2
$\begingroup$

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?

$\endgroup$
  • 7
    $\begingroup$ Possible duplicate of Lift and drag coefficients on other planets $\endgroup$ – honeste_vivere Oct 29 '15 at 0:14
  • $\begingroup$ Note both the velocity dependence and the drag coefficient change as the air gets thinner. $\endgroup$ – user10851 Oct 30 '15 at 14:06
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.