Some cool orbital mechanics I learned

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So there's a mod that adds a planet named Inaccessible, which rotates at a faster speed than its escape velocity.  This means that, in a counter clockwise orbit, decreasing your orbital velocity will increase your surface velocity, and decreasing your surface velocity to catch up with the planet's rotation will put you on an escape trajectory.  Though already hard to do, the easiest solution is to land on the poles, but you could also land on the equator with a constant Radial-In burn to bend your escape trajectory down to the planet's surface, and then clamp yourself to it with some kind of harpoon, like KAS, so that you don't fly off.  Because of the rotation, when you're on a planet you, for all intents and purposes, experience negative Gs, throwing you off the planet.

So I showed this video by Scott Manley to my dad:

And he brought up a really good point about it that the planet must have the shear strength to keep from tearing itself apart in order to stay as a planet, or else each layer would separate from another because they're spinning at different velocities.  And this gave me a really cool idea for some kind of crazy Star Wars planet;

What if the outer layers separated from the inner, creating a gap between the core(not the actual core, but the inner 2/3 or so of the planet) and the crust, and the core spun at a low enough rate for it to have gravity, while the crust spun fast enough to expand somewhat and create a few miles of gap between the two, and have artificial gravity on the inside, so that the planet could be colonized on the surface of the "core" and the inside of the crust?  Then, in the middle between the two, what if there was a ring station in orbit of the core that rotated in sync with the crust, and the gap between the two had an atmosphere?  So then they could drill giant holes in the crust for ships to get to the station.  How cool would it be from the perspective of the inside of the crust to see a Star Destroyer come through a big hole(with a shielded entrance to keep the atmosphere in) and dock at the station way above you?  That'd be awesome.

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If not connected the outer shell would not simply 'orbit' the inner core. Instead they would be two separate bodies orbiting each other. They would either drift apart and collide.

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There's no way it could happen naturally-- naturally occurring materials don't have that kind of strength, and will slump to a shape of hydrostatic equilibrium.  It's why planets are spheres, even though they're made out of hard rock.

Certainly you could posit a sci-fi situation where there's some sort of artificial structure around a planet, made of some science-fictional material with unreasonably high strength.

There's also the matter that a planet inside a spherical shell isn't a dynamically stable situation.  There's nothing holding the planet in the center, i.e. nothing stopping it from wandering until it collides with the shell.  (That's because a spherical shell exerts no gravity on its interior-- the interior space is a zero-gee environment.)  But of course, if you're positing a science-fictional artificial structure, you could also posit some sort of sci-fi tractor beam or something to keep it in place.

...In any case, we're talking about physics-type stuff rather than playing KSP, so this seems to belong over in Science & Spaceflight.  Moving.

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1 hour ago, Snark said:

There's also the matter that a planet inside a spherical shell isn't a dynamically stable situation.  There's nothing holding the planet in the center, i.e. nothing stopping it from wandering until it collides with the shell.  (That's because a spherical shell exerts no gravity on its interior-- the interior space is a zero-gee environment.)

Actually... it's worse than that, the situation is dynamically unstable - as soon as the outer shell gets off center, the planet's gravity starts to pull it further off center...

1 hour ago, Snark said:

But of course, if you're positing a science-fictional artificial structure, you could also posit some sort of sci-fi tractor beam or something to keep it in place.

Something like this happened after Larry Niven wrote Ringworld - somebody figured out that it was unstable (for the reasons given above)...  so he had to retcon into place an attitude control system.  (IIRC, Bussard ramjets mounted on the rim.)

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17 minutes ago, DerekL1963 said:

Actually... it's worse than that, the situation is dynamically unstable - as soon as the outer shell gets off center, the planet's gravity starts to pull it further off center...

Nope.  It's neither stable (gets pulled towards the center) nor unstable (pushed away from it).

The gravity field inside a hollow spherical shell is zero.  So the sphere inside the shell would basically be floating in free-fall, with nothing pushing it either towards or away from the shell.  If you stuck it in the middle at zero relative velocity, it'd probably stay there for a good while out of inertia, but over time there's nothing stopping it from drifting out of alignment.

17 minutes ago, DerekL1963 said:

Something like this happened after Larry Niven wrote Ringworld - somebody figured out that it was unstable (for the reasons given above)...  so he had to retcon into place an attitude control system.  (IIRC, Bussard ramjets mounted on the rim.)

