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(B) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.Since mass $M = 	ext{Volume} 	imes 	ext{Density} =

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$3$ times that on the surface of the earth

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(B) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.Since mass $M = 	ext{Volume} 	imes 	ext{Density} = \frac{4}{3} \pi R^3 ho$,we can substitute this into the formula:$g = \frac{G}{R^2} 	imes \frac{4}{3} \pi R^3 ho = \frac{4}{3} \pi G ho R$.This shows that $g \propto ho R$.Given $ho_p = 2 ho_e$ and $R_p = 1.5 R_e$,where $p$ denotes the planet and $e$ denotes the earth:$\frac{g_p}{g_e} = \frac{ho_p R_p}{ho_e R_e} = \frac{2 ho_e 	imes 1.5 R_e}{ho_e R_e} = 2 	imes 1.5 = 3$.Therefore,the acceleration due to gravity on the surface of the planet is $3$ times that on the surface of the earth.

If the density of a planet is double that of the earth and the radius is $1.5$ times that of the earth,the acceleration due to gravity on the surface of the planet is ........

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