Question

If the density of a small planet is the same as that of the earth while the radius of the planet is $0.2$ times that of the earth, the gravitational acceleration on the surface of the planet is _____.

• A. $0.2g$
• B. $0.4g$
• C. $2g$
• D. $4g$

Answer 1

The correct option is A

$0.2g$

Explanation for the correct option:

Step 1: State assumptions and known data

The density of Earth and the small planet is same.
Let the density of earth and the small planet be $\rho$

Let $R$ be the radius of Earth. Then, $0.2R$ is the radius of the small planet.

Step 2: Formulas used

Acceleration due to gravity is given as,

$g=\frac{GM}{R^2}$ $...\left(1\right)$

where $G$ is the gravitational constant, $M$ is the mass of the planet and $R$ is the radius of the planet.

Density of an object, $\rho =\frac{M}{V}$, where $V$ is its volume

Step 3: Derive expression for acceleration due to gravity in term of density of the earth

From the equation relating density and mass,
$M=\rho V$

The shape of a planet can be approximated to be that of a sphere (in fact, shape of the earth is geoid).

The volume of a sphere is $V=\frac{4}{3}\mathrm{\pi }{r}^3$, where $r$ is its radius.

Thus, the mass of the sphere is, $M=\frac{4}{3}\mathrm{\pi }{r}^3\rho$

Substituting the above expression in $\left(1\right)$,

$\begin{array}{rcl}g& =& \frac{4}{3}\frac{G\mathrm{\pi }{r}^3\rho }{r^2}\\ \mathit{\therefore }g& =& \frac{4}{3}G\mathrm{\pi r}\rho \end{array}$

Step 4: Calculate the acceleration due to gravity on the small planet

The acceleration due to gravity on Earth is,
$g=\frac{4}{3}G\mathrm{\pi }R\mathrm{\rho }$

The acceleration due to gravity on the small planet is,
$\begin{array}{ccc}& g\text{'}=\frac{4}{3}\mathrm{G\pi }\left(0.2R\right)\rho & \\ \Rightarrow & g\text{'}=0.2\times \frac{4}{3}\mathrm{\pi }R\rho & \\ \therefore & g\text{'}=0.2g& \left[\because g=\frac{4}{3}\mathrm{G\pi }R\rho \right]\end{array}$

Therefore, if the density of a small planet is the same as that of the earth while the radius of the planet is $0.2$ times that of the earth, the gravitational acceleration on the surface of the planet is $0.2g$.

Hence, option A is correct.