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In the climactic final battle Faction A engages Faction B on Faction B's planetary base, both in orbital and ground combat. Realizing they can't win, Faction A activates a device that transports and traps the entire planet in what can be compared to a pocket dimension.

So basically the entire planet vanishes from existence in (almost) an instant. Now there are two fleets of ships around the space where a planet used to be.

What is the effect (if any) on said ships?

Half of me thinks that because it's in space there won't be any effect, but of course it would be cool if there was some sort of displacement effect.

EDIT: A few answers mention the planet returning at some point. I'll have the trapped people escape several decades later, but the planet itself remains lost.

Several answers have been extremely useful, but unfortunately I can only pick one, so I went with the one that gave me the best alternative. The rest get upvotes though. Thanks to everyone that took the time to answer.

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    $\begingroup$ If the ships are orbiting the planet, they are, by definition, within the gravity well. $\endgroup$
    – Kevin
    Jan 10, 2017 at 20:49
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    $\begingroup$ Trapping the planet in a pocket dimension might have interesting thermal issues. One possible side effect: navigation computers will not be prepared for vanishing planets, so expect wrong results (trying to compensate for gravity of nonexistant planet) or outright crashes. $\endgroup$
    – pjc50
    Jan 11, 2017 at 11:00
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    $\begingroup$ Strap a toy starship to some rope. Spin it around fast, the let it go. That is what will happen to your starships. $\endgroup$
    – PlasmaHH
    Jan 11, 2017 at 11:21
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    $\begingroup$ Not a physicist so putting this as a comment rather than an answer, but as the planet vanishes, the momentum of the ships should change from a small planetary orbit to a solar orbit (likely just a line as the planetary scale at which you're viewing, i'd assume the curve to be minuscule). Kinda anti-climactic when looking at the effects from outside, but crews of these ships would likely be thrown into disarray. Navigators and bridge staff would now have to make fairly large corrections to remain in the cluster of ships, and prevent becoming scattered. $\endgroup$
    – lewis
    Jan 11, 2017 at 12:29
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    $\begingroup$ As an aside, I'd expect long term effects on the remaining planets in the system, if that matters. Also an inhabited planet is a delicate thing to transport - it wouldn't take much to mess up the tides or the seasonal weather patterns etc leading to all kinds of mass destruction. That's what struck me when the Daleks did this in The Stolen Earth. $\endgroup$ Jan 11, 2017 at 12:32

13 Answers 13

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Earth has a Schwarzschild radius of about 3 mm.

This means things at geostationary orbital distance have a time dilation of about $$\sqrt{1-\frac{9 mm}{35786 km}}$$ or one part in $10^{-10}$ roughly. (if they are actually orbiting this value changes slightly, as does the rotation of the Earth)

This also lines up with the length contraction factor.

When the Earth disappears "instantly", a gravitational wave of that magnitude is going to be produced. How much energy is that?

Well, 1 solar mass converted to gravitational waves and sent over 1.4 billion light years produced a $10^{-20}$ amplitude wave (LIGO observation). The energy in a gravitational wave is proprotional to amplutide squared.

So, per meter squared, the LIGO observation would carry:

(1 solar mass * c^2) / (1.4 billion light years*2)^2 / 4 pi * 1 m^2

2 * $10^{-5}$ J (apparently it was 1 solar mass of matter converted into a gravitational wave at a distance of 1.4 billion light years).

The gravitational wave from the Earth disappearing is going to be ${10^{10}}^{2}$ stronger than that, or 2*$10^{15}$ J. This is an insane amount of energy; however, very little of it actually deposits on normal matter.

Suppose we are 1 Jupiter-radius away from Jupiter instead.

Jupiter has a Schwarzschild radius of 2.2 m. Titan has an average orbit of 1,221,850 km. Then the gravitational wave would carry 500 times as much Energy.

The question becomes how well does it convert over to normal matter? Will it occur fast enough to disrupt an atomic nucleus?

The compression effect on molecular-level matter will only involve modest pressures. But the compression effect will occur all the way down the length scales, and I suspect it requires lots more pressure to compress a nucleus.


