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Due to the lack of an atmosphere on the Moon, the Apollo lunar flags had an upper horizontal pole in order to make them fully hoisted, looking like floating in the wind. Now Mars does have an atmosphere but quite a thin one, and I wonder whether Martian winds are strong enough to make a flag on a vertical pole float. Surely, the flag would float when in a strong dust storm but such storms don't happen that often, do they?

In the movie Mission to Mars there's an ongoing strong storm that allegedly is able to cover most of the planet, or something like that (it's a long time I've seen the movie). Is the movie right? There, the Stars and Stripes float like on Earth in the storm, but aside from storms, when it's sunny (and I suppose it's sunny almost all the time on the crew's landing location) would the winds still be strong enough to make the Stars and Stripes float or would NASA/SpaceX play it safe and attach the flag to a horizontal pole too like on the Moon? Are there any specific considerations by NASA or SpaceX?

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Short answer: Yes. Mars is not windy enough to properly wave most flags.

Long answer: In storm conditions, a flag constructed out of a very light material would be able to properly wave.

If we take a standard flag, say 3'x5' that's made of 200g Nylon

$\ell= 1.5$ meters

$h_f = 0.9$ meters

$W = A * 0.2 * g_{Mars}$, $W = 1$ Newton

Going off the calculations presented in this answer, that would give the flag:

$$C_{D \;@\frac {\ell}{h_f}=1.666} \approx 0.10$$

Now, calculating for Mars air density, we can use:

$$\rho = p / (0.1921 *(T + 273.1))$$

and at sub 7000-meter altitudes:

$$T = -31 -0.000998h$$ $$p = 0.699 * e^{-0.00009h}$$

where $\rho$ is in $kg/m^3$, $p$ in $KPa$, $T$ in Celcius, and $h$ in meters. So, at "sea level" where $h=0$ we get:

$$\rho = 0.015_{kg/m^3}$$

Now, maximum wind speeds during storms on Mars are around 60mph so, $v_{max} = 27_{m/s}$ and we can apply the formula for drag on the flag:

$$F_D=C_D(0.5\rho v^2) $$ $$F_{D_{max}}=0.55_N$$

Now, finally applying the formula where $\theta$ is the flag angle (from the horizontal):

$${\theta = \tan^{-1} (\frac {2W\ell}{C_{D}h_{f}\rho v^2})}$$

We get:

$$\theta = 71.9^{\circ}$$

Which means that a standard flag, at surface level during Martian storm conditions deflects around 18 degrees off of the flag pole--which is not very patriotic. If you really want to achieve the "flying flag" look during storm conditions, deflection angles of >80 degrees are possible if you use a super thin material (ballpark 10g/m^2) and on a windy day (10m/s winds) you would get flag deflection angles of around 40 degrees. Still, I suspect using an ultra lightweight material has disadvantages in durability and for photo-op purposes, PR will probably not want to wait for a dust storm to take photos and simply use a supported flag.

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    $\begingroup$ According to Mark Adler's answer to this question Mars winds may be 100 mph fast: space.stackexchange.com/questions/34412/… $\endgroup$ Jun 17, 2020 at 12:26
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    $\begingroup$ You can run the calculation yourself but even a 100mph sustained winds would only deflect the standard flag ~4 degrees and tissue-paper flag wouldn't "fly properly" relationship between flag angle and wind speed is not linear. See in the physics exchange answer I linked to get a feeling of what the curve looks like. $\endgroup$
    – Dragongeek
    Jun 17, 2020 at 13:16

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