# Question #61455

Oct 19, 2016

(b) $\frac{1}{3}$

#### Explanation:

As per Law of Universal Gravitation the force of attraction between two bodies is given as

${F}_{G} = G \frac{{M}_{1.} {M}_{2}}{r} ^ 2$

Where $G$ is gravitational constant.
It has the value $6.67408 \times {10}^{-} 11 {m}^{3} k {g}^{-} 1 {s}^{-} 2$.
Comparing it with the expression
$F \left(\text{weight}\right) = m g$
we get value of $g = G \frac{{M}_{e}}{r} _ {e}^{2}$ ......(1)
If $\rho$ is density of earth, then
$\rho = {M}_{e} / \left(V o l u m e\right)$
$\rho = {M}_{e} / \left(\frac{4}{3} \pi {r}_{e}^{3}\right)$
$\implies {M}_{e} = \frac{4}{3} \pi \rho {r}_{e}^{3}$
Inserting above value of mass of earth in (1) we get
$g = G \frac{\frac{4}{3} \pi \rho {r}_{e}^{3}}{r} _ {e}^{2}$
$\implies g = G \frac{4}{3} \pi \rho {r}_{e}$
$\implies g \propto \rho {r}_{e}$
From above we see that if radius of earth is increased by a factor of $3$, to keep value of $g$ same we need to decrease value of $\rho$ by a factor of $3$.
Hence we get $\frac{1}{3}$