A while back I saw this video talking about the habitability of double planets and Rocheworlds. I haven't seen any questions about the latter case here, so I decided to take a swing at it.

For some background, a Rocheworld is a double planet system that is so close together that the two planets share an atmosphere and have started to merge together. This would eventually result in the two planets violently becoming one large planet.

Hypothetically, it could be possible to travel from one of the planet components to the other over the region they overlap. I think that since these are two planets overlapping, this could lead to some intersting gravitational effects that could impede travel between the two sides.

How would a person feel gravity as they tried to walk from one part of a Rocheworld to the other?

(As in, would they feel a strong tug from either planet as they walked the overlapping region, or would it feel unanomalous?)

  • $\begingroup$ It's always awesome when someone mentions Isaac Arthur. $\endgroup$
    – GoingFTL
    Oct 12, 2017 at 23:48

1 Answer 1


First off, I should point out that, in order for this kind of arrangement to be even remotely stable, the planets will have to be be tidally locked, meaning that they do not rotate in relation to each other. No matter where you are on either planet, the other will always occupy the same place in the sky (if the other is above the horizon, that is). The planets will also be orbiting each other very quickly- on the order of hours, most likely- in order to keep from crashing into each other. This means that "days" on these planets will also be very short- again, probably only a few hours long.

Walking around on such a double planet would feel much the same as walking around on any other kind of planet, at least if you confine yourself to a reasonably small area. By the time this arrangement becomes stable, the planets will have reshaped themselves to be fairly oblong, with their "ends" pointed at each other. If you ignore mountains and continents and such, the ground will seem to be flat everywhere, in exactly the same way that the Earth's surface seems flat if you're standing on it, even though it's actually a sphere. In more technical terms, the surface of each planet will be at (roughly) constant gravitational potential.

Gravity will be noticeably weaker at the "ends" of the planets than around their middles, due to the tidal forces each exerts on the other. You can think of this as follows:

  • Where the other planet is directly overhead, gravity will seem weaker because the other planet is pulling you away from the surface. Not enough to lift you off the surface, but you'll definitely feel lighter.
  • On the opposite ends of the planets, gravity will seem weaker because the planets are spinning around each other and the centrifugal force is trying to sling you away. This isn't strictly accurate- it's more like the other planet is pulling the ground away from your feet faster than it's pulling on you, due to the inverse square law of gravity- but it gets you the right idea.
  • Around the middle of the planets, gravity will feel stronger for two reasons: One, because you're closer to the center of your planet; and two, because the other planet is pulling you toward the ground as well.

I should point out that this sort of arrangement, where two planets orbit each other close enough that their atmospheres touch, may not even be remotely possible. If a moon orbits too close to its planet, the tidal forces will rip it apart. On the side closer to the planet, the material at the surface of the moon will be lifted away, because the planet's gravitational attraction is stronger than the rest of the moon's; and on the opposite side, material will be slung out into space by the centrifugal force. The point where this happens is called the moon's Roche limit, and I have no idea how it applies if the "planet" and "moon" are the same size.

I also don't know what impact this arrangement would have on the planets' atmospheres. They'd certainly be much thicker between the planets and at the planets' "ends" than around the planets' middles (although the air pressure at the surface of each planet should be pretty much the same everywhere), but for all I know, the centrifugal force could sling all of the atmospheres out into space. You'd also need to watch out for that.

Update with Math

The formula for orbital period when the masses of both planets need to be accounted for is as follows: $$ T = 2\pi \sqrt{a^3 \over {G(M_1 + M_2)}} $$ where $a$ is the distance between the two planets and $M_1$ and $M_2$ are their masses.

For two spherical Earths orbiting just close enough that their atmospheres touch, this gives an orbital period of two hours and 49 minutes.

If I fudge the masses in an attempt to emulate the planets being ellipsoids with a 2:1 aspect ratio and put 0.455 Earth radii between them (see below), I get an orbital period of 7 hours, 40 minutes. I think this is about the shortest orbital period you'll be able to get in a system like this, if the system is going to be stable.

How did I get those numbers? The 2:1 aspect ratio and the 0.455 radii? Well.

