \sqrt{\frac{2G}{R} \cdot \frac{4}{3}\pi R^3 d} = \sqrt{\frac{8}{3}\pi G R^2 d} = R \sqrt{\frac{8}{3}\pi Gd}$.Equating $v = v_e$,we have $\sqrt{2gh} = R \sqrt{\frac{8}{3}\pi Gd}$.Squaring both sides: $2gh = R^2 (\frac{8}{3}\pi Gd)$.Solving for $R^2$: $R^2 = \frac{2gh \cdot 3}{8\pi Gd} = \frac{3gh}{4\pi Gd}$.Therefore,$R = [\frac{3}{4\pi} \frac{gh}{Gd}]^{1/2}$.

$A$ boy can jump to a height $h$ on the surface of a planet. If the density of the planet is $d$,what should be its radius $R$ so that the boy can escape from the gravitational pull of the planet?

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