- Home
- Standard 11
- Physics
If density of a planet is double that of the earth and the radius $1.5$ times that of the earth, the acceleration due to gravity on the surface of the planet is ........
$\frac{3}{4}$ times that on the surface of the earth
$3$ times that on the surface of the earth
$\frac{4}{3}$ times that on the surface of the earth
$6$ times that on the surface of the earth
Solution
(b)
Acceleration due to gravity on the surface of a planet is given by, $g=\frac{G M}{R^2}$
$M \rightarrow$ Mass of the planet
$R \rightarrow$ Radius of the planet
Also, $M=\frac{4}{3} \pi R^3 \times \rho$
$\Rightarrow g=\frac{G}{R^2} \times \frac{4}{3} \pi R^3 \rho=\frac{4}{3} \rho G \pi R$
$\rho \rightarrow$ Density of the planet.
$\Rightarrow$ Acceleration due to gravity $\alpha \rho R$
$\Rightarrow \frac{g_{\text {planet }}}{g_{\text {earth }}}=\frac{2 \rho_e \times 1.5 R_e}{\rho_e \times R_e}=3$
$\Rightarrow$ Acceleration due to gravity on the surface of planet is $3$ times that on the surface of earth.