But back up a second. We ripped the planet from our universe. One could argue that would involve forming an event horizon around the planet and making it disappear.

I mean, photons not coming from an area is the definition of event horizon. Stuff an event horizon somewhere, and you warp space. The volume we need to swallow is the planet. So, black hole the size of the planet in effect blinks in then poofs?

If we are 10 planet-radius away, and the event horizon forms tightly around the planet then disappears, this would generate two gravitational waves of impressive magnitude.

$$\sqrt{1 - \frac{1 r}{10 r}}$$ gives us a amplitude of 0.05. The energy carried by this wave is about 10^17 times greater than the ones we are describing above.

In effect, all matter would suddenly feel stretched by 5%, then compressed by 5%. This would occur all the way down to the molecules, quarks and nuclei. I'd be worried about fission events from this happening suddenly, let alone the amoung of energy released by compressing "incompressible" solids by 5%.

Compressing water by 5% would take 0.1 GPa, so we could estimate the effect would be akin to a pressure wave of that magnitude over humans. And 0.4 GPa for iron.

This level of pressure in a conventional blast is enough to blow limbs off.

I cannot believe the compressibility of EM mediated molecule-scale solids is in the same universe as that for a nucleus or a proton. So I would be very worried about atomic disintegration...

Ignoring that, matter slows down gravitational waves. The effect is extremely tiny, but we could use that to estimate how much impulse this would provide and how much energy deposited. I cannot find the correct equations for this case.


Calculating what exactly happens is going to be quite tricky. For a planet-sized event horizon, the effect will be explosive at "molecular" scales, blowing objects apart. At atomic scales, I don't know (will it cause fission?). At macroscopic scales, I don't know (will it impart a large radial impuse from the matter slowing the gravitational wave down?).

There is plenty of energy to work with. The kind of effect that energy density and flux could cause seem unbounded.

The above also neglects the power; how long the teleport takes determines the power the wave carries (how "sharp" it is, not just how much energy it carries).

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    $\begingroup$ So I suppose the game is this: does a disappearing planet produce an event horizon corresponding to the radius of the planet, or the mass of the planet? Mass is fine, radius is not fine. And since this is a made-up process anyway, it's author's choice. $\endgroup$ Jan 11, 2017 at 3:26
  • $\begingroup$ The sudden expansion and contraction of matter would make for an amazing effect. My only concern is that the same technology that I use to transfer the planet is used by the combat mechs of the protagonist's faction (for short distance A to B travel - "blinking"). But I can probably write the inconsistency off due to scale and and the sheer amount of energy used to remove a planet from existnce. $\endgroup$
    – Lu22
    Jan 11, 2017 at 11:16
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    $\begingroup$ @Lu22 Blinking is quite dangerous. Make sure you get the energy conservation right or you have a perpetual motion machine and an inertialess space drive :P $\endgroup$
    – Luaan
    Jan 11, 2017 at 11:41
  • $\begingroup$ @Luaan. I use it in a similar way to the movie "Clockstoppers". You enter a fake dimension with accelerated time but identical coordinates, walk to point b, and re enter the datum dimension. So you can travel an hour's distance in a second, at the cost of an hour of your life. Hence why they only use it for traveling a few meters at a time. The planet in question gets the reverse in treatment though. Power costs are also restrict it. $\endgroup$
    – Lu22
    Jan 11, 2017 at 13:36
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    $\begingroup$ It's worth noting that general relativity (the current theory we have that predicts for gravitational waves) doesn't allow for matter to "blink out of existence" at all, even if it "blinks into existence" somewhere else. This is explained in a bit more detail in this incredibly insightful answer by a totally humble guy over on Physics StackExchange. You would have to postulate new laws of physics to get around this, and at that point the gravitational wave amplitude could be anything you want. $\endgroup$ Jan 11, 2017 at 17:34
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You can have it affect them as much as you like. The device is magic, so the additional effects can be whatever you want.

On a more practical note if the mass of the planet really has disappeared then orbits are going to be thrown way out of whack - depending on what point of their orbit around the planet the ships were at they could be thrown in any direction as the acceleration towards the planet's center suddenly vanishes.

