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The altitude record for an unmanned balloon is 53.0 kilometers.

wikipedia

If we use this Earth based record for an estimate of the maximum possible height of a balloon started from Mars surface, what height would result?

What is the pressure at 53 km above Earth and what is the height above Mars with the same pressure?

Of course a balloon with a volume of 60 thousand cubic metres would be very difficult to realize on Mars.

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  • $\begingroup$ My old 1962 Standard Atmosphere book gives the pressure at 53,000 meters as .55 mb (.00054 x sea level pressure). $\endgroup$ – Organic Marble Mar 20 at 17:14
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    $\begingroup$ @OrganicMarble mb is mbar I think. So we get 0.55 mbar or hPa for 53 km. The surface pressure of Mars is 0.636 (0.4–0.87) kPa or 6.36 (4 to 8.7) hPa. Now we need a standard pressure to height formula for Mars. $\endgroup$ – Uwe Mar 20 at 17:37
  • $\begingroup$ Here is a Mars atmosphere model for metric units. $\endgroup$ – Uwe Mar 20 at 17:58
  • $\begingroup$ @uhoh I think the maximum height of balloon without rupture and the height record of a (non ruptured) balloon are two different things. If you fill the balloon at ground with very little gas, it could be lifted over its maximum height without rupture. $\endgroup$ – Uwe Mar 20 at 22:22
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    $\begingroup$ 53.7 km actually. $\endgroup$ – Mark Adler Mar 21 at 3:47
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In September of 2013, JAXA launched an $80,\!000\,\mathrm{m^3}$ zero-pressure Helium balloon from Hokkaido1. It reached a float altitude of $53.7\,\mathrm{km}$.

From the ICAO 1993 Standard Atmosphere2, the density at that altitude is $6.62\times 10^{-4}\,\mathrm{kg/m^3}$.

That same zero-pressure balloon would float at the same density at Mars. The altitude equivalent to that density will vary over the Martian year, since the mass of the Martian atmosphere varies $\approx\!25\%$ over that time! Since I have MER atmosphere models handy, I find that that density was at an altitude of $\bf 35.7\,\mathrm{\bf km}$, at an LS of $328^{\Large\circ}\!$, about two-thirds of the way into Southern Summer (the time of the Spirit landing).

Note that what matters here is density, not pressure. A zero-pressure balloon is an accurate densitometer at float altitude. Also, the difference in the compositions of the atmospheres, in particular the average molar mass, is already accounted for in the density.

plot of balloon altitude as a function of time, showing level flight at 53.7 km for 12 minutes

There is one other key difference that could affect the maximum altitude. That is temperature. At float altitude in Earth's atmosphere, the temperature was $264\,\mathrm{K}$. At the equivalent float altitude in Mars' atmosphere, the temperature would be about $178\,\mathrm{K}$. This would have no effect on the float altitude. That is, if the balloon works. However the fact that the Martian atmosphere is much colder than Earth's could change the behavior of the $2.8\,\mathrm{\mu m}$ thin polyethylene envelope. It might become brittle enough that it would not survive launch or ascent, and break when encountering aerodynamic forces, such as gusts of wind. Then the maximum altitude could be much lower, possibly not even making it off the surface of Mars.

By the way, $80,\!000\,\mathrm{m^3}$ may sound like a large balloon, but it's actually pretty small. The mass of the entire system, balloon, helium, gondola, parachute, and payload, was $41.1\,\mathrm{kg}$. Typical scientific high-altitude balloons are measured in the tens of millions of cubic meters, with thousands of kilograms of payload.

