# Is the wind's intensity on Mars similar to Earth?

I've read that in Mars' poles, the winds can be as fast as 400 km/h, when the poles are exposed to sunlight because the frozen $CO_2$ sublimes. I know that the Martian atmosphere is much thinner than Earth's atmosphere.

So, by knowing the wind speeds on Mars, is there any way to get an idea of its intensity, or in other words, the intensity of a wind of x speed in Mars, to which speed of wind of Earth is comparable, for them to have the same intensity?

• Related questions here: space.stackexchange.com/questions/9301/… and space.stackexchange.com/questions/2621/… The first link has math where the wind-force can be calculated. – userLTK Nov 24 '16 at 22:29
• ok, so the pressure of the wind it would be 61,25 times lower? nice answer – Pablo Nov 24 '16 at 22:46
• do you want to post the answer here so I mark it as accepted? – Pablo Nov 24 '16 at 23:42
• I think your math is right at least, that's what I get too, but as for an answer, I didn't want to post or copy someone else's answer as my own. – userLTK Nov 25 '16 at 2:38
• @com.prehensible The atmospheric pressure on top of Olympus Mons is 0.0007 x the normal pressure at sea level on Earth, or 0.7 millibar. For comparison, a vacuum pump that you could buy online for 125 USD makes 0.1 millibar, only 7 times better; a pump that costs 50 USD makes 0.2 millibar, or 3.5 times better. Colloquially, I would describe the pressure on top of Olympus Mons as "pretty lousy vacuum". Seems like there's room for a lot of wind speed there before it really becomes threatening. – Florin Andrei Dec 14 '17 at 0:28

Credit to this question for inspiration, though my calculation methods differ.

The dynamic pressure equation is $$q=0.5\rho v^2$$ where $$q$$ is the pressure, $$\rho$$ is the atmospheric density, and $$v$$ is the wind speed. If we want to know what wind speeds give us equivalent pressures on Earth and Mars, we simply generate dynamic pressure equations for each of them: $$q=0.5\rho_e v_e^2$$ and $$q=0.5\rho_m v_m^2$$, set them equal $$q=0.5\rho_e v_e^2=0.5\rho_m v_m^2$$, and solve for $$v_e$$ to get $$v_e=\sqrt{\frac{\rho_m}{\rho_e}}v_m$$ where $$\rho_m=0.020 \space kg/m^3$$ is the atmospheric density for Mars, $$\rho_e=1.225 \space kg/m^3$$ is the atmospheric density on Earth, $$v_m$$ is the wind speed on Mars, and $$v_e$$ is the equivalent wind speed on Earth.

With a velocity ratio of about 7.826 we can plug in a few values for wind speed in kilometers per hour for Mars to get:

v_mars    v_earth equivalent
10           1.28
50           6.39
100          12.8
200          25.6
400          51.1


These could be kph, or in fact, any units of velocity. screeenshot

and here's what hat looks like in a plot:

So the 400 kph gust on Mars only has equivalent pressure of a 51 kph gust here on Earth