Question

The atmospheric pressure on Mars is known to be equal o $1/200$ of the atmospheric pressure on
the Earth. The diameter of Mars is approximately equal to half the Earth's diameter, and the densities
${\rho }_E$ and ${\rho }_M$ of the planets are $5.5\times {10}^{3}kg/m^3$ and $4\times {10}^{3}kg/m^3$
Determine the ratio of the masses of the Martian and the Earth's atmospheres.

Answer 1

According to the law of universal gravitation, the force of attraction of the body of mass $m$ to Mars on its surface is $G\frac{M_{M^m}}{R_M^3}$, where $M_M$ is the mass of mars, and $R_M$ is its radius. This means that the free-fall acceleration on the surface of Mars is $g_M=G{M}_M/R_M^2$. If the mass of the Martian atmosphere is $m_M$, it is attracted to the surface of the planet with the force $m_{MgM}$, which is equal to the force of pressure of the atmosphere i.e. the pressure on the surface o f Mars is $PM=mMgM/\left(4\pi R_M^2\right)$.Similarity, for the corresponding parameters on the Earth, we obtain $PE=m_{EgE}/\left(4\pi R_E^2\right)$. The ratio of the masses of the Martian and the Earth's atmospheres is
$\frac{mM}{mE}=\frac{PM4\pi R_{MgE}^2}{PE.4\pi R_{EgM}^2}$
Considering that $M_M=\left(4/3\right)\pi R_{M\rho M}^3$ (a similar expression can be obtained for the earth) and substituting the given quantities we get
$\frac{m_M}{m_E}=\frac{PM}{PE}\frac{R_M}{R_E}\frac{\rho E}{\rho M}\simeq 3.4\times {10}^{-3}$
It should be noted that we assumed in fact that the atmosphere is near the surface of a planet. This is really so since the height of the atmosphere are much smaller than the radius of a planet (e.g.at an altitude of $10km$ above the surface of the Earth it is possible to breathe and the radius of the Earth is $R_E\simeq 6400km$)