I am planning a series of science fiction novels that take place on an imaginary binary planet system. Both planets have a lower surface gravity than the Earth and one has slightly more mass than the other.

If average temperature is the same as earth or cooler and the radius of the larger planet is allowed to vary, what is the minimum amount of gravity required for the larger planet to sustain a 50:50 Nitrogen/Oxygen atmosphere with a total pressure of 100 mmHg at 2000 m above sea level?

What law or theorem would I need to solve this problem.

Extra stuff about the planets, my motivations and other things I'm trying to figure out that you don't have to read if you don't want to:

I want the larger planet to have an atmosphere that is thinner and whose pressure varies to a greater degree by altitude than the atmosphere of the Earth, but is still breathable by humans at sea level. Basically I want Denver to feel more like Mount Everest and Mount Everest should be pretty much vacuum. Mount Everest has an oxygen partial pressure of about 43 mmHg, and Denver is at an altitude of 1.6 km. Let's say we want 50 mmHg of oxygen at 2km. I figure if oxygen concentration is higher on the planet then on Earth, then atmosphere can be thinner and overall pressure can be lower. I picked a 50:50 mix not knowing whether it is really feasible, but if it is then pressure could be 100 mmHg at 2km. I don't know how to calculate how big the radius and mass of the planet has to be to satisfy these or what the pressure would be at sea level and at what altitude pressure will be 0.

The smaller planet should have an atmosphere that is too thin for humans but could conceivably host sentient life.

Both planets should have lower gravity than the earth. Each planets has a different species of sentient beings and I would like them to be able to lift stuff into space for much cheaper than here on earth. Basically by the time they have something like a Saturn V rocket, there should already be some trade in manufactured goods between the two.

Both planets have awesome magnetic fields that are way better than Earth's at doing the stuff that magnetic fields do. I also would like the binary planets to be in a binary star system and have one or two small moons each, but this is probably pushing it. Even one small ice moon would serve as a handy plot device but I don't want to make it too improbable and I do want to get as much of the Physics right as I can and figuring out the orbits, climate map and the seasons will be difficult enough as it is with two planets, especially considering how little I know.

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    $\begingroup$ "Both planets have awesome magnetic fields that are way better than Earth's at doing the stuff that magnetic fields do." Just out of curiosity, what do you mean by this? $\endgroup$ Commented Oct 7, 2013 at 14:10

4 Answers 4


Some of your requirements are conflicting. The surface pressure of any atmosphere is the weight of the column of air (whatever mixture of gasses your atmosphere is made up of) of unit area extending upwards to the end of the atmosphere.

The surface pressure is therefore a function of the gravity and how much total gas is in the atmosphere. For example, Earth and Venus have about the same gravity, but the surface pressure of Venus is about 90x more because Venus has about 90x more stuff in its atmosphere.

You can generally assume that any planetary atmospheric layer will be thin enough that gravity is constant over that layer to a reasonable approximation. Therefore, for any chosen gravity, how fast the pressure decreases only has to do with the density of the gas. The denser it is, the faster you have a smaller weight of air column above you as you go up in altitude. To have a planet with 1 g surface gravity, for example, that has a more rapid falloff in pressure than earth, you need its atmosphere to be made of a more dense gas. The denser the gas, the thinner the atmospheric layer for the same gravity and surface pressure.

This is why the larger planet in your description just can't be. You say it has less gravity than Earth, a thinner atmosphere (presumably you mean lower surface pressure), but yet you want the pressure to fall off more rapidly than on Earth. This simply can't happen. Picture one inch square on the ground and the colum of air above it to the end of the atmosphere. With less gravity it will be less squished, so the column will be higher. This in turn makes the pressure fall off more slowly with altitude. If you also say it is a nitrogen-oxygen mixture, then it's density is pretty much defined over a narrow range (N2 has a molecular weight of 28 and O2 of 32).

  • $\begingroup$ @Kaan There are probably plenty of ways to make space travel between the two planets plausible without appealing to physics or geology at all. E.g. socioeconomic or religious factors. Or alien categories/motives for which no word even currently exists in the English language. Aliens don't have to make sense by human standards, after all. $\endgroup$ Commented Oct 7, 2013 at 14:14

This is to answer the comment ``at least name the formula''.

What is probably the simplest model of an atmosphere goes like this: The density $\rho$, temperature $T$, and pressure $p$ depend only on altitude $y$. The weight $g$ of a unit mass does not depend on $y$ (atmospheres are not thick compared to the planet radius, say). A balance of vertical forces on a horizontal slab of area $A$ between $y$ and $y+\Delta y$ gives $-p(y+\Delta y)A+p(y)A-\rho g\Delta y A = 0$, which says that the pressure just balances the weight. Divide by $\Delta y$ and take the limit as $\Delta y$ tends to 0 to get the basic formula $p'(y) = -\rho g$. Where you go from here depends on what else you assume about the atmosphere.

Case 1) constant density: $p(y) = p(0)-\rho g y$.

Case 2) ideal gas with constant temperature: then $R\rho'(y)T = -\rho g$, and you get density decreasing exponentially with altitude, and $p(y) = p(0)e^{-gy/RT}$.

Case 3) ideal gas with temperature decreasing linearly with $y$ (as in our troposphere): $T(y) = T_0-ky$, then you have a harder differential equation to solve, and find $p(y) = p(0)(1-\frac{ky}{T_0})^{2-g/(kR)}$.


For such a small planet (such as Mars) that is sufficiently warm for liquid water (inside the frost line) many of the less-massive gases (such as hydrogen, nitrogen, oxygen, etc.) seep out of the atmosphere into space. To prevent this, the planet needs to have sufficient surface gravitational field strength and sufficient radius such that the atmosphere (thickness) can extend many miles above the surface.

Further, to retain oxygen in large quantities, the planet must be far enough from its star such that water remains in its liquid form, absorbs the $CO_2$, allows the sea floor to leach the $C$ out of the $CO_2$, leaving oxygen as a by-product. Otherwise, like Venus, $CO_2$ (the most prevalent heavy gas in planet formation) will make up most of the atmosphere, causing runaway Green-house effect (very high surface temperatures). These high temperatures cause water to disassociate. The extreme amounts of $CO_2$ buoy the hydrogen and oxygen sufficiently high in the atmosphere that they escape.

Thus, the planet size (radius) is definitely a concern, though the field strength at the surface is not inappropriate. The planet also needs to be at an average temperature considerably lower than the boiling point of water at the planet surface.

  • $\begingroup$ This wasn't really the kind of answer I was expecting but has still helped me a lot. I would like to up vote your answer but I need a few more reputation points before I can do that. It would be great if you could at least name the formula(s) I can use to calculate the pressure as I have indicated in the revised question. $\endgroup$ Commented Oct 5, 2013 at 10:38

You can use the centrifugal force to counteract the gravitational one. Speed up the rotation of the planet to lower the pressure at the equator. The maximum pressure is at the poles (On Earth it is so).

It is strange that I could not find a reference to this effect.

  • $\begingroup$ Jupiter rotates very fast. The gravitational acceleration at the equator is significantly lower than at the poles: Acceleration (eq., 1 bar) (m/s2) 23.12 Acceleration (pole, 1 bar) (m/s2) 27.01. This does deform the planet, of course, so we specify it at the radii of constant atmospheric pressure. $\endgroup$ Commented Jan 28 at 12:55

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