# If the Sun disappeared, could some planets form a new orbital system?

If the sun were to suddenly disappear, the planets would continue to travel tangentially to their former orbits. (This I know from this answer to a somewhat related question here.)

In such a scenario, is it possible (however unlikely) that a new orbital system would form from two or more of the now-rogue planets? For example, is it physically and mathematically possible that Earth might end up getting captured by Jupiter and thereby become one of its moons? Or do the size and speed of the planets completely rule out such a scenario, even if the sun disappeared at such a time that the planets' new trajectories would cross?

• Our best current theory of gravity is Einstein's theory of general relativity (GR). According to GR, mass-energy is locally conserved, and therefore the sun can't suddenly disappear. Therefore our current theories can't say anything about what would happen if the sun suddenly disappeared. – Ben Crowell Feb 23 at 20:23
• @ben But it is trivially easy to make a hypothesis and run a simulation based upon it. – Michael Richardson Feb 23 at 21:28
• @BenCrowell: "Our best current theory of the speed of light is that it varies depending on the movement of the aether. Therefore we can't say anything about what would happen if it was constant, and no use speculating what would happen if it was" -- Someone in 1880. Thought experiments, even starting from seemingly nonsensical initial conditions, can be useful for learning and understanding the physical world. – whatsisname Feb 23 at 21:52
• Those objecting to the premise of the sun disappearing are missing the point of the question. You could just as easily assume that the question is about pairs or sets of completely independent rogue planets that happened to meet in a neighbourhood of space, on any of the same set of possible trajectories envisaged in the OP. This scenario is certainly realistic insofar as it doesn't violate the laws of physics (though of course it's vanishingly unlikely). – Psychonaut Feb 24 at 8:30
• @Ben your line of reasoning is a completely general argument against any kind of counterfactual reasoning. The laws of physics are deterministic so there is only ever one possible future (ignoring quantum randomness, but then quantum jumps also make it possible for the sun to disappear with a vanishing chance). A what-if scenario is thus in general asking about a situation that cannot actually happen. Nevertheless there is usually a strictly speaking not physical model describing the situation that is still useful. – JanKanis Feb 24 at 9:14

The issue here is whether pairs of planets can become gravitationally bound to each other.

In the two-body problem the trajectories or orbits are ellipses (bound orbits), parabolas and hyperbolas (unbound). For all practical purposes, an encounter looks like they start out at infinity with some finite speed, approach each other, and then maybe fly away or remain bound. This situation (finite speed when far away) corresponds to a hyperbola, so the generic case will not form a new bound system: some of that energy needs to be shed somehow.

One way this can happen is an inelastic collision, i.e. a merger. A less extreme case is a close encounter where tidal forces dissipate some of the kinetic energy. But the kinetic energy of planets is large, so a single encounter is unlikely to dissipate enough to make them bound.

Another important possibility is a three-body encounter. This is how comets get into short-periodic orbits when falling into the solar system: an encounter with Jupiter (or some other planet) leads to a gravitational "assist" that makes it bound with the sun.

In your scenario where the rogue planets fly off along their tangents it is very unlikely that any start out bound to each other since planetary orbital velocities around the sun are much bigger than the orbital velocities of a bound planet-planet pair at a large distance. Hence they will just fly apart. In theory they might have an encounter before leaving the solar system vicinity (e.g. Mercury by sheer chance sweeping past Jupiter) and that leaves a tiny chance of a three-body interaction with moons or a merger, but the chances are truly tiny. The generic case is just scattering.

For a body to be bound to another body of mass $$M$$ it need to move slower relative to it than escape velocity $$v_{esc}=\sqrt{2GM/r}$$. If we assume planets move in circles we can then take their orbital velocities for all possible positions along the orbit, calculate the relative velocity, and compare it to the escape velocity.

The result is this diagram, where I have plotted the separation on the x-axis and how many times escape velocity the relative velocity is on the y-axis. The numbers denote which pair of planets I am comparing. The blue horizontal line is $$\Delta v=v_{esc}$$.

Looking just at the lowest relative velocities shows that the closest to being potentially bound is Saturn to Jupiter were the sun vanish just as they are moving in parallel, but even in this case Saturn has several times Jupiter escape velocity.

One could complicate it with inclinations and elliptic orbits, but it will not change the qualitative picture: the planets are not even close to being bound to each other in the absence of the sun.

