<p>Credit to this <a href="https://space.stackexchange.com/questions/9301/could-you-feel-the-wind-on-mars?noredirect=1&amp;lq=1">question</a> for inspiration, though my calculation methods differ.</p>
<p>The dynamic pressure equation is <span class="math-container">$q=0.5\rho v^2$</span> where <span class="math-container">$q$</span> is the pressure, <span class="math-container">$\rho$</span> is the atmospheric density, and <span class="math-container">$v$</span> 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: <span class="math-container">$q=0.5\rho_e v_e^2$</span> and <span class="math-container">$q=0.5\rho_m v_m^2$</span>, set them equal <span class="math-container">$q=0.5\rho_e v_e^2=0.5\rho_m v_m^2$</span>, and solve for <span class="math-container">$v_e$</span> to get <span class="math-container">$$v_e=\sqrt{\frac{\rho_m}{\rho_e}}v_m$$</span> where <span class="math-container">$\rho_m=0.020 \space kg/m^3$</span> is the atmospheric density for Mars, <span class="math-container">$\rho_e=1.225 \space kg/m^3$</span> is the atmospheric density on Earth, <span class="math-container">$v_m$</span> is the wind speed on Mars, and <span class="math-container">$v_e$</span> is the equivalent wind speed on Earth.</p>
<p>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:</p>
<pre><code>v_mars    v_earth equivalent
   10           1.28
   50           6.39
  100          12.8
  200          25.6
  400          51.1
</code></pre>
<p>These could be kph, or in fact, any units of velocity. <a href="https://i.sstatic.net/aaOBA.png" rel="nofollow noreferrer">screeenshot</a></p>
<p>and here's what hat looks like in a plot:</p>
<p><a href="https://i.sstatic.net/CAAHR.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/CAAHR.png" alt="enter image description here" /></a></p>
<p>So the 400 kph gust on Mars only has equivalent pressure of a 51 kph gust here on Earth</p>

CC BY-SA 4.0
2020-12-01T21:58:02.497Z
2020-12-01T23:20:22.220Z
2020-12-01T23:20:22.220Z
Connor Garcia
<p>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.</p>

<p>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?</p>

CC BY-SA 3.0
2016-11-24T22:29:15.453Z
Related questions here:   http://space.stackexchange.com/questions/9301/could-you-feel-the-wind-on-mars?noredirect=1&lq=1  and  http://space.stackexchange.com/questions/2621/what-would-it-feel-like-to-be-in-a-martian-dust-storm  The first link has math where the wind-force can be calculated.
userLTK
CC BY-SA 3.0
2016-11-24T22:46:01.870Z
ok, so the pressure of the wind it would be 61,25 times lower? nice answer
Pablo
CC BY-SA 3.0
2016-11-24T23:42:59.780Z
do you want to post the answer here so I mark it as accepted?
Pablo
CC BY-SA 3.0
2016-11-24T23:49:14.763Z
BTW, if I calculated correctly it would be the equivalent of 51,11 kms/h wind in Earth, a strong wind, but not in the level of a tornado
Pablo
CC BY-SA 3.0
2016-11-25T02:38:56.117Z
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
CC BY-SA 3.0
2016-12-22T06:03:10.653Z
they are the same within an order of magnitude, the summit of Everest reached 280km/h in 2004. Olympus Mons is 21km high, so the speed of wind there may be much higher, although it's not as forceful.
bandybabboon
CC BY-SA 3.0
2017-12-14T00:28:11.243Z
@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
CC BY-SA 3.0
2017-12-14T16:56:58.860Z
@Florin Andrei you can say, 10 meters of water above a hose or a tap is 1 bar, so you can imagine holding your thumb on a hose with 10 meters of uplift, it's very easy. If you had to use your entire body to hold back a large pipe with 10 meters of pressure, it would be a lot difficult!!! so 0.7 millibar is like trying to hold back 7cm of water bag covering all your body. If that wind was moving past you at 100 / 1000kph, I would be kindof strong pressure? 10m tap = 1 bar, I know because we have a syphon of one.
bandybabboon
CC BY-SA 4.0
2020-11-02T08:59:56.323Z
You can find an answer to your question here: https://space.stackexchange.com/questions/12237/are-martian-winds-as-strong-as-portrayed-in-the-martian
Stefano Freiesleben Sias
<planetary-atmosphere><mars>
Is the wind's intensity on Mars similar to Earth?
Pablo
# Is the wind's intensity on Mars similar to Earth?

(Asked by: username_0 on 2016-11-23)

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?

**Tags:** planetary-atmosphere, mars
> **Comments:**
> **username_1**: Related questions here:   http://space.stackexchange.com/questions/9301/could-you-feel-the-wind-on-mars?noredirect=1&lq=1  and  http://space.stackexchange.com/questions/2621/what-would-it-feel-like-to-be-in-a-martian-dust-storm  The first link has math where the wind-force can be calculated.  
> **username_0**: ok, so the pressure of the wind it would be 61,25 times lower? nice answer  
> **username_0**: do you want to post the answer here so I mark it as accepted?  
> **username_0**: BTW, if I calculated correctly it would be the equivalent of 51,11 kms/h wind in Earth, a strong wind, but not in the level of a tornado  
> **username_1**: 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.  
> **username_2**: they are the same within an order of magnitude, the summit of Everest reached 280km/h in 2004. Olympus Mons is 21km high, so the speed of wind there may be much higher, although it's not as forceful.  
> **username_3**: @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.  
> **username_2**: @Florin Andrei you can say, 10 meters of water above a hose or a tap is 1 bar, so you can imagine holding your thumb on a hose with 10 meters of uplift, it's very easy. If you had to use your entire body to hold back a large pipe with 10 meters of pressure, it would be a lot difficult!!! so 0.7 millibar is like trying to hold back 7cm of water bag covering all your body. If that wind was moving past you at 100 / 1000kph, I would be kindof strong pressure? 10m tap = 1 bar, I know because we have a syphon of one.  
> **username_4**: You can find an answer to your question here: https://space.stackexchange.com/questions/12237/are-martian-winds-as-strong-as-portrayed-in-the-martian  

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## 1 answers

### Answer by username_5 on 2020-12-01. Score: 6 | Accepted answer

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


---

