# The Roche motel

One of the staples of SF art is images of alien worlds with satellites or planetary twins hanging low and huge in the daylight sky. This blog post brings he trope home by simulating what the Earth’s Moon would look like if it orbited the Earth at the distance of the International Space Station.

The author correctly notes that a Moon that close would play hell with the Earth’s tides. I can’t be the only SF fan who looks at images like that and thinks “But what about Roche’s limit”…in fact I know I’m not because Instapundit linked to it with the line “Calling Mr. Roche! Mr. Roche to the white courtesy phone!”

Roche’s limit is a constraint on how close a primary and satellite can be before the satellite is actually torn apart by tidal forces. The rigid-body version, applying to planets and moons but not rubble piles like comets, is

d = 1.26 * R1 * (d1 / d2)**(1/3)

where R1 is the radius of the primary (larger) body, d1 is its density, and d2 is the secondary’s density (derivation at Wikipedia).

And, in fact, the 254-mile orbit of the ISS is well inside the Roche limit for the Earth-moon system, which is 5932.5 miles.

The question for today is: just how large can your satellite loom in the sky before either your viewpoint planet or the satellite goes kablooie? To put it more precisely, what is the maximum angle a satellite can reasonably subtend?

We need to constrain the conditions a bit more, actually. By holding the mass of the satellite constant but decreasing its density we can push up its angular diameter. It’s not easy, mind you; because volume increases with the cube of radius, it will take roughly an eightfold fall in density to double the diameter.

But the artistically interesting cases are terrestroid rocky worlds or moons orbiting each other and it turns out their densities don’t vary by a lot. Earth’s is 5.51g/cm**3, the Moon’s is 3.43g/cm**3 and Mars’s is 3.93g/cm**3. For comparison the Sun’s density is 1.41 and Jupiter’s is 1.33.

Let’s start by setting the primary and secondary densities equal, then. This is convenient, because in the formula above we actually minimize d (and thus maximize subtended angle) when the density ratio is 1 (if it goes below 1 the primary and secondary switch roles). Our simplest case for minimizing d – two worlds of equal mass/density/radius – is actually pretty plausible.

OK, computation time. By hypothesis our worlds have the same radius R. The minimum distance from a point on the surface of the primary to a point on the surface of the secondary has to be 1.26 R. We must add R to get to the secondary’s center, because the visual angle subtended by the secondary will be that of a disk passing through the center of the secondary at right angle to the line of vision to it [[Edit: This turns out not to be quite right – see the comment thread – but close enough]].

The implied triangle has one corner at the surface of the primary world, another corner at the center of the secondary, and a third on the secondary’s limb subtending the maximum angle. x = 2.26R, y = R. Elementary trigonometry gives us the following formula:

a = 2 * arctan(R / 2.26 * R) = 2 * 0.41 radians = 46 degrees.

That’s actually pretty dramatic – 80 times the size of a full moon. More importantly the human visual field is about 150 degrees; our loomingest possible satellite could take up almost a fifth of it. Looks like those SF illos are fairly plausible after all.

Oh, and why did I title this “The Roche motel”? Because of the destructive effects when a hapless celestial body wanders inside the limit. “Satellites check in…but they don’t check out.”

(Two mistakes in the original posting have been corrected.)

1. ams says:

I thought I originally heard “roche limit” used in the context of a spinning loosely bound asteroid: It’s the limit where the outer edge of the pie-plate shaped asteroid ends up unbound due to centrifugal forces (it’s in its own orbit.) Depends on mass, radius, and spin rate.

1. esr says:

>the limit where the outer edge of the pie-plate shaped asteroid ends up unbound due to centrifugal forces

A special case of the more general Roche effect. You can sort of think of it as centrifugal forces, but it’s actually tidal stress due to the differential in gravitational forces on matter just inside and just outside the Roche limit of the primary.

There are exactly two kinds of people who grok this: astronomers and hard-SF fans. The latter group is probably larger.

2. Edward Cree says:

Uhh, I think you added 1.26 to 0.5 and got 1.31, instead of 1.76…?

Using the latter, 2? = 2 * arctan(0.5 / 1.76) = 0.554 rad = 31.7°. Since the body appears as a disk, the solid angle is 2? (1 – cos ?) = 0.239sr; our visual field (? = 75°) is 4.66sr by the same formula, so the Rochemoon fills about 5% of the sky. (Linearly it stretches a little over a fifth of the way across our view.)

