Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is g and that on the surface of the new planet is g', then

g' = 3g

$g' = \frac{g}{9}$

g' = 9g

g' = 27g

Solution

g' = 3g

The acceleration due to gravity on the planet can be found using the relation

$g = \frac{GM}{R^{2}} but M = \frac{4}{3} {πR}^{3} ρ ρ - density g = \frac{G \times \frac{4}{3} {πR}^{3} ρ}{R^{2}} g = G \times \frac{4}{3} πRρ$

⇒ g ∝ R

$∴ \frac{g'}{R} = \frac{R'}{R} \Rightarrow \frac{g'}{g} = \frac{3 R}{R} \Rightarrow \frac{g'}{g} = 3 \Rightarrow g' = 3 g$

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