Yup, loved that.  He put the attitude control system into Ringworld Engineers, and actually made it a significant plot device.

Not quite literally for the reasons given above-- I was speaking of a hollow spherical shell, which is at the "neutral stability" point.  The Ringworld case involves a ring, not a sphere, and it's not mathematically the same.

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Well by the time you get to a shell around a core you're deep into sci-fi.  Especially if it was used in Star Wars you wouldn't have much to worry about as far as orbital mechanics are concerned since everything flies around like a fighter.

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What everybody says, I suppose.

- Spinning faster than orbital velocity -> periodic mass-loss from flung materials

- Rigid body rotation w/ such speed -> the planet will resize

One interesting note : the planet should've differentiated to a "core" planet w/ valleys + a strip of low lying "rings". Because after some time the rotation rate will be lower than the orbital velocity, the mass loss would slow down; but ejecta from previous ones would make a ring. Quite bizarre there.

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4 hours ago, Snark said:

Nope.  It's neither stable (gets pulled towards the center) nor unstable (pushed away from it).

The gravity field inside a hollow spherical shell is zero.  So the sphere inside the shell would basically be floating in free-fall, with nothing pushing it either towards or away from the shell.  If you stuck it in the middle at zero relative velocity, it'd probably stay there for a good while out of inertia, but over time there's nothing stopping it from drifting out of alignment.

The hollow shell wouldn't impart any change due to gravity, but the inner sphere would accelerate the shell if it got off center.

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6 hours ago, Snark said:

There's no way it could happen naturally-- naturally occurring materials don't have that kind of strength, and will slump to a shape of hydrostatic equilibrium.  It's why planets are spheres, even though they're made out of hard rock.

Certainly you could posit a sci-fi situation where there's some sort of artificial structure around a planet, made of some science-fictional material with unreasonably high strength.

There's also the matter that a planet inside a spherical shell isn't a dynamically stable situation.  There's nothing holding the planet in the center, i.e. nothing stopping it from wandering until it collides with the shell.  (That's because a spherical shell exerts no gravity on its interior-- the interior space is a zero-gee environment.)  But of course, if you're positing a science-fictional artificial structure, you could also posit some sort of sci-fi tractor beam or something to keep it in place.

...In any case, we're talking about physics-type stuff rather than playing KSP, so this seems to belong over in Science & Spaceflight.  Moving.

Some asteroids have rotation so fast but they are solid rocks. it would not work for anything who is rounded because of gravity.

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12 minutes ago, Xeorm said:

The hollow shell wouldn't impart any change due to gravity, but the inner sphere would accelerate the shell if it got off center.

Actually, no, it wouldn't.  See my earlier post:  there's no gravitational field inside a hollow shell.  Gravity, like all forces, is by definition bidirectional.  "The shell exerts no force on objects within" is exactly tantamount to "objects within exert no force on the shell."

If you don't believe it, do the integral-- it works out.

To be clear:  For any bit of matter inside a hollow spherical shell, there is net zero gravitational force between the shell and the object inside.

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I'm not sure that is true. If you have a sphere 100m in radius, and you are 1m from the surface of the sphere (on the inside), since the mass of the sphere at the far side is much farther away than the mass at the near side (1m away), wouldn't there be a net attraction to the edge?

From 1m outside the sphere, there would be gravitational attraction to the sphere that woudl get greater as the distance to the sphere approaches 0. Is it really going to become exactly zero and stay there as soon as you cross the boundary?

Lets go 2D:

If we made the ring 10,000km thick it would be of comparable thickness to the Earth. If we then made the diameter equal to 1 AU - if you were place 1 m from the inside surface, would you really expect to float over the surface of the sphere?.

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1 hour ago, KerikBalm said:

I'm not sure that is true. If you have a sphere 100m in radius, and you are 1m from the surface of the sphere (on the inside), since the mass of the sphere at the far side is much farther away than the mass at the near side (1m away), wouldn't there be a net attraction to the edge?

From 1m outside the sphere, there would be gravitational attraction to the sphere that woudl get greater as the distance to the sphere approaches 0. Is it really going to become exactly zero and stay there as soon as you cross the boundary?