1. Edward Cree says:

And it looks like WordPress ate my thetas :(

2. esr says:

>(Linearly it stretches a little over a fifth of the way across our view.)

You’re right about my arithmetic error but that’s close to the same result I got anyway.

I question the relevance of solid angles here, though. I think most peoples’ sense of “how wide” has to do with linear angles of subtention parallel to the horizon line.

3. Ian Bruene says:

Having a giant moon peering down at you is one thing, but the real fun begins when you realize that such a system must be tidally locked to be stable. Then you pull out the mega-engineering toolkit.

Starting with a single inter-world elevator you can slowly build up a pseudo-solid link between the bodies. Properly engineered you can even use this to circularize the orbits over a very long period of time. In the long run you have a solid cylinder of enclosed space, with planets for end-caps.

Hey, I did say *mega*engineering. Small ain’t part of the deal.

1. darkestkhan says:

As a fun of SFIA (Science and Futurism with Isaac Arthur) you are not thinking big enough… what about moving entire galaxies in the direction desired by us?

4. Peter Donis says:

d = 1.26 * R1 * (d1 / d2)**(1/3)

I don’t think this formula is valid for the case where the two bodies are the same radius and density. The approximation in the Wikipedia article is that the radius r of the satellite is much less than d (which Wikipedia defines as the center-to-center distance between satellite and primary). But if the satellite radius r is equal to R1, the primary radius, and d1 = d2, then we have d = 1.26 R1 = 1.26 r, which doesn’t justify the approximation the Wikipedia article uses.

Since it is evident that d1 = d2 minimizes d, to realistically use the above formula, we would have to figure out a reasonable value for the ratio r / R1 of satellite to primary radius for which the approximation is still valid. Alternatively, we could try to work with the full quartic formula (or expand it out to more orders than just the leading order approximation in the Wikipedia article). My guess is that the latter procedure would be necessary for the question at issue here, since the approximation in the Wikipedia article basically requires r to be very small, and therefore the satellite to subtend a small angle.

5. Jon Brase says:

>And, in fact, the 254-mile orbit of the ISS is well inside the Roche limit for the Earth-moon system, which is 5932.5 miles.

>The minimum distance from a point on the surface of the primary to a point on the surace of the secondary has to be 1.26 R.

Ummmm… I believe the Roche limit is a radius (center to center), whereas the orbital altitude of the ISS is an altitude (surface to surface), and likewise, in the second bit I quoted, it’s the minimum distance between centers, not surfaces, that has to be 1.26 R, which means that two objects of equal mass and density will be in contact before reaching their Roche limit with respect to each other (but that particular formula is valid only for rigid bodies that remain spheres until they break up, and will underestimate the Roche limit in this scenario, though bodies of equal mass and density will still end up in contact with each other without disruption).

When you take into account the fact that a planet will have very little rigidity and basically behave as a fluid droplet, the equipotential surface defining each planet’s surface will become elongated as you move them toward each other, until one or both surfaces come in contact with the L1 point between the two. If they come in contact with L1 at about the same time, you get a contact binary. If one comes into contact with L1 significantly before the other, it will leak mass through L1 until it shrinks to where it no longer contacts L1, until it is totally gone, or until the other body absorbs enough material to grow until it also contacts L1.

See:

https://en.wikipedia.org/wiki/Roche_lobe

https://en.wikipedia.org/wiki/Contact_binary

1. Peter Donis says:

two objects of equal mass and density will be in contact before reaching their Roche limit with respect to each other

This is what I’m getting by using my spreadsheet’s solver to find values of d that fit the non-approximated equation from the Wikipedia page. Setting F_T = F_G using the expressions there without making the small r assumption leads, after some algebra, to the equation

d [ (1 – r/d)^2 / (2 – r/d) ]^1/3 = R ( d1 / d2 )^1/3

where r is the radius of the satellite and R is the radius of the primary. This leads to values of d smaller than R + r for d1 = d2, not only if r = R, but for values of r somewhat smaller than R.

