Question

The density of a planet is double that of the earth and the radius of it is $1.5$ times that of the earth, if $g$ is the acceleration due to gravity on the surface of earth then the acceleration due to gravity on the surface of the planet is

• A. $\frac{3}{4}g$
• B. $3g$
• C. $\frac{4}{3}g$
• D. $6g$

Answer 1

The correct option is B $3g$

Let, radius of the earth be, $R$ and its density be $\rho$

So, radius of the planet, $R_p=1.5R$ and its density ${\rho }_p=2\rho$

Acceleration due to gravity on the surface of a planet is given by, $$g=\frac{GM}{R^2}...\left(1\right)$$ Where,
$M\to$ Mass of the planet
$R\to$ Radius of the planet

Also,
$M=\frac{4}{3}\pi R^3\times \rho$

Using equation $\left(1\right)$,

$g=\frac{G}{R^2}\times \frac{4}{3}\pi R^3\rho =\frac{4}{3}\rho G\pi R$

$\rho \to$ Density of the planet.

Thus, $g\propto \rho R$

$\Rightarrow \frac{g_{planet}}{g}=\frac{{\rho }_p\times R_p}{{\rho }_e\times R_e}$

Substituting the values,

$\Rightarrow \frac{g_{planet}}{g}=\frac{2\rho \times 1.5R}{\rho \times R}$

$\therefore g_{planet}=3g$

Thus, Accelaration due to gravity on the surface of planet is $3$ times that on the surface of the earth.

Hence, option (b) is the correct answer.

Key Concept: Acceleration due to gravity on the surface of a planet is given by $$g=\frac{GM}{R^2}$$Why this question: To make students understand the parameters on which the acceleration due to gravity depends.