Question

Mars has about ${\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}^{th}$mass of earth and about $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ the diameter of earth. The acceleration due to gravity at the surface of mars is about $\dots \dots \dots$

Answer 1

Acceleration due to gravity

1. It is the acceleration of a body under the influence of gravity during free fall.
2. The value of g on earth is $9.8{\text{ms}}^{-2}$
3. It depends on the location of the body.
4. It can be calculated using the formula $g=\frac{GM}{r^2}$; where $M$is the mass of earth , $G$ is the gravitational constant and $R$ is the radius of earth.

Given data and calculations:

Mass of mars $M_m$ =${\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}^{th}M$, where $M$is the mass of earth.

The radius of mars $R_m=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.R$, where $R$ is the radius of the earth

Hence, we can calculate acceleration due to gravity on Mars, $g_m$ ,as follows:

$\begin{array}{rcl}{g}_m& =& \frac{G{M}_m}{{R_m}^2}\\ & =& \frac{G({\displaystyle \raisebox{1ex}{$M$}\!\left/ \!\raisebox{-1ex}{$10)$}\right.}}{{\left({\displaystyle \raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}^2}\\ & =& \frac{4}{10}\frac{GM}{R^2}\\ & =& \frac{4}{10}{g}_{earth}\\ & =& \frac{4}{10}\times 9.8\\ & =& 3.92{\text{ms}}^{-2}\end{array}$

Thus, acceleration due to gravity at the surface of Mars is $g_m=3.92{\text{ms}}^{-2}$.