The ships are not "outside" the gravity well, in fact there is no such thing as "outside" a gravity well, although there is a point at which it's insignificant. The ships are in free-fall, but at the same time they are moving sideways so fast that even though they are falling towards the planet all the time they move far enough that they never actually hit the atmosphere.

If you're writing space-based sci-fi you should definitely read up on some basic physics and orbital dynamics (simple Newtonian stuff would be plenty) if you want the results to be plausible.

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    $\begingroup$ Read or better play Kerbal Space Program :D This game can fix basic misconceptions really fast. $\endgroup$
    – Mołot
    Jan 10, 2017 at 13:16
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    $\begingroup$ @Samuel gravitational orbit is free fall. $\endgroup$
    – Ruslan
    Jan 10, 2017 at 20:07
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    $\begingroup$ @Samuel I never said free fall implied lateral movement. I said the converse: orbiting without thrust is free fall. Any object following its geodesic is in free fall by definition. $\endgroup$
    – Ruslan
    Jan 10, 2017 at 20:14
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    $\begingroup$ "depending on what point of their orbit around the planet the ships were at they could be thrown in any direction" I think I'm misreading this. If the planet were to disappear, the object would continue on its current straight-line trajectory, entirely predictable. It wouldn't be "thrown" in any direction - it would only follow a single vector. Can I suggest an edit to clarify this point? $\endgroup$
    – user21726
    Jan 10, 2017 at 21:12
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    $\begingroup$ @Amy The moment of planets disappearance is random (at least with respect to ships' orbits and positions). Orbits can only be predicted after that "random" event. $\endgroup$
    – hyde
    Jan 10, 2017 at 23:00
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Gravitational acceleration is continually changing the direction of a body in orbit around it, tugging it in a circle (we will assume) around it. If this tug disappears objects will just carry on in a straight line with their original velocity. The directions will just be along the tangent of the orbit they were on.

So nothing too dramatic so far.

However it really depends on how your planet disappearing act works, if they are tangled up in different but still present dimensions the mass may still be present and our 3D space might still act as though the planet is still there. This, however, is all speculative as we have no real idea. Another possible speculation is that, if the mass does disappear, our space will spring back and the space-ships will experience the stretch and squash of gravitational waves caused by such a dramatic difference in the dip.

But in the end, we don't know, you get to decide how the machine works to best fit your story. Do they know how to return? (This would imply they were still linked to the original position) Or perhaps they can return anywhere in the universe (possibly very messy) which would do wonders for space-travel.

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    $\begingroup$ I think it is much better to still exert gravitational attraction (even as singularity) as disappearing mass means that laws of physics are not invariant due to translation via Noether's theorem. $\endgroup$ Jan 11, 2017 at 8:18
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The orbiting ships would stop moving in the curved (by gravity) line of their orbit and suddenly (continue to) move in a straight line in whatever direction they were moving at the time of the disappearance.

At worst, this would be inconvenient, as to stay in the vicinity of where the planet was, they would have to decelerate then turn around and come back, at which time staying in the vicinity would be simply adopting the speed and velocity of the former planet, as their new stable orbit would be around the host star.

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If the planet goes away, the gravity well goes away, and the well is no longer "turning" their orbits. Wherever they are in their orbits, they now go in a straight line. That's not the problem.

The planet comes back. Uh-oh.

The problem is when the spaceships come back and loiter in open space to await the return of the planet. (their science teams figure out pretty quick that's a highly improbable way to destroy a planet.)

The ships have nothing to orbit, so they must "sit there" in the general vicinity of where the planet is expected to return. When the planet returns, they will suddenly find themselves in a gravity well, with no sideward velocity for an orbit. They will Fall Straight Down while the planet turns under them.

Their engines will need to generate somewhat more than one local "gee" of thrust to escape - even more if they're late - and that may not be practicable for a slow, large interstellar capital ship. Could they merely bend their fall sideways into an orbital velocity? Maybe, I haven't run the numbers.

They might have a better shot if their ship is designed for reentry.

The critical delay

Keep in mind what Sully Sullenberger talked about at the NTSB hearing. He pointed out that all the simulations had the pilots knowing exactly what to do, zero seconds into the event. He said what actually happens in real life is some time figuring what's going on and what to do about it - when they added 30 seconds of "figuring out options" time, every simulation crashed.

