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(B) The acceleration due to gravity $g$ on the surface of the Earth is given by the formula: $g = \frac{GM}{R^2}$.Since mass $M = 	ext{density} (ho

Vedclass · Accepted Answer

$19.6$

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(B) The acceleration due to gravity $g$ on the surface of the Earth is given by the formula: $g = \frac{GM}{R^2}$.Since mass $M = 	ext{density} (ho) 	imes 	ext{volume} (V) = ho 	imes \frac{4}{3} \pi R^3$,we can substitute this into the formula for $g$:$g = \frac{G (ho \cdot \frac{4}{3} \pi R^3)}{R^2} = \frac{4}{3} \pi ho G R$.From this expression,it is clear that $g \propto ho$ when the radius $R$ is kept constant.Therefore,$\frac{g_2}{g_1} = \frac{ho_2}{ho_1}$.Given that the density is doubled,$ho_2 = 2ho_1$.Thus,$g_2 = 2 	imes g_1 = 2 	imes 9.8 \ m/s^2 = 19.6 \ m/s^2$.

If the density of the earth is doubled keeping the radius constant,find the new acceleration due to gravity (in $m/s^2$)? $(g = 9.8 \ m/s^2)$

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