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  • $\begingroup$ excellent! They call it a "zero-pressure balloon" but it's not a vacuum, I can't figure this out. $\endgroup$ – uhoh Mar 22 at 4:28
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    $\begingroup$ @uhoh The "zero-pressure" part refers to zero pressure differential between inside and outside the envelope. This is something of a misnomer, because ultimately the buoyancy of a balloon results from a difference in pressure between the interior and exterior of the balloon, on the parts that have an upward component to the local normal vector (to the balloon surface). This pressure differential stems from the different densities of the internal and external gases. In zero-pressure balloons, the part of the balloon hanging down from the "bubble" is at the same pressure as the external air... $\endgroup$ – Tom Spilker Mar 22 at 4:43
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    $\begingroup$ @uhoh If you look at the launch of a zero-pressure balloon, you see a lot of seemingly useless envelope plastic hanging from a "bubble" of helium (or whatever gas they're using) at the very top. As the balloon rises and the external pressure decreases that bubble expands. If the balloon design has insufficient envelope volume, that bubble will eventually take up the entire maximum volume of the envelope, and further ascent causes the envelope to burst. $\endgroup$ – Tom Spilker Mar 22 at 4:52
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    $\begingroup$ @uhoh It should probably be called "zero gauge pressure", where the gauge pressure is the difference between the inside and outside pressure. The "gauge" is, for example, your tire gauge, which measures and reads out that difference in pressure — not the absolute pressure inside the tire. The approximately zero gauge pressure is important, in order to apply the least possible stress to the extremely thin and fragile balloon material. $\endgroup$ – Mark Adler Mar 22 at 5:06
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    $\begingroup$ Nope. Gravitational acceleration on the displaced air and on the balloon system are the same. $\endgroup$ – Mark Adler Mar 22 at 14:14
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What is the pressure at 53 km above Earth and what is the height above Mars with the same pressure?

According to Wolfram Alpha, the pressure on Earth at 53km is 55 Pascals compared to 100,000 Pa at sea level. Quite the difference.

Wolfram Alpha gives an atmospheric pressure on the surface of Mars as 620 to 1100 Pa. Presumably due to wide seasonal variations in temperature and the problem of not having a "sea level" to agree upon.

We can use the NASA supplied equations to figure out when the Martian atmosphere hits 55 Pa. NASA uses 699 Pa as pressure at the surface.

$$p = 699 Pa/m * e^{-0.00009 * h}$$

We have p, the pressure at 55 Pa. We need to solve for height, h.

$$55 Pa = 699 Pa/m * e^{-0.00009 * h}$$ $$0.0787 m = e^{-0.00009 * h}$$ $$\ln(0.0787) m = -0.00009 * h$$ $$-2.542 m = -0.00009 * h$$ $$28,245 m = h$$

28,245 m or a bit above 28km. That's a lot higher than I expected!

We can check our work by plugging the height into the original equation. We should get 55 Pa.

$$p = 699 Pa/m * e^{-0.00009 * h}$$ $$p = 699 Pa/m * e^{-0.00009 * 28,245 m}$$ $$p = 699 Pa/m * e^{-2.54205 m}$$ $$p = 699 Pa/m * 0.0787049 m$$ $$p = 55 Pa$$

Whether this means that's the highest we could fly a balloon on Mars, I can't say for sure.

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  • $\begingroup$ Nice math and result. But what about the gravity of Mars (smaller than that of Earth) and the carbon dioxide atmosphere (about 60 % more dense than air)? $\endgroup$ – Uwe Mar 20 at 21:15
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    $\begingroup$ @Uwe NASA supplies a density equation, r = p / [.1921 * (T + 273.1)], which would take into account the density of carbon dioxide. We have the pressure. The temperature at that altitude can be calculated with T = -23.4 - 0.00222 * h giving -86.1C (and I'd imagine varies with latitude and season). I'll leave the density calculation as an exercise. You'd have to ask a balloonist whether a balloon will fly in these conditions. $\endgroup$ – Schwern Mar 20 at 21:19
  • $\begingroup$ The pressure does take into account the density of carbon dioxide, but at the given pressure, the density inside and outside the balloon is essential. There is a difference between a balloon filled with hydrogen or helium in an atmosphere of air or mainly carbon dioxide at the same low pressure. $\endgroup$ – Uwe Mar 20 at 21:49
  • $\begingroup$ This also seems to answer How high could a weather balloon be used on Mars without rupturing?. $\endgroup$ – uhoh Mar 20 at 22:08
  • $\begingroup$ I got the same result of 28 km without density correction. But with a correction for 60 % more density of CO2 instead of air, I get 33 km. As long as air, carbon dioxide and hydrogen or helium may be treated as an ideal gas and temperature inside and outside the balloon is equal, the effect of temperature on density of these gases is the same. $\endgroup$ – Uwe Mar 20 at 22:29

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