The Roche limit page details two different Roche limits: one for a rigid satellite, and one for a fluid satellite. The rigid limit is the orbital radius at which loosely-bound material on the satellite's surface (or planet's surface, in your case) will be lifted off and ejected into space. Soil, water, and air all count as "loosely-bound", as there's nothing beyond the gravity of the planet holding them down, so that's something to keep in mind.

The fluid Roche limit accounts for how the tidal forces will distort the satellite and make it more vulnerable to being torn apart. It serves as a good upper bound on how close two bodies can orbit with no risk of being torn apart, although, in practice, many moons in our Solar System orbit well within their fluid Roche limits. They are, in part, held together by tensile forces. They're made of ice or rock, not water. However, I'm fairly sure that none of those moons have atmospheres or any of the fluid-like inner workings that allow Earth's plate tectonics and other geological activity to function. If you want Earth-like planets, you'll need to adhere to this limit more strictly.

The formula for a fluid Roche limit on that Wikipedia page that takes the most things into account (and also happens to give the largest upper bound for where the planets will begin to break apart) is this:

$$ d = 2.455R \root 3 \of {\rho_M \over \rho_m} $$

where $R$ is the long-axis radius of one of the planets, and $\rho_M$ and $\rho_m$ are their densities. I'm not certain which planet $R$ is supposed to be the radius of, but if the Rocheworld planets are the same size, it shouldn't matter.

This tells us that the distance between the planets' centers of mass can be no less than 2.455 times their long-axis radii without risking the planets losing their atmospheres or breaking up entirely. This means that, if the planets are in a stable orbit, there will be at least 0.455 long-axis radii between them. If they're roughly Earth-sized, that'll be about 3000 kilometers. There's not going to be any appreciable atmosphere between them at all, much less enough to fly a plane from one to the other.

It is also worth pointing out that a fluid satellite (or an Earth-like, geologically active Rocheworld) under these conditions on the verge of breaking up will have a 1:1.95 ratio between its long and short axes. If the tidal forces stretch the planet out any more, they will exceed the gravitational forces holding the planet together, and cause it to disintegrate. So your planets should not be any more elongated than that. Also note that if the planets are elongated that much, there will be almost no gravity at their ends. You may need to keep them a bit farther apart and a bit less oblong, just to keep their atmospheres in check.

  • $\begingroup$ Nice in-depth answer! $\endgroup$ Oct 13, 2017 at 3:14
  • 1
    $\begingroup$ The arrangement works as the Roche limit varies based on the mass and density of the objects--two equal-sized, equal-density objects would only be in trouble if they were touching. My understanding is that a shared atmosphere would be in trouble, though--the plane between the two worlds isn't in either's gravitational control and thus the atmosphere would bleed out around the point where the atmospheres touch. $\endgroup$ Oct 15, 2017 at 5:59
  • $\begingroup$ @LorenPechtel They would be in danger before they were touching. The rigid body calculation with equal density of the two objects is 1.26 r. en.wikipedia.org/wiki/Roche_limit#Rigid-satellite_calculation Applying that to a two body calculation gets tricky but I think it's likely that the trouble area would be well before touching. That 1.26 doesn't factor in rotation, which would be fairly rapid or tidal stretching, which would be significant. I think you'd want a bare minimum of a full radius between the two objects, probably a bit more. $\endgroup$
    – userLTK
    Oct 15, 2017 at 6:13
  • $\begingroup$ Small point, but when you say the objects would orbit each other in the order of minutes, that's not true. An object at Earth's surface would have an orbital period of 1.41 hours. The space shuttle is a few minutes longer than that being a couple hundred miles up. That's about the maximum period of rotation velocity cause that's where things begin to fly off the planet. With some distance between the two planets, you're looking at least a couple hours to complete one orbit around each other. Minutes wouldn't be possible unless your planets were super dense like white dwarfs. $\endgroup$
    – userLTK
    Oct 15, 2017 at 6:42
  • $\begingroup$ @userLTK Good points; I hadn't bothered to go through with the calculations. I'll have to do that sometime. An hour and a half, like the ISS, is probably a good estimate for the orbital period. While the other planet does orbit much farther away than the ISS, increasing the orbital period, it also has much more mass, which will decrease the orbital period. I'm not sure offhand to what extent those two effects will cancel out, but it'll still most likely be in the range of one hour to a few hours. $\endgroup$ Oct 16, 2017 at 21:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .