# Could a helium balloon on Mars and on Triton float at air pressures lower than it could on Earth due to the bodies' low gravities?

The highest altitude ever reached by a balloon above the Earth's surface is about 33 mi (53 km), unmanned above Japan. The 2nd-highest one reached 51.8 km above California. Both reached the lower mesosphere. The Earth's gravity in the mesosphere is about 0.98 g. I wonder how high such balloons could go on Mars where, due to Mars' low gravity, they might get to altitudes of even lower air pressure than on Earth (Mars' upper atmosphere gravity is about 0.37 g).

As for Neptune's moon Triton, its surface gravity is very weak at 0.08 g. Triton's surface air pressure is around 1.4 Pa which is a lower pressure than the one at 33 mi / 53 km above the Earth's surface (Triton's one corresponds to the air pressure about 48 mi / 77.5 km above the Earth, still within the mesosphere) but due to Triton's very low gravity a helium balloon might rise to a certain altitude despite the thin atmosphere.

This question has answers concerning fixed-wing aircraft only and while it also asks for airships it doesn't do for helium balloons.

Wikipedia: Flight altitude record

• See the related question maximum height for a mars balloon? – Uwe Jun 18 '20 at 8:39
• While the link above and another duplicate question of it are helpful, they don't seem to consider the probability that a balloon may go higher to even lower air pressure due to Mars' low gravity. – LoveForChrist Jun 18 '20 at 9:30
• Neutral point for buoyancy is determined by density. The only way you can go higher is to use less helium in the same volume, or make the container lighter. – John Dvorak Jun 18 '20 at 10:04
• @LoveForChrist The balloon stops rising when the average densities are equal i.e. when the weights of the balloon and an identical volume of atmosphere are equal. It doesn't matter what the actual weights are. I think that a careful, thoughtful read of both answers there will show that the difference in gravity between the two planets are accounted for there. – uhoh Jun 18 '20 at 10:08
• @uhoh Thank you. I'd appreciate if you or someone post it as an answer perhaps. – LoveForChrist Jun 18 '20 at 11:06

No, it would float lower if anything. To see this think about the forces on the balloon:

• the acceleration due to gravity is $$g$$ and I assume this is constant (the planet is large, the balloon isn't getting very far up: this is a good assumption);
• the density of the gas inside the balloon is $$\rho_H$$, the density of the atmosphere is $$\rho_A$$.

If the 'radius' of the balloon is $$r$$ – this really means some characteristic linear size – then:

• the volume of the balloon is $$k_1 r^3$$;
• the surface area of the balloon goes like $$r^2$$ and I will assume that the mass of the balloon's structure can be assumed also to go like $$r^2$$ and we'll call it $$k_2 r^2$$.

$$k_1$$ and $$k_2$$ are just fudge factors to be determined, which depend on what the balloon is made of etc etc. Significantly $$k_1 > 0$$ and $$k_2 > 0$$.

So, OK now we can write an expression for the lift force generated by the balloon:

\begin{aligned} L &= g\left((\rho_A - \rho_H)k_1 r^3 - k_2 r^2\right)\\ &= g r^2\left((\rho_A - \rho_H)k_1 r - k_2\right) \end{aligned}

OK so what can we say about this? The first thing is that $$g$$ doesn't matter: it's just a factor outside the whole expression (in fact, $$g$$ does matter because it controls $$P$$, but it controls it the wrong way: other things being equal lower $$g$$ means lower $$P$$ and this is going to hurt us). The second thing is that we can assume $$\rho \sim P$$ where $$P$$ is atmospheric pressure to some reasonably good approximation (this comes from the ideal gas law: $$PV - nRT$$), so let's write $$\rho_H \equiv \rho_{H,0}P$$ & similarly for $$\rho_A$$:

$$L = gr^2 \left((\rho_{A,0} - \rho_{H,0})k_1rP - k_2\right)$$

And look at the expression $$(\rho_{A,0} - \rho_{H,0})k_1rP - k_2$$: in the first term there are a bunch of constants multiplied by $$rP$$ and in the second there is a constant.

This means that the lower $$P$$ is the smaller the lift is, and the more the fixed mass of the balloon starts to matter. In other words as $$P$$ goes down we get to fly less high. And, finally, as I said above, other things being equal, lower $$g$$ means lower $$P$$.

So balloons fly less high in lower gravity (or they need to be bigger so the factor of $$r$$ between the lift from the gas and the mass of the balloon's structure helps you more).

In fact, for hack value, we can go further than this: in the previous expression we had $$P$$, the atmospheric pressure, and $$g$$ the acceleration due to gravity. Well, if the mass of the atmosphere is $$M_A$$, the mass of the planet is $$M_P$$, and the radius of the planet is $$R$$, and the atmosphere is a thin layer (it is for rocky planets), then we get an expression for $$g$$:

$$g = \frac{G M_P}{R^2}$$

and also for $$P$$:

\begin{aligned} P &= g\frac{M_A}{4\pi R^2}\\ &= \frac{G M_P M_A}{4\pi R^4} \end{aligned}

And so the lift expression turns into

$$L = r^2\frac{G M_P}{R^2} \left((\rho_{A,0} - \rho_{H,0}) k_1 r \frac{G M_P M_A}{4\pi R^4} - k_2\right)$$

And I think that this expression is more useful than the previous one: it tells you that

• a bigger balloon helps (we know this);
• a more massive atmosphere helps;
• a more massive planet helps;
• a bigger planet hurts you quite badly;
• a larger value of $$G$$ helps.

This last thing I found extraordinary and I'm quite worried it means I've made a mistake.

• G is a constant defined by nature. It is no use arguing a larger value of G would help. – Uwe Jun 20 '20 at 12:11
• @Uwe: obviously you can't vary $G$: but in a universe with a higher $G$ then balloons would be able to fly on lower-mass planets, if I'm right. – tfb Jun 20 '20 at 13:32