Question

Radius of earth is $6400\,km$ and that of mars is $3200\,km$. Mass of mars is 0.1 that of earth’s mass. Then the acceleration due to gravity on mars is nearly.
A) $1\,m{s^{ - 2}}$

B) $2.5\,m{s^{ - 2}}$

C) $4\,m{s^{ - 2}}$

D) $5\,m{s^{ - 2}}$

Answer 1

Hint:Here we have to apply the formula and concept of acceleration due to gravity.
Acceleration due to gravity is the acceleration of the mass due to gravitational force. The SI unit is given by $m{s^{ - 2}}$. It is a vector quantity since it has both magnitude and direction. Acceleration due to gravity is given by $g$ .

Complete step by step solution:
Gravity is the power by which the planet draws the body to its core. Let us consider two bodies of masses ${m_a}$ and ${m_b}$. With regard to the application of equivalent forces on two bodies, the mass in terms of body $b$
 is given as:
${m_b} = {m_a}\left( {\dfrac{{{a_A}}}{{{a_B}}}} \right)$

The above mass is called the inertial mass of a body.
Under the influence of gravity the mass is given by:
${m_B} = \left( {\dfrac{{{F_B}}}{{{F_A}}}} \right) \times {m_A}$

The above mass is called the gravitational mass.
We know that according to universal law of gravitation:
$F = \dfrac{{GMm}}{{{r^2}}}$

Also,
$F = ma$

Equating the two equations we get:
$a = \dfrac{{GM}}{{{r^2}}}$

If $g$b is the acceleration due to gravity then:
$g = \dfrac{{GM}}{{{r^2}}}$

The acceleration due to gravity depends on the mass and radius.
This helps us to understand:
All bodies undergo the same acceleration due to gravity, regardless of their mass. Its value on earth depends on the density of the earth, not the density of the material.
Let us consider ${M_e},\,{R_e}$ be the mass and radius of the earth and ${M_m},\,{R_m}$
 be the corresponding quantities of mars, then acceleration due to gravity on earth is:
${g_e} = \dfrac{{G{M_e}}}{{{R_e}^2}}$ ...... (i)
Acceleration due to gravity on mars is:
${g_m} = \dfrac{{G{M_m}}}{{{R_m}^2}}$ ...... (ii)
Also according to question:
$
  {M_m} = 0.1{M_e} \\
  {R_e} = 6400\,km \\
  {R_m} = 3200\,km \\
  {g_e} = 9.8\,m{s^{ - 2}} \\
$

Dividing equation (i) by (ii), we get:
$
  \dfrac{{{g_m}}}
{{{g_e}}} = \left( {\dfrac{{{M_m}}}
{{{M_e}}}} \right){\left( {\dfrac{{{R_e}}}
{{{R_m}}}} \right)^2} \\
  {g_m} = {g_e}\left( {\dfrac{{{M_m}}}
{{{M_e}}}} \right){\left( {\dfrac{{{R_e}}}
{{{R_m}}}} \right)^2} \\
  {g_m} = \left( {9.8\,m{s^{ - 2}}} \right)\left( {\dfrac{{0.1{M_e}}}
{{{M_e}}}} \right){\left( {\dfrac{{6400\,km}}
{{3200\,km}}} \right)^2} \\
   = \left( {9.8\,m{s^{ - 2}}} \right) \times 0.1 \times 4 \approx 4\,m{s^{ - 2}} \\
$

Hence, option C is correct.

Note:Here we have to pay attention while putting the values in the formula. If we put a wrong value or change the value we may not get the correct answer.