Wednesday, July 20, 2011

Gravity inside solid (and hollow) bodies

I have in the past talked about gravity on the surface of a planet and ways of creating artificial gravity on things like space stations. Today, I thought I'd talk about what happens when you start to move into a planet, particularly deep beneath the surface (perhaps more applicable to asteroids, small moons and other bodies without plate tectonics, although I can envisage a story taking place inside a gas giant that might need to take this sort of thing into account). Alternatively, this also applies to giant hollow spheres, y'know, if that's your Big Dumb Object of choice.

Alternate titles for today's post could be "Why Jules Verne's Journey to the Centre of the Earth was inaccurate from a physical point of view (as well as biologically), even taking the physics of the day into account" or "The Shell Theorem". I thought I'd stick with something both short and explanatory.

Hollow Shell

OK, this is the simplest case so I'm going to start here. Consider a large hollow sphere. Make it massive enough that it has an appreciable gravitational attraction from the outside (Hmm... no one really talks about the gravitational field of the Death Star, probably because Star Wars is full of unexplained artificial gravity, but that sort of structure would be ideal to keep in mind).

From the outside, it would behave the same way as a non-hollow sphere of the same mass. Basically, with anything roundish you can just pretend that all the mass is at the centre and do your gravitational force equations from there. That's what we do for planets. And it works fine if you're floating around outside or standing on the surface, but things are a little different once we move inside. Below, I've put a little doodle of a shell with some labels to make it clearer what I'm talking about.

The grey circle is our spherical shell. R is the radius of the shell, r is the distance from the centre to the location of interest. I used snowmen with jaunty hats as stand-ins for stick figures because I found them in the list of symbols in Helvetica. You can pretend they're space-suited people with comms devices on their heads (or brain slugs) if you prefer.
So outside the sphere, at positions A and B, a regular calculation of the gravitational field based only on the mass, M, of the sphere and the distance, r, of the snowman from the centre of the sphere. The acceleration due to gravity is then:

g is acceleration due to gravity, G is a constant, M is the mass of the sphere and r is the distance from the centre. The equation applies for objects outside of the sphere (ie positions A and B) in the image above.
The only difference, really, between A and B is that at B the distance from the centre is equal to the radius of the sphere (r = R) so you can just substitute R for r.

Let's look at C now. In the centre of the sphere, the gravitational pull from each bit of shell is exactly balanced by the gravitational pull of the bit of shell directly opposite. Easy. All the gravitational forces from all the bits of the shell cancel out and overall there is no gravitational pull in any direction so the centre snowman would feel weightless. (Technically not free-fall as we could use in orbit since there is no falling involved. Just saying.)

OK, what about snowman D then? This is the really cool part. As with C, every bit of the shell pulls on him in different directions. You might think that D would then fall towards the nearest edge of the shell, but this isn't the case. Although D is closer to some edges than others, the forces still all cancel out. How nifty is that? The actual mathematical derivation is a bit complicated but you can read about it on wiki if you are interested. Instead, I present a graphic that qualitatively explains it in a very hand-wavey way:

I highlighted two bits of shell above which would cancel each other out. Remember that as well as mass, the gravitational pull also depends on proximity and in fact drops off with the inverse square of proximity (inverse square means 1/r2). There is less mass in the closer bit of shell but the larger blob of shell is larger in such a way that the further distance cancels out the increased mass exactly. Also my drawing is fairly rough but hopefully you get the idea.

Wow, that turned out even more vague an explanation than I'd hoped. Oh well.

One more thing I offer without any explanation (other than Gauss's Law mentioned and linked-to in brackets) is this: The shell theorem is not confined to only spherical shells. Within any shape of enclosure, there will be no gravitational force due to the enclosure so long as no part of the enclosure extends inside itself. What I mean is, it has to be hollow. Also, it has to be closed, meaning it has no holes in it (but I think that holes with a symmetric counterpart on the opposite side would be OK within reason).

Inside solid bodies

Now for a solid sphere. Hopefully, it is obvious by now that outside the solid sphere gravity behaves normally (the same as points A and B on the first diagram). The next diagram only includes points C and D.

Again, R is the radius of the now solid sphere and r is the distance from the centre to the snowman/brain slug of interest.
For the sake of plausibility, let's just assume our snowmen/brain slugs are in a little spherical cavity inside the planet (and not buried under a tonne of rock). For C at the centre, we can just pretend he's inside a particularly thick shell. As with the thin shell considered earlier, this gives us a weightless snowslug at the centre.

D is a little bit more complicated and we need to consider it in two parts. First the section of the sphere that's outside of D's distance from the centre. That is, the shell of thickness R - r, if D is r from the centre, can be treated just like a shell from the first example and hence cancels out. That leaves us with the bit of sphere closer to the centre than D is. This bit we can treat the same way as if it were a regular sphere we were standing on top of:

As above, but here M(r) is the enclosed mass; the mass within r, the distance from the centre.
Easy! Even better, if we assume that the sphere is of uniform density (lots of things—Earth, gas giants, stars—aren't but asteroids probably are), then we can simplify this even further since the enclosed mass then becomes a simple function of the volume and, skipping the maths, we end up with a linear relationship: a straight line that runs from the surface gravity at r = R to zero at r = 0. Nifty!

Remember, R is the radius of the sphere (which shouldn't change!) and r is the distance from the centre.
So now, if you have a sizeable asteroid and you want to bury a base half way to the centre, you can work out what acceleration due to gravity your colonists/soldiers/evil genii will experience. Huzzah!

What was with that Jules Verne reference at the start?
Oxygen does not get denser closer to the centre of the Earth. Really, I shouldn't complain too much since he tried (outside of, y'know, the dinosaurs and the ability to walk the long distance to the centre of the Earth) but in some ways it's the trying not hard enough that sometimes makes it worse. But props for using the speed of sound to work out how far apart they were. -1 mark for not also taking air density into account.

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