Lets go 2D:

If we made the ring 10,000km thick it would be of comparable thickness to the Earth. If we then made the diameter equal to 1 AU - if you were place 1 m from the inside surface, would you really expect to float over the surface of the sphere?.

It is true. I've done the integrals and, as weird as it sounds, it is true.

If you don't fancy doing it yourself, see below:

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7 hours ago, KerikBalm said:

I'm not sure that is true.

That's okay, even if you're not, Isaac Newton was  (Thanks to @Steel for the wiki link, that's an excellently written article, and even walks through the mathematical proof.)

The fun part about this theorem is that it's actually pretty simple to prove, as long as you know basic trigonometry and high-school calculus.  It's a fairly straightforward combination of geometry with a simple integral-- no fancy differential equations or the like are needed.

Maybe you have to be an Isaac Newton to invent the calculus, but you don't have to be him to actually do the math.  Ordinary mortals can do it.  Really cool!

7 hours ago, KerikBalm said:

If you have a sphere 100m in radius, and you are 1m from the surface of the sphere (on the inside), since the mass of the sphere at the far side is much farther away than the mass at the near side (1m away), wouldn't there be a net attraction to the edge?

Well, the short answer would be "go click the link, the math says it won't."    However, if you'd like a more hand-wavy, less mathy explanation:  Yes, the mass of the sphere at the far side is much farther away.  But there's also a lot more of it-- i.e. the distant part is bigger and therefore more massive than the close-by part.  And it turns out, through a happy mathematical "coincidence" (if such a thing can even be said to exist, in math) that those two effects precisely cancel out, for a spherical shell.  If you sum up all the contributions from all the pieces of the shell (which can be done with a simple integral), it turns out that the combination of distance, size (therefore mass), and trigonometric angle all cancel out, precisely.

Note that this same math is what makes it so that we can treat the mass of a planet as being a point mass located at its center, even if you don't know anything about its internal density distribution other than the assumption that it's spherically symmetric.  That's not a convenient approximation-- it's actually, literally mathematically true.  It's the same math.  If there were gravitational attraction inside a spherical shell, then you also couldn't treat spheres as point masses when you're outside them.

Note that this particular bit of physics and mathematics is a direct consequence of the fact that gravity follows an inverse-square law with distance.  Change the exponent, and you change the result.  For example, let's say you have a hollow spherical shell.  If gravity went with the inverse distance (instead of inverse squared), then there would be gravity inside a hollow spherical shell:  you would get pulled toward the center of the shell.  And it gravity were an inverse cube instead of an inverse square, you would also have gravity inside:  you would get pulled away from the center, to the inner surface of the sphere.  But with an inverse square, it turns out that the force is precisely, mathematically zero everywhere inside the shell.

7 hours ago, KerikBalm said:

From 1m outside the sphere, there would be gravitational attraction to the sphere that woudl get greater as the distance to the sphere approaches 0.

Yes.

Though it's worth noting that saying "distance to the sphere approaches 0" is mathematically misleading, here, since the distance to the sphere's surface is pretty much irrelevant.  What matters is the distance to the sphere's center.  The gravitational attraction towards the sphere when you're 1m from its surface would be only microscopically bigger than the attraction when you're 1000m from its surface, because that's the difference between being  150,000,000,001 meters from the center, versus 150,000,001,000 meters.  A difference of less than 0.000001%.

To find the attraction towards the sphere, you'd take the total mass of the entire spherical shell, multiply by the universal gravitational constant, and divide by the square of your distance from the center.

7 hours ago, KerikBalm said:

Is it really going to become exactly zero and stay there as soon as you cross the boundary?

Yes.

7 hours ago, KerikBalm said:

Lets go 2D:

Sure, if you want, but then the discussion is moot, because the math is completely different.  A 2D ring is not the same as a 3D spherical shell.  You will get different mathematical results, and I believe that the net gravity inside a hollow ring is non-zero, even though it's zero inside a spherical shell.

7 hours ago, KerikBalm said:

If we made the ring 10,000km thick ...  would you really expect to float over the surface of the sphere?.

Sorry, you're mixing apples and oranges-- you just referred to a ring and a sphere (which are completely different things) as interchangeable.

However, if you'd like to keep the analogy mathematically relevant, let's talk about a sphere, not a ring.  I'll re-phrase your statement, using a sphere instead of a ring.