Even if they expect it, the vigilance could be hard if there's a long wait for the planet's return. Guard duty is boring. It might be the third-watch crew who's least competent, and things might not get sorted until the captain comes out in his pajamas.

So you have to figure 30-90 seconds minimum for crews to discover the planet has returned, realize their peril, make a plan to thrust away from it, and get the engines spun up / ship physically oriented engines-down. They may not know exactly where the planet will reappear.

That's going to depend a lot on the distance from the planet. Hopefully their captains have thought this through (probably not, it doesn't ever happen so how would they be prepared for it?) What if it's a "race" by competing factions, where you need to be closer than the other guy?

Assuming Earth pops up 300km away, in Sully's 30 seconds, things have gotten rather worse - you have fallen 4.5km and gain 300m/s velocity straight down - that's a kilometer in 3 seconds, a mile in 5. At 90 seconds you've fallen 40km with 900m/s velocity, a kilometer per second! Since you hit atmo' in 150 seconds (2:30), the point of non-recoverability may be sooner than you think.

Damn, that'll be an exciting chapter. Several chapters if you spend time on several ship bridges watching different captains deal with it in a different way.

Maybe there's even a spot for unexpected heroism, where the Vollchon ship throws a tractor beam or grapple to help a Hegemon ship get enough escape velocity.

Why so close?

What? And risk my ship? So much easier to sit 100 million km away with a bag of popcorn. But consider the drama.

Whatever made the planet disappear, something else is going on. It's not that simple. The situation may justify putting ships at risk.

  • A siege doesn't happen in peace. What's the point of the blockading ships sitting at a nice safe distance, when the blockade runners simply position themselves much closer? (and the capital ships may presume the blockade runners may have insider information.)
  • A Mad, mad, mad world treasure-hunt doesn't have one competitor. (at least not ones that get into science-fiction novels). You're safe, you lose.
  • The waiting ships may have a reason to want the planet inside weapons range the moment it returns.

It may be worse. Their science teams may have error or faulty presumptions in their calculations. (this is a new phenomenon, right?) Rogues don't even have science teams.

Lastly, what if the planet is be able to control where it reappears? If so, that is trivial to weaponize, the waiting ships would have no choice whether the planet appears 200km under them. That will surely occur to their science teams. And they may not know whether the planet can control that, if this situation is novel to them.

Speaking of novel, that's up to the writer. Of course, the power of science is that we can avoid drama, but if science fiction novels wanted to avoid drama, they too would stay in spacedock.

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    $\begingroup$ (-1) "When the planet returns, they will suddenly find themselves in a gravity well, with no sideward velocity for an orbit. They will Fall Straight Down while the planet turns under them." This answer demonstrates a very poor understanding of velocity, or the effects of gravity at distance. Ships in the "general vicinity" would not "fall straight down", they would probably experience a minor alteration on their general orbit around the star. Unless they were pretty darn close to where the planet itself materialises, they're unlikely to suffer any drastic effects. $\endgroup$ Jan 11, 2017 at 11:06
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    $\begingroup$ What!? And risk my ship!? Sure, if it was my ship, and nothing was at stake. But with drama in mind, that could matter. If the siege force keeps a fearful distance, the blockade runners position much closer. Or treasure-hunting competitors would vie for "closer than the other guy, but not too close" and we know how that ends. All will have error in their "where will it reappear" calculations, if they have a science team at all. What if the planet can control (weaponize) their reappearance location? The campers might not know how this works. $\endgroup$ Jan 11, 2017 at 17:33
  • $\begingroup$ Well at this stage you risk going well out beyond the implausible to the impossible: moving a planet in and out of a 'pocket dimension' is completely different from altering its physical location and velocity. If you hide something at a specific point travelling at 30km/s in a given direction, you need to bring it back at that specific point and velocity. Anybody with a calendar and its orbital path knows exactly where it's coming back, and can't hang around in that spot for any length of time (ironically, they would fall into the sun). So no, this is a Fantasy answer, not Sci-Fi. $\endgroup$ Jan 12, 2017 at 14:53
  • $\begingroup$ Raise your objection with OP, Your "impossible" was his given of the exercise. I will not debate labeling of soft sci-fi vs fantasy, on a semiconducting supernetwork of devices that fit in my pocket, powered by energy sent over far distances, from magic rocks pushed together, whilst metal birds fly over my head. $\endgroup$ Jan 12, 2017 at 16:12
  • $\begingroup$ No, his given was it going into the pocket and coming back again. It's only you that's added the fantasy elements... $\endgroup$ Jan 12, 2017 at 16:14
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The basic idea behind orbit in Newtonian physics is that the centripetal acceleration required for the satellite to go in a circle around the planet (or other body) is perfectly fulfilled by gravitational acceleration. That is,

