Question

Acceleration due to gravity at surface of a planet is equal to that at surface of the earth and density is $1.5$ times that of earth. if radius of earth is $R$, radius of planet is

• A. $\frac{R}{1.5}$
• B. $\frac{2}{3}R$
• C. $\frac{9}{4}R$
• D. $\frac{4}{9}R$

Answer 1

Answer: (B)

$\frac{2}{3}R$

Gravity at surface of a planet $=$ gravity at surface of the earth.

$g_p=g_e$

$P_1={1.5P}_e$

Radius of earth $=R$

$g=\frac{G(P\times \frac{4}{3}\pi R^3)}{R^2}$

It's given, $P_P=1.5P_E$

$\frac{G\times \frac{4}{3}\pi R^3\times P_e}{R^2}=\frac{G\times \frac{4}{3}\pi \times {\left(R^1\right)}^3\times P_P}{{\left(R^1\right)}^2}$

or, $R.P_e=R^1{P}_P$

or, $R^1=R\times \frac{P_P}{P_P}=\frac{R}{1.5}=\frac{2R}{3}$.