"If we made a spherical shell 10,000km thick, it would be of comparable thickness to the Earth. If we then made the radius equal to 1 AU - if you were place 1 m from the inside surface, would you really expect to float over the inner surface of the sphere?"

And the answer is:  Yes, that's exactly correct.  That's what @Steel , Isaac Newton, Wikipedia, and I are all telling you.

That's what "zero gravity inside a hollow spherical shell" means.  Neat, huh?

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If you want to move from a 3D world to a 2D one, you would - I guess - have to change gravity to a 2D variant as well, and my intuition says it would likely fall off linearly (not squared!) with distance, in which case the same mathematical proof would turn out that in this 2D world gravity between an object inside the ring and the ring would always be a net zero.

Now, this is my intuition, not a proof of any kind, and therefore is likely wrong. It depends a lot on how gravitation works in detail, especially if the number of spatial dimensions play a role in how gravitation crosses distances.

For all I know the thingies that transmit gravity could be like soap bubbles growing into every dimension, but expire like radioactive decay, causing a distance squared effective strength, no matter how many dimensions there are.

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The supergiant planet Mesklin in Hal Clement's Mission of Gravity is a similar example: it has a solid surface with a very high rotation speed, resulting in roughly 3 gees of gravity at the equator and several hundred gees at the poles. Clement depicts the planet as an extremely oblate spheroid, but in reality it would be almost a flat rotating disc.

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Posted (edited)

So, OK, it's clearly impossible with real materials.

And, for those who don't want to do the integral proving that the outer shell imparts no gravity on anything inside it, you can get halfway there by thinking about it in terms of subtending angles. Choose some spot to look out from the inner body, and some angle of deviation for your gaze (say, 10 degrees). Then, when the inner body is in the centre, the cone of your gaze will cover (subtend) a certain amount of the outer shell. Now move the inner body further away, by having it drift in the opposite direction of your gaze. Then, the outer shell on that side will be further away, and therefore impart less gravitational force - but the same angular cone will subtend more of it, meaning that more of it will pull along that cone. It so happens that the amount of less pull because of distance and amount of more pull because of greater subtension exactly cancel one another.

Also, for those who have done undergrad E&M: gravitational potential works exactly the same way as electric field potential, so you've already done this proof.

Edited by Jovus

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7 hours ago, weissel said:

If you want to move from a 3D world to a 2D one, you would - I guess - have to change gravity to a 2D variant as well, and my intuition says it would likely fall off linearly (not squared!) with distance

Yes, that seems like a pretty reasonable assumption.  That's all it is, of course, an assumption, and can't be definitively "proved" one way or the other because there is in fact no such thing as a 2D universe, and we can't literally apply any of our physics to such a situation because the laws of real physics are inherently three-dimensional.  We can make reasonable extrapolations, but it's important to realize that when we do so, we're constructing plausible fiction rather than what "really would" happen.

By far the most thoughtful, detailed, "what would really happen in a 2D universe" account I've seen was the book Planiverse, by A.K. Dewdney.  This was back in the 80's, not sure how possible it would be to track down these days.  It wasn't exactly a NYT bestseller.

7 hours ago, weissel said:

in which case the same mathematical proof would turn out that in this 2D world gravity between an object inside the ring and the ring would always be a net zero.

Nope.  The math is different in 2D than it is in 3D.  Unlike "what would the physics be"-- which is a matter of plausible fiction-- this is math, which means we can come up with a definitive answer.  Once you've specified your assumed laws of physics for your 2D world-- e.g. "gravity decreases linearly with distance"-- then you can unambiguously work out the mathematical consequences thereof.

7 hours ago, weissel said:

Now, this is my intuition, not a proof of any kind

Correct, it's not a proof, and I don't think anyone can prove such a thing.  No one can "prove" you either right or wrong if you say "in a 2D universe, gravity would be 1/R instead of 1/R2."  But we can certainly just stipulate it and see where the math takes us from there.

One thing that would fall out of such an assumption is that general relativity is seriously horked.  Why?  Because if gravity is 1/R, that means that escape velocity is infinite.  Imagine that you're standing on the surface of a 2D planet-- no matter how small, no matter how weak its gravity is.  You fire a projectile upwards.  No matter how fast you fire it... it will eventually slow down, stop, and fall back.  Might take a really really really long time and distance to do so, but it will never escape.