6. Peter Donis says:

a = 2 * arctan(0.5 * R / 1.76 * R)

I’m not sure this formula is right. When I draw the diagram of the right triangle, with angle 1/2 theta at the surface of the primary (because we want one corner of the triangle to be at the center of the satellite), and the right angle at the limb of the satellite (because at that point the line of sight is perpendicular to the radius of the satellite), I get

theta = 2 * arcsin(r / (d – R))

where r is the radius of the satellite, R is the radius of the primary, and d is the center to center distance (per the definition on the Wikipedia page).

7. Geoffrey Tobin says:

How large is Jupiter’s subtended angle from its closest moon?

8. Geoffrey Tobin says:

Trigonometry implies that from Metis, which orbits two radii from the planet’s centre, Jupiter subtends an angle of 60 degrees.

9. Anon says:

Skeptical of the density variable – couldn’t we transcend that by eg building the satellite out of some tensegrous metamaterial or such?

10. Dave says:

OK, for a real challenge, in ESR’s posited system of dual-earth @ ~6000 miles, what happens to the Van Allen belts? How seriously interacting are the bodies magnetic fields? What happens when 1 body undergoes a magnetic field reversal? Aurora?

What’s the orbital decay due to gravitational radiation here?

If both bodies are tidal locked, do the 2 equal body Trojan points become geosyncronous?

How stable are the Lagrange points in this kind of system?

11. d5xtgr says:

Minor reasoning error when you formed your equation:

the visual angle subtended by the secondary will be that of a disk passing through the center of the secondary at right angle to the line of vision to it

At small angles, i.e. lunar orbits approaching infinite radius, this would be correct. But you’re talking about observer-to-moon distances of the same order of magnitude as the moon’s diameter. So you should recognise that the horizon seen by an observer at finite distance from a sphere isn’t an equator – from the ground you clearly can’t see at the same time a point on the Earth and its antipode.

Your sightlines to the lunar horizon form a cone with its vertex at your eye and each ray tangent to the lunar surface – so the lunar radius passing through the point of tangency is orthogonal to the sightline. This means that the math doesn’t change much: you merely change the arctangent to an arcsine; i.e. your 1.76 dimension is the hypotenuse rather than the long leg. I calculated an extra degree and a half by accounting for this, for a total of just over 33.

1. esr says:

>Your sightlines to the lunar horizon form a cone with its vertex at your eye and each ray tangent to the lunar surface – so the lunar radius passing through the point of tangency is orthogonal to the sightline.

Ah, yes. You are right. I should have noticed that.

12. Keith Lynch says:

I’m surprised nobody has yet mentioned Robert L. Forward’s
_Rocheworld_, aka _Flight of the Dragonfly_. The world(s) consist of
two raindrop-shaped planets which almost but not quite touch at their
pointy ends. It’s possible to fly an airplane from one planet to the
other, and the occasional rogue wave splashes water from the wetter
planet to the drier one.

It can last for “only” a few hundred million years, due to tidal
effects of the system’s sun and other planets.

Given who Forward was, I’m sure he got the physics right. I’m also
pretty sure he used numerical simulation. Most of the classical
closed-form orbital mechanics known to Kepler, Newton, Lagrange,
and Roche requires idealizations such as various quantities being
very small compared to various other quantities.

Energy loss by gravitational wave radiation is utterly negligible for
objects of typical planetary masses and velocities. The Earth-Sun

1. esr says:

>I’m surprised nobody has yet mentioned Robert L. Forward’s
_Rocheworld_,

I thought of it, of course, but it didn’t seem directly relevant to the problem I was trying to model.

1. Jon Brase says:

Well, it puts a bit of a sanity check on your result: Most drawings I’ve seen of the shape of Roche Lobes look to have the equipotential surface passing through L1 meeting itself at a 60° to 90° angle, so that should be the maximum possible angle of subtention.

1. esr says:

Yeah, but you don’t get a Roche lobe unless the bodies are fluid at a very close approach. Not the case under consideration here.

2. Troutwaxer says:

There’s also a Robert Sawyer book where the Roche limit is a plot point. One of his Dinosaur books.

13. Jon Brase says:

The bodies may as well be fluid: it’s not my understanding that a rocky surface has the strength to maintain a potential between its highest and lowest points equivalent to more than a few 10s of km under Earth gravity (which is to say a few hundred thousand (m/s)^2).

