Question 290: Gravitation

Question

The density of a new planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If $R$ is the radius of earth, then radius of the planet would be

Accepted Answer

Given, $ho_{\mathrm{p}}=2 ho_{\mathrm{e}}, \mathrm{g}_{\mathrm{p}}=\mathrm{g}_{\mathrm{e}}$
$g=\frac{4}{3} \pi ho G R$
$$\begin{aligned}
& \begin{array}{l}
	herefore \quad \frac{R_p}{R_e}=\left(\frac{g_p}{g_e}ight)\left(\frac{ho_e}{ho_p}ight)=(1) 	imes\left(\frac{1}{2}ight) \
\end{array} \
	herefore& \quad R_p=\frac{R_e}{2}=\frac{R}{2}
\end{aligned}$$

Answer

4 R

Answer

$\mathrm{R} / 2$

Answer

$\frac{\mathrm{R}}{4}$

Answer

$2 R$

Gravitation

The density of a new planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If R is the radius of earth, then radius of the planet would be

Choose the correct answer: