The escape velocity of an object from a plane | vedclass

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(B) The escape velocity $v_e$ is given by the formula:$v_e = \sqrt{\frac{2GM}{R}}$Since the mass $M$ of a planet can be expressed in terms of its dens

Vedclass · Accepted Answer

$48$

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(B) The escape velocity $v_e$ is given by the formula:$v_e = \sqrt{\frac{2GM}{R}}$Since the mass $M$ of a planet can be expressed in terms of its density $d$ and radius $R$ as $M = d 	imes \frac{4}{3} \pi R^3$,we substitute this into the formula:$v_e = \sqrt{\frac{2G}{R} 	imes \frac{4}{3} \pi R^3 d} = \sqrt{\frac{8}{3} G \pi R^2 d} = R \sqrt{\frac{8}{3} G \pi d}$For the first planet $(A)$:$v_1 = 16 \ km/s = R \sqrt{\frac{8}{3} G \pi d}$For the second planet $(B)$ with radius $R' = 3R$ and density $d' = 2d$:$v_2 = (3R) \sqrt{\frac{8}{3} G \pi (2d)} = 3R \sqrt{2} \sqrt{\frac{8}{3} G \pi d} = 3 \sqrt{2} 	imes v_1$Substituting $v_1 = 16 \ km/s$:$v_2 = 3 \sqrt{2} 	imes 16 = 48 \sqrt{2} \ km/s$Given $v_2 = v \sqrt{2} \ km/s$,we find:$v = 48$

The escape velocity of an object from a planet is $16 \ km/s$. If the escape velocity of the object from another planet having twice the density and three times the radius of the planet is $v \sqrt{2} \ km/s$,then the value of $v$ is

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