The mean diameters of Mars and Earth are$\mathbf{6}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathit{k}\mathit{m}$ and$\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{k}\mathit{m}$ , respectively. The mass of Mars is$\mathbf{0}\mathbf{.}\mathbf{11}$ timesEarth’s mass. (a) What is the ratio of the mean density (mass perunit volume) of Mars to that of Earth? (b) What is the value of thegravitational acceleration on Mars? (c) What is the escape speedon Mars?

The mean diameter of mars is${\text{6.9×10}}^{\text{3}}\text{km}$.

The mean diameter of the earth is${\text{1.3×10}}^{\text{4}}\text{km}$.

The mass of mars is 0.11 times the mass of earth.

Find the ratio of the density of earth and mars using the formula for density in terms of mass and radius. Using this ratio, find the gravitational acceleration of mars. Also, find the escape velocity using the mass of mars in terms of the mass of earth.

Formulae are as follows:

$$\begin{gathered} \rho =\frac{3}{4}\times M\pi R^3 \\ g=\frac{GM}{R^2} \\ V=\sqrt{\frac{GM}{R}} \end{gathered}$$

where g is an acceleration due to gravity, M is mass, R is the radius, G is the gravitational constant,and V is volume and p is density.

The density of the earth can be written as,

${\rho }_e=\frac{M_e}{\frac{4}{3}\pi R_e^3}$

The density of mars can be written as,

${\rho }_m=\frac{M_m}{\frac{4}{3}\pi R_m^3}$

Now, take the ratio as,

$\frac{{\rho }_m}{{\rho }_e}=\frac{M_m{R}_e^3}{M_e{R}_m^3}$

So,

$$\begin{gathered} \frac{{\rho }_m}{{\rho }_e}=0.11{\left(0.65\times \frac{{10}^{4}km}{3.45\times {10}^{3}km}\right)}^3 \\ =0.74 \end{gathered}$$

Hence, the ratio of the mean density of Mars to Earth is $0.74$.

Gravitational acceleration for mars,

$\begin{array}{rcl}{a}_{gm}& =& \frac{G{M}_m}{R_m^2}\\ & =& \frac{M_m{R}_e^2}{M_e{R}_e^2}\left(\frac{G{M}_e}{R_e^2}\right)\\ & =& \frac{M_m{R}_e^2}{M_e{R}_m^2}{a}_{ge}\\ & & \end{array}$

$\begin{array}{rcl}{a}_{gm}& =& 0.11{\left(0.65\times \frac{{10}^{4}km}{3.45\times {10}^{3}km}\right)}^2\times 9.8{\text{m/s}}^{\text{2}}\\ & =& 3.8 {\text{m/s}}^{\text{2}}\\ & & \end{array}$

Hence, the value of the gravitational acceleration on Mars is $3.8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.

Escape speed can be found as,

$\begin{array}{rcl}\text{v}& =& \sqrt{\frac{\text{2GM}}{{\text{R}}_{\text{m}}}}\\ v& =& \sqrt{\frac{2\times 6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}\times 0.11\times 5.98\times {10}^{24}kg}{3.45\times {10}^{6}m}}\\ & =& 5\times {10}^3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\\ & & \end{array}$

Hence, the escape speed on Mars is $=5\times {10}^3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Therefore, the gravitational acceleration on mars can be found by comparing the densities of earth and the moon. The escape velocity can also be found using the mass and radius of the orbit.