1. Keith Lynch says:

Right. Every planet is pretty much fluid, by the IAU definition.
The Roche limit doesn’t apply to objects with significant mechanical
strength. The ISS isn’t torn apart despite being well below the
limit, since it’s held together by welds and rivets, not by the mutual
gravitation of its parts.

1. esr says:

>Every planet is pretty much fluid, by the IAU definition.

Now waitasecond. I get it about hydrostatic equilbrium – in fact, by a weird coincidence I actually had a minor role in writing that part of the new IAU definition of “planet” back during the Pluto flap.

The way that happened is that I’m on a mailing list called General Technics; one of the other members is Guy Consolmagno, who was briefly famous as the “Pope’s Astronomer” (he’s a Jesuit who runs the observatory at Castel’ Sant-Angelo in Rome. At the time of the Pluto flap Guy was on the IAU Small Bodies committee – chairing it, I think, though I could be misremembering.

So Guy posts to GT one fine day that there’s a controversy developing over the definition of “planet” for nomenclatural purposes, and that he’s been tasked by IAU to collect some discussion of various proposed elements of a definition with intelligent laypeople. This was a completely appropriate thing to do – GT is so stuffed with scientifically and astronomically literate SF fans that I’d be hard put to think of any way to find a better sample of “intelligent laypeople” if IAU asked me to do it..

In the ensuing discussion I was one of the earliest and strongest advocates of “spherical under self-gravitation” as a necessary criterion. So yeah, I totally get it about about hydrostatic equilibrium. But rigidity is a matter of timescale, dammit. It may be technically true that even rock is a fluid and that “hydrostatic equilibrium” describes the long-term condition of a terrestroid planet, but rock is not fluid on the same timescale that water is.

That’s why there’s a rigid-body version of the Roche equation to begin with. Using the fluid form of the Roche equation for a rocky or terrestroid planet will give grossly unphysical results unless you’re willing to wait one hell of a long time, probably a longer scale than the orbital dynamics remain approximately stable.

1. Keith Lynch says:

I have to disagree with you there. If you subject something the size
of the Moon to negative net gravity, it *will* come apart like a pizza
being fought over by two large dogs. And I don’t just mean on a
geological time scale.

I got in argument with Dr. Stephen B. Harris, one of the smartest
people I know. I insisted that if the Earth were to drop into the
sun, it would be completely torn apart by the sun’s tidal effect
before it got close enough to melt, much less vaporize. Its solid
core is its strongest part, but in this context it can be modeled as
an iron cable, since the strength of a cable of given length under
tension increases linearly with its cross-sectional area, but so does
its mass, so the cross-sectional area drops out.

Speaking of which, a fun problem is how long a steel rod or cable can
exist in interstellar — or better yet intergalactic — space, before
it collapses due to self-gravitation. It’s the only gravitational
problem I know where the Riemann zeta function is relevant. You’re
allowed to counteract gravity by having it rotate end-over-end, but
that’s of limited utility since it doesn’t scale right.

A related question is whether a long cable in intergalactic space
would experience any tension from the expansion of the universe.
I’m not sure anyone knows the answer to that one, but I think the

2. Jon Brase says:

My gut reaction is that the problem with your line of reasoning is:

1) If the bodies are brought that close together on a short timescale, well before you get to the Roche limit, tidal forces are going to fracture the rock, and you’ll have lots of chunks that will, on a planetary scale, be fluid on short timescales.

2) Also at some time before the Roche limit is reached, (and my gut feeling is that it will likely be around or before the point of fracture) tidal forces may well deposit sufficient energy to melt rock, at which point it will behave as a fluid in more or less the same timescale as water.

3) The earthquakes and magma floods attendant with 1) and 2) are not likely to be healthy for the continued existence of cameras, painters, or other elements of the scenic postcard industry on the planet in question, especially given that:

4) Bringing the two bodies together quickly implies either a flyby on an escape trajectory, which will be a one-time event (so the window for making scenic postcards of the view of the other planet at maximum subtension will be short) and will involve rapid changes in the tidal forces exerted, which will likely contribute to lots of fracturing and heating, or else applying braking forces to move two planets in a wide orbit into a progressively narrower one, which, if done on a timescale short enough for rock to be solid, will likely heat the planet even more than tidal forces will.