When escape velocity is infinite, it means that it's therefore faster than the speed of light (however fast you posit that is in our hypothetical 2D universe... and never mind that the very concept of "light" is problematic, since it's an inherently 3D phenomenon in our universe, since it consists of an electric field and a magnetic field at right angles to each other and to the direction of propagation).

Since every individual 2D "atom" therefore has an infinite escape velocity, that means that it is, in effect, a black hole of infinite size, which in terms of general relativity means causality is out the window and it's all cloud-cuckoo land.

Doesn't mean we can't have interesting speculation about what a 2D universe might be like, of course... just that we're going to have to make a lot of arbitrary stipulations, because you simply can't migrate our physics there, intact.  The laws of physics are heavily inter-related, and you can't pick and choose.  Either you have to take the whole package, or make stuff up from scratch.

7 hours ago, weissel said:

For all I know the thingies that transmit gravity could be like soap bubbles growing into every dimension, but expire like radioactive decay, causing a distance squared effective strength, no matter how many dimensions there are.

Or, indeed, for all anyone knows, because there's no authoritative right answer for the physics.  So your speculation is as good as anyone's.

Once a particular set of physics laws have been stipulated, however, the math then will be unambiguous.

43 minutes ago, sevenperforce said:

The supergiant planet Mesklin in Hal Clement's Mission of Gravity is a similar example: it has a solid surface with a very high rotation speed, resulting in roughly 3 gees of gravity at the equator and several hundred gees at the poles. Clement depicts the planet as an extremely oblate spheroid, but in reality it would be almost a flat rotating disc.

Excellent book and highly recommended, to anyone who may be reading this.

As for what shape a really fast-spinning planet would be, that's a hard problem.  Not just that it's really difficult math, but also that it depends a lot on what the internal density distribution is, which in turn depends a lot on the properties of the materials that it's made of.  For example, a planet made of near-uniform density material wouldn't have the same shape as one that has a hyper-dense nucleus surrounded by an envelope of much less dense stuff.

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38 minutes ago, Snark said:

As for what shape a really fast-spinning planet would be, that's a hard problem.  Not just that it's really difficult math, but also that it depends a lot on what the internal density distribution is, which in turn depends a lot on the properties of the materials that it's made of.  For example, a planet made of near-uniform density material wouldn't have the same shape as one that has a hyper-dense nucleus surrounded by an envelope of much less dense stuff.

You pretty much have to start with the formation of the planet and come up with a way for it to be spinning that fast, and allow the planetary evolution to guide you from there. Because composition, density gradients, and rotation are all interdependent and arise based on planetary formation route.

That itself is an interesting question. How could you get a planet spinning ridiculously, ridiculously fast? What mechanisms would produce that result? Conservation of angular momentum is a cold and inflexible mistress.

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Posted (edited)

Exactly. Such a planet would have no chance to form. And to answer you question, @sevenperforce, sure, contraction. Cooling would lead to higher rotation speed. If theoretically a body formed from a fast rotating disc with already a high momentum and cooling and contraction would bring it over the limit then it would fly apart edit: or loose mass from the equatorial parts until equilibrium is regained. But planets don't form like that, they from inside a disc by collisions and the disc usually has a sun in the centre.

A central body from a disc could go through such a process, but i know of none and even white dwarfs or neutron stars don't rotate fast enough to overcome their gravity through centrifugal force. Edit: in exchange for the momentum gain they lost their atmospheres in the process of contraction.

Btw. our gas planets rotate quite fast, which makes Saturn remarkably and observably flat.

Now someone could claim that a series of well timed impacts adds to the angular momentum of a rotating body, but such a process is improbably, to say the least. A single impact that resets a planet forming process is far more likely. Earth almost had it, Mars maybe as well i read a few weeks ago from a paper that dealt with Mars' moons.

But where star destroyers can fly through a big hole in a shell around a core with a ribbon in between there surely can happen strange things as well :-)

Edited by Green Baron

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I'm surprised no one mentioned the words "Dyson Sphere" in this conversation.

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Even if a "sphere-within-a-hollow-sphere" thing was achieved somehow, I doubt that the core and the shell could rotate at different rates. At least not for long. Wouldn't the atmosphere between them create quite a lot of friction between the two, eventually slowing the shell or accelerating the core until their speeds were somewhat matching?