• The only planet likely to get a three-body capture is Mercury. Everything else is just too large relative to the available moons. You might be able to capture Mars by ejecting all four of the Galilean moons, but that's a real longshot. – Mark Feb 22 at 21:28
• What about orbital inclination? From Jupiter, Mars is less than a degree off; Mercury is over five. But even at only .54 degrees off, I'd assume that's too much over the distances we're talking about. Also, what speeds are they all going to be going at? Who's going to be playing catch-up to who? (assuming you pick the exact right time to send two of them off in the same direction - not that you really can because of their inclinations....) – Mazura Feb 22 at 22:30
• Something to consider is that the planets may not even be starting on Hyperbolic orbits relative to each other. Mars and Jupiter have their lowest relative velocities at the same time as they have their closest approach. If this orbit isn't already hyperbolic then the interaction with moons becomes largely irrelevant. I'm at work so can't spend the time to work it out right now, but I'd be interested to see someone do that calculation. – Turksarama Feb 22 at 23:15
• The formula you're using for escape velocity $v_{esc} &= \sqrt{2GM/r}$ is for the case where the mass of the escaping body can be ignored. For planets that's not valid, and the escape velocities will be higher due to the mass of the other planet that this equation ignores. – JanKanis Feb 23 at 11:35
• @AndersSandberg I might suggest using the mutual escape velocity $v_e=\sqrt{\frac{2G(M+m)}{R}}$ derived here: physics.stackexchange.com/questions/121679/…. It will dispel this argument and cut the number of pairs you need to graph in half. It should increase rather than decrease $v_e$. By the way, I think your graphs are a genius way of presenting this problem! Love this answer! – Connor Garcia Feb 23 at 16:43

No. The 8 planets would go into 8 different directions. It is because their relative velocity to each other is much higher than the escape velocity, even from their smallest distance. If it would not be so, their orbits could disturb each other significantly, even today with the Sun. Thus, the Solar System would become chaotic.

However, the moon systems of the planets would remain practically intact.

There is a very little chance that a 3-body or more-body system from the planets is created for a short time. Then some of them escapes, leaving a gravitationally bound (most likely) 2-body system. Similarly possible, but very unlikely, that an impact would happen. Both of these have a very small probability, you can safely expect that all the 8 planets go into 8 directions.

• Did you really mean 9? – Spencer Feb 23 at 18:22
• @Spencer 1. Mercury, 2. Venus, 3. Earth, 4. Mars, 5. Jupiter, 6. Saturn, 7. Uranus, 8. Neptune. Wow! You are right. I fixed the post. – peterh Feb 23 at 19:18
• @JasonGoemaat Right, thanks :-) – peterh Feb 23 at 19:42
• Just saying....it should be 9 . – Spencer Feb 23 at 22:47
• @Spencer Well we know it shouldn't be 9. Maybe 10, 19, or many more depending on how you want to define a planet and what we've discovered in our solar system. – Jason Goemaat Feb 23 at 23:18

I gave a try at refining Anders Sandberg's answer with the equations for a two body system.

Basically, a two body system is bound if the sum of the total kinetic energy and the (negative) gravitational potential energy is still negative. (To escape, kinetic energy needs to cancel out the potential energy.)

Gravitational potential energy:

$$U = -{GMm\over{R}}$$ where the gravitational constant $$G = 6.67430×10^{−11}\ m^3kg^{–1}s^{–2}$$, $$M$$ and $$m$$ are the masses of the two bodies, and $$R$$ is the distance between them.

The total kinetic energy is the sum of the kinetic energies ($$E={1\over{2}}mv^2$$) for each body relative to the common center of mass.

According to Anders' plot Jupiter and Saturn are the closest to being bound, so I made a calculation for that pair using data from these nasa pages, ignoring excentricity and inclination.

Jupiter:
$$M = 1.898×10^{27}\ \text{kg}$$
$$\mathrm{semimajor\ axis} = 778.570×10^{9}\ \text{m}$$
$$v = 13060\ \text{m}⋅\text{s}^{-1}$$

Saturn:
$$M = 5.683×10^{26}\ \text{kg}$$
$$\mathrm{semimajor\ axis} = 1433.529×10^{9}\ \text{m}$$
$$v = 9680\ \text{m}⋅\text{s}^{-1}$$

At closest approach, the distance $$R = 654.96×10^{9}\ \text{m}$$. The gravitational potential energy then becomes $$-1.099×10^{32}$$ joules.