5) So the only way you’re going to get an inhabited planet with a long-term scenic view of a high-subtension sattelite is if it has arrived in that situation slowly, in which case the timescale is long enough for solid rock to reach hydrostatic equilibrium. All the short term scenarios also result in hydrostatic equilibrium, but you have to pay more for your scenic postcard, because the camera drone needs to be engineered for extreme conditions, and you can’t enjoy the view from a local bed and breakfast ’till the magma solidifies.

14. Alrenous says:

I want to know where he got the hi-res render of the moon.

The math works out if you want to bring the moon closer while reducing its mass, to keep the tides reasonable.

g_force(r) ~ sqrt(r), but r(m) ~ cbrt(m). If you halve the distance, the mass needs to quarter, but this only reduces the radius to roughly 63%. arc_subtended(r_orbit) is inverse linear in r_orbit. 2*0.630 = 1.26, so we’re gaining apparent radius. To double apparent radius, we need to go through 1.26^x = 2, x = 3 such halvings. 1/8th, 2.99*10^4 miles.

Can go up to quadruple size, then you’re at 1/64th, though that’s inside the Roche limit at 3.73*10^3.

Exactly on the line, the moon would be 3.43 times its current apparent size.

However, to get a full-on anime moon without breaking the tides you’re talking about getting so close that, never mind the Roche limit, you’re skimming atmosphere.

The other weird physics question I have is what it would be like to be standing on a planet which is rotating faster than its own escape velocity. I assume the surface would break up, and thus drop it in altitude until it is at least slower, but is there any reason the debris couldn’t stay near the surface?

1. Keith Lynch says:

Right. As a result, the tidal effect of an spherical object depends
only on its density and its angular size. The moon, which is about
the same angular size as the sun, has about three times the tidal
effect on Earth because it has about three times the density.

I like to point to my (personally autographed) copy of Misner,
Thorne, and Wheeler’s _Gravitation_, and ask visitors whether they’re
experiencing more tidal effect from that book or from the moon. :-)

1. Jon Brase says:

Eh, it is presumably not a spherical book, so I’m not sure that works.

15. grobber says:

Something that’s been bothering me about the original post:

>By hypothesis our worlds have the same radius R. The minimum
>distance from a point on the surface of the primary to a point
>on the surace of the secondary has to be 1.26 R. We must add
>0.5R to get to the secondary’s center

Re: that second quoted sentence: why *half* a radius? The distance from the center to the surface is the radius (or half the *diameter*).

So assuming your framework sound (i.e. ignoring the issues noted by others of whether or not the approximation is reasonable in the regime you assume), I believe both legs of your right-angled triangle should be larger by 0.5R: x=2.26R and y=R.

Am I missing anything?

1. esr says:

>Am I missing anything?

LOL. How did that get past both me and the people who have found other mistakes?

Will correct.

1. grobber says:

Ah, thanks for confirming! I thought I was going insane there for a bit.

For what it’s worth, looks like the correction helps the case (again, ceteris paribus):

If 0<y<x and 0<a then (y+a)/(x+a) > y/x, so you get a larger angle as the arctan (large angle being the point to begin with).

16. TRX says:

> pie shaped

Which brings to mind Hal Clement’s classic “Mission of Gravity”, with a disc-shaped planet with a spin so fast that it had 700g at the poles and 3g at the equator.

Later computer simulation at MIT said the polar gravity would likely have been closer to 275g, but the story still rocks.

1. Keith Lynch says:

To his dying day, Clement apologized for getting the shape of that
planet (Mesklin) wrong, even though there was no way he could have
gotten it right without a computer. Computers were in short supply
when he wrote it in the 1940s.

Actually, the shape depends on the planet’s density profile. That’s
how we know Earth’s density profile, and how gravity varies with depth
on our planet. It does not increase linearly with distance from the
core, as Newton proved would be the case on a planet with uniform
density.

How can I preserve paragraph breaks here? If I can’t, can we please
move this conversation to Usenet?

17. MP says:

“volume increases with the cube root of radius” ?

Gonna guess that needs rewording. Otherwise, a very nifty essay.

18. Daniel Franke says:

Eric, you screwed up the math here. Peter Donis’s comment correctly explains how, but here’s an intuition pump to make it clear that your answer is wrong.