$$G\frac{M}{r^2} = \frac{v^2}{r}$$

This means that these satellites (space ships) are not supplying any power if they're in orbit, but "coasting" (so to speak). It also means that ships at the same altitude must have the same speed.

For an in-orbit dogfight, what this means is that maneuvering is something that won't happen much, and when it does, changes in speed must either be accompanied by changes in altitude or fuel spent to maintain altitude. For example, speeding up means decreasing altitude and slowing down means increasing it. Or to keep altitude constant, speeding up means directing thrust away from the planet and slowing down means directing thrust toward the planet. Saving fuel is important, so we'll likely see more of the former than the latter.

If we suddenly take away the planet, the conditions of orbit are no longer satisfied. But the basic equation of physics,

$$F = ma$$

should still apply. As such, we know that since no force is applied to the space ships (most will be "coasting", remember?), they will simply continue on their present trajectories. They will no longer be moving in a "circle," though, but a straight line, just like a ball on the end of a string when the string breaks.

If one ship is chasing another, they will both head off in ever so slightly different directions. Any kinematic shots that were in-transit when the planet disappeared will likely still make contact unless they're fired from particularly long range. In order for that kind of attack to be effective, it has to outrun the ship it's trying to hit, which means it's shot in a trajectory closer to straight-line than the orbit anyway.

But as these ships are floating off on slightly different trajectories, no doubt the pilot or navigation system will notice and make corrections to either continue dogfighting or retreat because what the #@*$ just happened.

I'm no expert on gravitational waves, but I don't think they should be too much of a consideration. With how weak gravitational force is and how quickly the waves propagate, it would take a much larger object disappearing and a much larger satellite for this to make too much of a difference. Geosynchronous orbit on Earth is around 35 786 km. Propagating at lightspeed, gravitational waves would cover that distance in roughly $\frac{1}8$ of a second, and cover the distance across the ship's y axis in no time. There will be some distortion of space but nothing intense enough to be noticed.

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Let's talk side effects.

As noted by others, the ships in orbit would cease orbiting and start moving in a straight line, the exact motion dependent on how far they were from the planet. The ISS for instance would have a considerable straight line speed and would complete an axial rotation once every 90 minutes. Something the distance of the Moon away wouldn't experience a particularly notable difference in orbit until minutes or hours have passed. That bit is boring.

First interesting side effect: Gravity, or sudden lack of, propagates at the speed of light. Objects will continue to feel the pull of gravity from the now missing planet until the difference has propagated the distance between them. Mostly irrelevant due to the tiny magnitude of action during that period of time, but still kind of interesting. Would be more of an interesting point if the ships were more like 1 AU or more from a star and the star disappeared at a known time.

Second interesting side effect: How far out does this disappearance affect? Does the planet have anything in orbit? In the case of the Earth, if the effect were confined to the right distance, the planet and it's atmosphere could be "removed" while leaving objects in orbit in place. Nearby ships would then have the fun of thousands of high speed metal objects suddenly moving at tangents to the planet's previous position. I imagine it could be considered a solar fragmentation grenade.

Third interesting side effect: How does the planet disappear? If the disappearance involves stretching of space time around the planet then you have some interesting relativistic consequences to consider. If the mass of the planet disappears but space remains the same you instead have a large hard vacuum in a spot where the intra-solar vacuum wasn't all that empty, I'd expect this possibility to result in something like an implosion that would affect anything close by though I don't know how much by.

