Question

Calculate the acceleration due to gravity at the surface of mars if its diameter is 6760km and mass one tenth that of the earth. The diameter of earth is 12742km and acceleration due to gravity on earth is $9.8m/{{s}^{2}}$.

Answer 1

Hint: We have been provided with the value of mass diameter of earth and mars. Value of g earth is $9.8m/{{s}^{2}}$ and value of acceleration due to gravity on moon need to be calculate to calculate acceleration due to gravity on mars, use expression of acceleration due to gravity which gives the relation between the diameter mass and acceleration due to gravity calculate value of g for earth and then for mars. Take the ratio of both acceleration due to gravity and calculate acceleration due to gravity on the surface of mars.
Formula used:
 $g=\dfrac{Gm}{{{R}^{2}}}$
Where, G= universal constant, m = mass of the object, R = distance between the object.

Complete answer:
We have been provided with the diameter of earth which is 12742km. and it is given by, ${{D}_{e}}=12742$
The acceleration due to gravity on earth is given by, ${{g}_{e}}=9.8m/{{s}^{2}}$ now, we have provided with the diameter of mars which is given by ${{D}_{m}}=6760km$
Mass of mars is one-tenth of mass of earth which can be written as,
${{M}_{m}}=\dfrac{1}{10}{{M}_{e}}$
Where, ${{M}_{e}}$ is the mass of earth
${{M}_{m}}$ is the mass of mars
Now we need to calculate the value of acceleration due to gravity on mars denoted by${{g}_{m}}$ we know that the weight of the object is equal to gravitational force is acting on body so the expression of acceleration due to gravity is given by, $g=\dfrac{Gm}{{{R}^{2}}}$
We have provided with the diameter of earth and mars and we know that twice of radius is diameter i.e. \[\begin{align}
  & D=2R \\
 & \Rightarrow R=\dfrac{D}{2} \\
\end{align}\]
Then,
$g=\dfrac{4Gm}{{{D}^{2}}}$
Thus acceleration due to gravity on mars is given by,
${{g}_{m}}=\dfrac{4G{{M}_{m}}}{D_{m}^{2}}......\left( 1 \right)$
And the acceleration due to gravity on earth is given by,
${{g}_{e}}=\dfrac{4G{{M}_{e}}}{D_{e}^{2}}......\left( 2 \right)$
Take ratio of (1) and (2) we get,
$\dfrac{{{g}_{m}}}{{{g}_{e}}}=\dfrac{\dfrac{\dfrac{4G{{M}_{m}}}{D_{m}^{2}}}{4G{{M}_{e}}}}{{{D}_{e}}^{2}}.........\left( 3 \right)$
We know that value of G is $6.673\times {{10}^{-11}}$ which a constant quantity is, therefore equation (3) can be written as,
$\begin{align}
  & \dfrac{{{g}_{m}}}{{{g}_{e}}}=\dfrac{{{M}_{m}}}{{{M}_{e}}}\times {{\left( \dfrac{{{D}_{e}}}{{{D}_{m}}} \right)}^{2}} \\
 & \dfrac{{{g}_{m}}}{9.8}=\dfrac{1}{10}\times {{\left( \dfrac{12742}{6760} \right)}^{2}} \\
\end{align}$
By solving we get,
${{g}_{m}}=3.48m/{{s}^{2}}$

Hence, acceleration due to gravity at the surface of mars is $3.48m/{{s}^{2}}$.

Note:
Acceleration due to gravity depends upon the height of the object above the surface of the earth or any other planet. The mass of the object value of acceleration due to gravity varies with altitude, depth and latitude at the center of earth. The value of g is equal to zero.