# Comparing atmospheric drag in different atmospheres

The book and film The Martian begins with a dust storm on Mars. Is it possible to calculate and compare the effects of a 150mph wind on Mars with winds on Earth? How fast would the wind need to be on Earth to have the same effect as a 150mph storm on Mars

I found the "drag equation" $$D = C_d \rho V^2A/2.$$ If I want to compare Mars with Earth, it seems I need to know the density of the atmosphere. $\mathrm{CO}_2$ has a density of 1.98kg/m3 at standard temperature and pressure, but on Mars it is colder (I estimated night-time temperature at 180K) and lower in pressure (at low elevations I estimated about 0.1 atm or 10000pa) Putting these values into the Gas Laws gives me a density of Mars's atmosphere of about 0.3 kg/m3, which seems quite a lot, compared with 1.225 for air on Earth.

150 mph is about 67ms-1 so with these values, and after cancelling, the drag equation becomes $$1.225\times V^2 = 0.3\times 67^2.$$ When solved that gives a value of the Earth wind speed of 33ms-1 or about 70mph: Hurricane force.

This seems quite high. Is the approach I've taken here correct? In particular is the use of the drag equation reasonable? A comment in A related question suggests that for turbulent flow this formula breaks down. Is the calculation of the density of the Martian atmosphere correct?

• Possible duplicate of Lift and drag coefficients on other planets – honeste_vivere Oct 29 '15 at 0:14
• Note both the velocity dependence and the drag coefficient change as the air gets thinner. – user10851 Oct 30 '15 at 14:06

This suggests that the pressure at the lowest altitudes of Mars is around 0.1 psi, not 0.1 atm, and this shows an average daily temperature range of about 200K-270K. From these two pieces of info, I get:

$\rho = \frac{pM}{R^*T}$

where $p = 1 kPa$, $M = 0.044 \frac{kg}{mol}$, $R = 8.314 \frac{J}{K\cdot mol}$, and $T = 200K$

hence a density of about $2.65\cdot10^{-2}\frac{kg}{m^3}$.

Using your equation for comparing the wind speeds:

$2.65\cdot10^{-2} * 67^2 = 1.225 * V^2$,

which yields $V \approx 9.9 \frac{m}{s}$, so not that fast

EDIT: Wolfram Alpha also claims that $CO_2$ at that temperature and pressure is a supercritical fluid, so it may behave differently. That being said, this is a worst case scenario of temperature and pressure for Mars, along with some of the highest speeds.