Picture two spherical, equal-sized pebbles in circular orbit around each other, such that their surfaces come just barely short of touching. In this case, viewed from the center of one body, the other body fills 60 degrees. (To get this without calculation, picture how oranges are stacked on a grocery stand). Viewed by a flea on the surface of one body at the point of near-tangency, the other body fills fully 180 degrees: look down at all, you see one body, look up at all, you see the other. Yet this arrangement is obviously stable for a simple reason: the surface gravitational force of a pebble is not nearly sufficient to tear apart another pebble!

1. esr says:

Daniel, your counterargument proves too much. There must be some mass scale effect that’s important, otherwise you could make the same argument about two Earth-sized pebbles in near-tangent contact and Roche-limit disintegration would never occur.

I guess what this tells us is that the Roche’s-law approximations I got from Wikipedia are just wrong – they must need some absolute mass term rather than simply a mass ratio. Either that or there’s some interplay with the material’s strength under tension and the magnitude of the tidal shear isn’t being captured in the approximation…ah, yes that’s probably it.

Your pebble case doesn’t actually have a Roche limit at all, because the magnitude of the tidal shear is small compared to covalent/crystalline bond strengths in the pebble. But as the mass scale of the system rises, tidal shear forces increases while bond strength remains constant. You only start getting effects (and a consequent distance limit) when the shear force exceeds the sum of the breaking strain of the material and the self-gravitation of each body.

So no, in fact, you haven’t shown that the max subtended example is can be 60 degrees. You have to model a system where tidal stress can exceed breaking strain to do that. Your pebbles aren’t it.

Now I grok that this must be why there are both rigid-body and fluid versions of the limit formula. The fluid version models a body with no tensile strength, just self-gravitation.

1. Jon Brase says:

Indeed, I must confess some confusion over the usefulness of the rigid body formula (which doesn’t actually model the tensile strength of the material, just the point where loose objects on the surface will be lifted off), beyond just being simpler math, but I think there’s one thing that it’s useful for: a tensile-bound body inside the rigid-body Roche Limit will experience accelerated impact erosion: not only the “prompt” debris (i.e, that which is accelerated to escape velocity by the impact itself), but also any material simply pulverized and left in place by the impact will be lost.

1. esr says:

>Indeed, I must confess some confusion over the usefulness of the rigid body formula

I’m now thinking about it this way: rigid-body applies when the sat’s reponse to stress is dominated by its tensile strength. fluid-body when (much weaker) self-gravitation dominates.

19. The Monster says:

> (Two mistakes in the original posting has been corrected.)
Thus introducing yet another mistake:

Two mistakes have been corrected.
(One mistake has yet to be corrected

1. The Monster says:

And now it has been corrected.

20. simon says:

It looks to me that you are assuming that d refers to the distance between the surfaces of the objects, but the derivation seems to use d as the distance from the center of the primary to the surface of the secondary.

So: The minimum distance from a point on the surface of the primary to a point on the surface of the secondary has to be 0.26 R, not 1.26 R.

x = 1.26R, not 2.26 R, y = R.

a = 2 * arctan(R / 1.26 * R) = 2 * 0.67 radians = 77 degrees.

Although, actually shouldn’t it be arcsin?

a = 2 * arcsin(R / 1.26 * R) = 105 degrees

1. simon says:

Actually, it seems it is even referring to the distance between the centers! Which would imply that the bodies can come into contact for the rigid version.

But there’s also an assumption that the secondary is much smaller than the primary, which is obviously broken with them being the same size.

Still, I think two equal size and density rigid bodies should still be able to (just barely) come into contact, since you’d expect the gravity at the point of contact to exactly cancel (so, stuff isn’t falling off but just at the point of falling off).

1. simon says:

The symmetry argument should also apply to non-rigid bodies, as long as the deformation is also symmetrical.

I’m assuming the bodies are tidally locked, and in a circular orbit around each other. Also, that the problem point where the Roche breakdown occurs will be at the point of an object facing the other object (this should be the case because tidal forces are stronger closer to an object).

When moving in a line between the two objects the relevant forces (centrifugal, gravity including tidal forces) pull more in a direction the more you move in that direction. So, as long as there is a neutral point between the two objects, the forces should be away from that neutral point at other points on the line between the objects. So no Roche breakdown.