Fourth interesting side effect: Anything mechanism resulting in the disappearance of a planet should require a massive amount of energy, which in turn should result in a massive amount of waste energy being emitted. I think it would be justifiable to assume the space previously occupied by the planet would now contain energy equivalent to some portion of the planet's mass. How much energy depends on how strict you want to be but I'd expect a pretty substantial amount of gamma and X-ray emission. If you wanted to be very strict you could plausibly say that the planet was replaced with an amount of energy equivalent to the entire planet's mass, that's an explosion of considerable magnitude.

Fifth interesting side effect: 3 words: Van Allen Belts. All that lovely radiation is confined by the influence of the planet. No more planet, no more confining influence.

Hope that gives you something to work with.

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The ships will fly off at their current orbital velocity, as others have explained.

However, you can easily have the effect justifiably be more damaging: in orbit, the ships posses a large angular momentum. We might suppose that the defensive mechanism goes to pains to make sure the planet is treated gently enough, and in any case it has a huge moment of rotational inertia so the imposed change has little reaction.

But the disappearance causes each ship’s orbital angular momentum to be dumped into the ship itself! And in a fairly random haphazard dumping: so various parts of the ship suddenly are sent spinning with great force, tearing the ship apart.

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  • $\begingroup$ If I gave the friendly ships warning before vanishing the planet, could they counteract this in any way? $\endgroup$
    – Lu22
    Jan 11, 2017 at 11:19
  • $\begingroup$ I suppose. It’s just a suggestion for your plot. $\endgroup$
    – JDługosz
    Jan 11, 2017 at 11:34
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As other answers have noted, if you think it would be cool for your planet eraser to cause a displacement effect then go for it; no one could dispute it without knowing the exact science behind the superweapon, which means you get a lot of license. (And let's not forget the Rule Of Cool!)

I don't think that you'd necessarily see any effect in 3D space, because the orbiting ships were all being accelerated "downwards" by gravity and so the planet doesn't play much of a role at all in their relative motion unless the direction of that "downwardness" is significantly different for each of them. If a ship is approaching another ship, it will still do so once neither is being pulled downward. If however your ships are so far apart that their vectors are significantly different, so that they're effectively fighting around the planet rather than above it, then with the planet's gravity removed the ships would start to drift apart if neither corrected their course. In other words, if ships A & B are dogfighting somewhere and ships C & D are dogfighting a quarter-orbit away, neither of the dogfights would be noticeably affected by the loss of the planet, but the two dogfights would drift apart. This could easily be corrected for, just as if the other ships had simply changed direction.

So far, so unsexy. Your superweapon could still have side effects, however, depending on how it works -- which is up to you! You don't have to simply replace the planet with vacuum and call it a day. If the entire section of 3D space that contained the planet has now been "pinched off" by the superweapon, then the space itself is now inaccessible. In other words, if you have ship A and ship B who were very close to the planet on opposite sides, those ships would now find themselves right next to each other as the space that separated them has been pocketed off. There isn't just a big empty space between them, because that space is where the planet still is. The whole thing has just been pulled out of our plane of 3D space.

By this logic, if you had a dense shell of spaceships around the planet, on the event horizon, they would find themselves smooshed together into a small dense blob. Either handwave this away, or don't draw attention to it.

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  • $\begingroup$ This story tends to contain more Rule of Cool than science at this point. As much as I'd like a collapsing space effect, most of my protagonist army has a personal device such as the one used, so that would complicate things immensely. $\endgroup$
    – Lu22
    Jan 11, 2017 at 10:52
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I don't think anything would happen to them at all, but I'm not a physicist.

If they turn on their engines, then it's a moot point. They can go where they want, since they already possess enough power to overcome solar system gravity.

If they don't turn the engines on, and since the planet was orbiting a sun, the ships are, too. They will still orbit the sun. For a time, their orbits will be unstable because they were primarily orbiting something that's now gone. Their orbits must adjust to a new focal point.

They would probably settle into a stable orbit around the sun. But it's possible that one or both would be flung out into the solar system, or inward. They could become captured by another celestial mass and slowly settle into orbit there.

