Question

If the diameter of Mars is $6760\mathrm{km}$ and mass one-tenth that of the earth. The diameter of the earth is $12742\mathrm{km}$. If acceleration due to gravity on earth is $9.8m{s}^{-2}$, the acceleration due to gravity on Mars is

• A. $34.8m{s}^{-2}$
• B. $2.48m{s}^{-2}$
• C. $3.48m{s}^{-2}$
• D. $28.4m{s}^{-2}$

Answer 1

The correct option is C

$3.48m{s}^{-2}$

Explanation for the correct option:

Step 1: Given data

Diameter of Mars $=6760\mathrm{km}$.

Diameter of the earth $=12742\mathrm{km}$.

Let $M$ be the mass of Earth. Then the mass of Mars is $\frac{M}{10}$.

Acceleration due to gravity on Earth $g_e=9.8{\mathrm{ms}}^{-2}$.

Step 2: Formula used:

Acceleration due to gravity is given as,

$g=\frac{GM}{R^2}$

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

Step 3: Finding acceleration due to gravity on mars

Acceleration due to gravity on Earth is,
$g_e=\frac{GM}{{12742}^2}$ $...\left(1\right)$

Acceleration due to gravity on Mars is,
$\begin{array}{cc}& g_m=\frac{G{\displaystyle \frac{M}{10}}}{{6760}^2}\\ \Rightarrow & g_m=\frac{GM}{10\times {6760}^2}...\left(2\right)\end{array}$

Dividing $\left(2\right)$ by $\left(1\right)$,

$\begin{array}{cc}\Rightarrow & \frac{g_m}{g_e}=\frac{{\displaystyle \frac{GM}{10\times {6760}^2}}}{{\displaystyle \frac{GM}{{12742}^2}}}\\ \Rightarrow & \frac{g_m}{g_e}=\frac{{12740}^2}{10\times {6760}^2}\\ \Rightarrow & g_m=9.8\frac{{12740}^2}{10\times {6760}^2}\\ \therefore & g_m=3.48{\mathrm{ms}}^{-2}\end{array}$

Therefore, the acceleration due to gravity on Mars is $3.48{\mathrm{ms}}^{-2}$.

Hence, option C is correct.