Now the symmetry between the objects implies that there is such a neutral point between the objects, even with (symmetrical) deformation. So, you can get symmetrical objects arbitrarily close without breaking down, and they start to break down when they touch, same as with the rigid case.

But, this does not necessarily mean that you can get 180 degrees of another object in the sky, in the case of fluid bodies. That’s because the lobes of the objects that face each other might get pointier and pointier as they get closer and closer, so their view angle from the other body never exceeds some angle threshold less than 180 degrees.

I’ll try to estimate this.

Imagine we have two bodies just touching. They are completely fluid so that their surfaces are equipotential surfaces, in the rotating reference frame of the tidally locked bodies.

An equipotential surface has the net force of gravity+centrifugal force normal to it. But, at the point of touching the net force is zero. So we will have to consider the force at points at a small displacement from that point. I’ll only consider the forces outside the bodies.

For simplicity lets assume the gravity of the objects is like that from a point mass, although it obviously isn’t. Hopefully this won’t change the qualitative conclusions.

Let x be the direction between the bodies, z be the direction of the axis of rotation, and y be the other direction. Origin at the symmetry point between the bodies.

On the x axis near the origin, first let’s imagine there’s some space in between the bodies. This is because we are only considering the forces outside the bodies. The forces on the x axis are in the x direction, positive for positive x positions, negative for negative x positions. Both the centrifugal force and the gravity of the objects behaves like this and it should be approximately proportional to displacement from the origin for small x.

Now on the z axis, the forces will be in the z direction, positive for negative z positions, negative for positive z positions. The amount of z axis force should again be approximately linear with distance from the origin.

On the xz plane near the origin, the forces should be a superposition of the x and z axis components. If you imagine a point off the x and z axes (but still close to the origin) and imagine drawing a line through it that at each point is aligned with the net force in this rotating reference frame, you’ll get something that is near parallel the x axis at one end and near parallel the y axis at the other and in between at some point it will cross a line through the origin perpendicularly. Since near the origin the force is approximately linear, all other points on the line will also have the force perpendicular and this is an equipotential surface.

So, in the x-z plane, when the bodies touch, their surfaces will be converging to a point at an angle (not 180 degrees)

On the y axis, you have both centrifugal and gravity related components, but now they act in opposite directions. If the gravity near the origin along the y axis is stronger than the centrifugal force, then the situation on the xy axis will be qualitatively the same as on the xz axis, except that the surfaces will converge at a wider angle (still less than 180 degrees). If gravity is weaker, then my assumptions break down and I don’t know what happens. If they are the same, then either they form a 180 degree angle or my assumptions break down, depending on whether the next order effects are attractive or repulsive.

However, I am confident that the gravity effect will be stronger than the centrifugal force. The reason is, imagine you place some test masses on the y axis at plus and minus epsilon. Now consider the effect of just one of the bodies on the test masses. If the masses are in orbit around the body I expect them to remain stationary. If they are closer than orbit for this amount of rotation, they should converge to the center of the body (and thus also to each other), and if they are farther than orbit, they should diverge from the center of the body (and thus also from each other). Now of course at this rotation speed, the orbit distance should be around the center of the other body, so the test masses are much closer than orbit and should converge.

So, in conclusion:

Symmetric bodies (symmetric in size, density and shape) should always be able to approach arbitrarily close to touching without breaking the Roche limit. BUT if the bodies are fluid, they don’t actually take up a 180 degree angle anyway, they will instead, when they touch, converge at smaller angles. Actual calculation of angles I leave to someone less lazy.

1. Jon Brase says:

See the discussion of Roche Lobes upthread. I’ve generally seen the contact angle for bodies just barely touching illustrated as something that looks to be in the 60-90 degree range.

>Symmetric bodies (symmetric in size, density and shape) should always be able to approach arbitrarily close to touching without breaking the Roche limit.

Technically, they are exactly at the Roche limit with respect to each other at exactly the point where they are just barely touching, but no disruption occurs, even if they are within the Roche limit as long as their surfaces are at the same potential (if not, mass will flow from the body with a surface at higher potential to the one with lower potential until the potentials match, the higher body loses sufficient mass to no longer be within its Roche limit with respect to the other, or the higher body is completely disrupted.