Question

Assuming the mass of Earth to be ten times the mass of Mars, its radius to be twice the radius of Mars and the acceleration due to gravity on the surface of Earth is $10\,\,m{s^{ - 2}}$. Then the acceleration due to gravity on the surface of Mars is given by:
(A) $0.2\,\,m{s^{ - 2}}$
(B) $0.4\,\,m{s^{ - 2}}$
(C) $2\,\,m{s^{ - 2}}$
(D) $4\,\,m{s^{ - 2}}$
(E) $5\,\,m{s^{ - 2}}$

Answer 1

The acceleration due to gravity on the surface of mars can be calculated by using the formula derived from the acceleration due to gravity on the surface of the earth which consist of radius of the earth and mass of the earth.

Formulae Used:
The acceleration due to gravity is;
$g = \dfrac{{GM}}{{{R^2}}}$
Where, $g$ denotes the acceleration due to gravity, $G$ denotes the gravitational of the earth, $M$ denotes the mass of the earth, $R$ denotes the radius of the earth.

Complete step-by-step solution:
The data given in the problems are;
Mass of the earth, ${M_e} = 10 \times {M_m}$
Mass of mars, ${M_m}$,
Acceleration due to gravity on the surface of the earth, $g = 10\,\,m{s^{ - 2}}$
Radius of the earth, $2{R_m}$,
The acceleration due to gravity on the surface of mars is, ${g_m}$;
Let’s take that $\dfrac{M}{{{R^2}}} = \left[ {\dfrac{{\left( {\dfrac{{{M_m}}}{{{M_e}}}} \right)}}{{{{\left( {\dfrac{{{R_m}}}{{{R_e}}}} \right)}^2}}}} \right]$ and $g = \dfrac{{{g_m}}}{{{g_e}}}$;
Where, ${M_m}$ denotes the mass of the mars, ${M_e}$ denotes the mass of the earth, ${R_m}$ denotes the radius of the mars, ${R_e}$ denotes the radius of the earth, ${g_m}$ denotes the acceleration due to gravity on the surface of mars.
Substitute the values;
$\dfrac{{{g_m}}}{{{g_e}}} = \left[ {\dfrac{{\left( {\dfrac{{{M_m}}}{{{M_e}}}} \right)}}{{{{\left( {\dfrac{{{R_m}}}{{{R_e}}}} \right)}^2}}}} \right]$
since we only need the acceleration due to gravity on the surface of mars;
${g_m} = {g_e} \times \left( {\dfrac{{{M_m}}}{{{M_e}}}} \right) \times {\left( {\dfrac{{{R_e}}}{{{R_m}}}} \right)^2}$
Substitute the given values;
${g_m} = 10\,\,m{s^{ - 2}} \times \left( {\dfrac{{{M_m}}}{{10 \times {M_m}}}} \right) \times {\left( {\dfrac{{2 \times {R_m}}}{{{R_m}}}} \right)^2}$
By simplifying the above equation;
${g_m} = 10 \times \left( {\dfrac{1}{{10}}} \right) \times 4$
${g_m} = 4\,\,m{s^{ - 2}}$
Therefore, the acceleration due to gravity on the surface of mars is ${g_m} = 4\,\,m{s^{ - 2}}$.
Hence, the option (D) ${g_m} = 4\,\,m{s^{ - 2}}$ is the correct answer.

Note:- Acceleration due to gravity is the speed gained by an object due to gravity force. The SI unit is $m{s^{ - 2}}$. It represents both magnitude and direction; therefore, it is a vector quantity. The common value of gravity on the plane of the earth at sea is $9.8\,\,m{s^{ - 2}}$.