A newly discovered planet has a density eight | vedclass

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Question

(A) Let the density of Earth be $d$ and its radius be $R$.Then the density of the planet is $8d$ and its radius is $2R$.Mass of Earth $M = \frac{4}{3}

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(A) Let the density of Earth be $d$ and its radius be $R$.Then the density of the planet is $8d$ and its radius is $2R$.Mass of Earth $M = \frac{4}{3} \pi R^3 d$.Mass of the planet $M' = \frac{4}{3} \pi (2R)^3 (8d) = \frac{4}{3} \pi (8R^3) (8d) = 64 M$.Acceleration due to gravity on Earth is $g = \frac{GM}{R^2}$.Acceleration due to gravity on the planet is $g' = \frac{GM'}{(2R)^2} = \frac{G(64M)}{4R^2} = 16 \frac{GM}{R^2} = 16g$.For a body falling freely from rest,the distance $S$ covered in time $t$ is $S = \frac{1}{2}at^2$.On Earth: $S = \frac{1}{2}g(1)^2 = \frac{g}{2}$.On the planet: $S = \frac{1}{2}g't^2 = \frac{1}{2}(16g)t^2 = 8gt^2$.Equating the two expressions for $S$: $\frac{g}{2} = 8gt^2$.$t^2 = \frac{1}{16} \implies t = \frac{1}{4} = 0.25 \, 	ext{sec}$.Note: The mass of the object does not affect the time of free fall.

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