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  • $\begingroup$ I don't think there's really an adjustment to the new focal point: they were in free fall with the planet as the most significant local body. Now they're still in free fall, with a velocity that depends exactly where they were in their planetary orbit at the time. Barring some other planet having an effect, this new velocity gives them either a collision course with the star, or an orbit around it, just like anything else in free fall below escape velocity. If it's a near miss then the new orbit is highly elliptical. If they don't like the new course then, as you say, they use their engines. $\endgroup$ Jan 11, 2017 at 3:22
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Assuming the planet has no moons nearby which might substantially alter things, they would just go into a solar orbit of roughly the same path as the former planet.

Their position in the orbit of the planet will affect slightly their final orbit. Assuming an equatorial and prograde orbit, if the craft were on the opposite side of the sun from the planet when the planet disappeared, that tangential velocity would be added to their orbital velocity, resulting in an aphelion slightly higher than that of the former planet. If the craft were on the side of the planet closer to the sun, its tangential velocity would be subtracted from the orbital velocity, resulting in a perihelion slightly lower than the planet had.

You could even get into a situation where the tangential velocity cancels out the eccentricity of the planet's original orbit, and puts the craft into a more circular orbit than the planet had.

But really, the end result is basically nothing will happen, unless the planet is of a huge mass such that the tangential velocity of the orbiting fleet is large relative to the orbital velocity of the planet around its star.

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  • $\begingroup$ This answer would greatly benefit from actual numbers, such as Earth's satellites' different orbital velocities (from LEO to the Moon) vs. Earth's own orbital velocity around the Sun. $\endgroup$
    – hyde
    Jan 10, 2017 at 23:06
  • $\begingroup$ Earth's orbital speed (around Sol): 108,000 km/h. ISS orbital speed around Earth: 27,578 km/h (higher orbits = slower speed) So ships definitely won't just go into a solar orbit - assuming all motion is co-planar, they'll immediately start moving on more elliptical orbits that diverge from the planet's (previous) orbit. $\endgroup$
    – Spike0xff
    Jan 11, 2017 at 16:29
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As has been at least hinted at more or less precisely/coherently, the moment the planet vanishes then any object actually orbiting that planet will continue moving in a straight line* in the direction they were moving when the planet disappeared.

What happens next depends on a lot of things. If they are flung towards another planet, they may get trapped in an orbit around that planet, or "slingshot" around it. They may smack straight into it. They might get flung into an orbit around the sun the planet is orbiting - assuming the planet is orbiting a sun. They may - and could be the most likely outcome - leave the solar system altogether, maybe bending round a few of the other bodies in the system on the way.

But that's objects. What about ships? Well, Star Trek-y, Culture-y, or Star Wars-y ships that basically scoot about all over the place at ridiculous speed with effectively infinite fuel won't care much at all, they'll just scoot off somewhere else.

But for a more realistic ship, as you might find in Arthur C Clark, or The Expanse, the effect would be profound. What if their planned itinerary required refuelling at that planet and they didn't get the chance? What about an orbital shuttle or space station? Even under ideal conditions suddenly hurtling into the dark at thousands of kph with no other planet closer than a few months away, and maybe with not even enough fuel on board to "stop"** let alone figure out how to reach safety, is Bad News. And if all other craft in the same boat are similarly realistic in their limits, the chance of rescue or aid is slim even from someone fully stocked up.

And there in you might find the real drama. How to combine scarce resources - time, fuel, food, oxygen - with orbital mechanics to weave an improbable path to safety.

*Very very Straight-ish anyway. Gravitational influence from other bodies will be extremely slight at typical solar system distances.

**"Stop? Relative to what?" you cry. Yeah I know, the point is you're instantly in the situation of having a lot velocity that may well be very unhelpful.

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Gravitational effects of said planet will disappear instantly. However, orbiting vehicles will experience the change in gravity effect based on their orbital distance since gravity travels at the speed of light.

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  • $\begingroup$ This is perfectly true and a good start to an answer, but it doesn't yet fully answer the question. Would you be able to expand this so that it explains how this (the change in gravity) actually affects the orbiting starships? Thanks $\endgroup$ Jan 11, 2017 at 22:28

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