The difference in speed $$\Delta v = 3380\ \text{m}⋅\text{s}^{-1}$$. Weighted by the relative masses that velocity is attributed to Jupiter for 23% and to Saturn for 77%. This gives Jupiter a kinetic energy of $$5.76×10^{32}$$ and Saturn $$19.23×10^{32}$$

So in conclusion, the kinetic energy is about 25 times too high for Jupiter and Saturn to be bound, so they will fly apart.

• It occurs to me that, even without these detailed calculations, there's a more intuitive reason they couldn't be close to being gravitationally bound - if they were, the same rules of motion would apply even in the presence of the sun, so they'd already be orbiting one another and we'd call the smaller one a "moon", just like the earth/moon pair; or at least significantly disrupting each other's orbits. – Steve Feb 24 at 8:44
• @Steve indeed. Peterh makes the same argument in his answer. – JanKanis Feb 24 at 8:51

I love Anders's answer — well done.

But you could have got the idea already by looking at orbit perturbations the planets inflict on each other: They are mostly negligible.

Come to think of it — it cannot be otherwise in a stable planetary system: During the evolution of planetary systems the orbits of planets which are too close to each other will be changed exactly because they are too close (relative to their mass). Their orbits will become irregular; they may be ejected from the system, collide with other bodies or come to rest in a new, more stable orbit. The stable state of a planetary system, and hence the look of any "mature" system, is one where the planets don't perturb each other's orbits.

In such a system, like ours, the central star's gravitational impact is orders of magnitude stronger than the planets'. (Some residual perturbations have an effect when accumulated over astronomical times and lead to resonances etc., but we are looking at one-time effects here.)

Even strictly empirically we can observe that Earth's trajectory does not depend significantly on the relative position e.g. of Jupiter. That fact is independent of the Sun: If it went AWOL Earth's trajectory would still not depend on Jupiter. it would look significantly different, namely mostly straight, because the Sun does have a significant impact on Earth's trajectory, as long as it's around. But as with an orbit, that straight trajectory would simply continue imperturbed.

Anders' computations showed that the same is true for the other planet pairings.

I see one possible way for this to sort-of happen. Anders Sandberg did a good job of showing why a simple capture is impossible, but there's an extremely unlikely but not impossible scenario he missed.

Consider Jupiter and Saturn. They're on the same side of the sun, Saturn is ahead of Jupiter. The black hats kidnap the sun and they're free flying. Jupiter comes barreling up on Saturn, way too fast for a capture--but what if we have an offset impact event? Jupiter will steal mass of Saturn but if exactly the right amount of energy gets transferred Saturn ends up with ever so slightly more energy than it needs to escape. It then needs to run over a Jovian moon on the way back out, losing a bit of speed but changing the shape of it's orbit so its periapsis isn't within Jupiter's Roche limit.

The results are cataclysmic, whether you should even call the results the same planets is questionable.

Without the sun... at current orbital velocities... The planets would fly apart... But exactly how far would they fly? Would they continue indefinitely into interstellar space? Or would they fly out into the Kuiper belt or Ooort cloud and subsequently fall back toward some barycenter.

Removing the sun removes 99% of the mass of the solar system... so this could dramatically alter the barycenter of the system. Assume all/most planets are in one quadrant at time of sun's disappearance. Then presumably those planets with roughly similar trajectory would not have escape velocity relative to their shared barycenter.

Take for example the recent alignment of Jupiter and Saturn. There was definitely a window of time this year where if the sun had disappeared, Jupiter and Saturn would have formed an orbital system between the two, and had Earth or some other planet been behind them in orbit, overtaking them like a rock in a sling that, released at the precise moment, find its mark, that planet might get caught up in their system. (especially because our inner planets have higher velocities than more outer planets)

But if you could, with your hand of god, place the planets just so that they were least optimally arranged for the formation of an orbital system in the sun's absence... how far would they fly? Would they escape into interstellar space? I would think in 99% of planetary arrangements, this would happen since these planets are so tiny relative to the space between them and relative to their velocities and relative to the mass of the sun which is 99x the total mass of everything else in the solar system.

• Escape velocity depends on distance, and at a distance of 4.4 AU, the mutual Jupiter-Saturn escape velocity is near zero. – Mark Feb 23 at 23:15
• Wikipedia actually says Sun constitues 99.86% of the solar system mass... that's a difference from 99%. Then where the "99% of planetary arrangements" number comes from? – user1079505 Feb 23 at 23:22
• The whole point of escape velocity is that the objects in question will not orbit each other at any distance if the escape velocity is exceeded, no? Also, there will be no Oort Cloud or Kuiper Belt, as all of those objects will fly apart absent the sun's gravity, too. – reirab Feb 24 at 22:54