Gravity Radius of Earth

Let $g$ be the acceleration by Earth gravity, $R$ , $M$ , $\rho$ and $V$ be the radius, mass, density and volume of Earth respectively and those after the radius has been increased by $3$ be $R'$ , $M'$ , $\rho '$ and $V'$ .

We know that:

$g \propto \dfrac {M}{R^2} = \dfrac {\rho V}{R^2} \propto \dfrac {\rho R^3}{R^2} = \rho R$

For the same $g$ before and after the radius change, we have:

$\Rightarrow \rho ' R' = \rho R \quad \Rightarrow \rho ' = \dfrac {\rho R}{R'} = \dfrac {\rho R}{3R} = \boxed{\frac{1}{3}}\rho$

Here we know , $g = \frac {GM}{R^2}$

Now As our Earth is A Sphere , then V = 4/3 $\pi$ $R^3$ And lets say that the mean density is $\rho$ .

Hence Mass = Volume x Density or = 4/3 $\pi$ $R^3$ $\rho$ . Now putting values we will get.

Hence Now we get $\rho$ / $\rho'$ = $3$

Hence we get $\rho'$ = 1/3 $\rho$ = factor is = $\boxed {0.33333}$