Living on the Inside of a Hollowed-Out Planet: Gravity, Crust Thickness & More

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Living inside a hollowed-out planet would result in no gravitational force acting on occupants due to Gauss's Law, which states that a spherical shell exerts no gravitational pull on objects within it. The crust's thickness would need to be significant to create usable gravity on the inside surface, but even then, gravitational effects would vary based on the shell's density and shape. If the crust were unevenly thick, occupants might experience varying gravitational pulls, potentially allowing them to walk on the thicker side while becoming lighter or weightless on the thinner side. Additionally, if the planet rotated, occupants could feel a centrifugal force similar to gravity, but this effect would diminish away from the equator. Overall, the scenario raises intriguing questions about gravity, movement, and living conditions in such an environment.
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This seemed teh best sub-forum for this question.
Feel free to move it if it is not.

What would be the gravitational effects of living on the inside plane of a hollowed-out planet with a huge crust?
How thick would the crust have to be in order to have a great enough gravity to allow you to use the inside surface of the planet as a living surface?

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How would I determine what the difference between gravitational forces on a body at points A, B and C?

Other than having no light from an eternal source (and the requirement of the planet not having a liquid core), what concerns/difference from living on a planet's outside surface would there be with this scenario?

As you head more towards the center, gravity would exert less of a force on you, right?

If I based a story on a scenario such as this, what other things would I have to address?
Thanks
 
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Actually, if you assume that the Earth is a perfectly spherical shell, anything inside the crust would feel no force. Gauss's Law applied to gravity confirms this, as there is no mass contained in a Gaussian surface that is tucked against the inner surface of the shell, and hence there is no gravitational field.

The person outside the shell would feel the gravity as if it were all centered at the center of the shell (although they still couldn't go through the ground, naturally).

cookiemonster
 
But isn't gravity the attraction of two massive bodies?
If you hang a rock from a string next to a mountain, teh rock will be drawn to the mountain.
If that's true, why would people not be drawn to a sufficiently thick crust?
Let's say the crust were several hundred miles thick.
Wouldn;t the sheer mass of the rock be enough to exert a gravitational force on somthing on the inside surface?
If not, why not, because you lost me with this:
there is no mass contained in a Gaussian surface that is tucked against the inner surface of the shell

Also, what did you mean when you said:
although they still couldn't go through the ground, naturally
 
It's a tricky balance. Let's start with the easy one. Imagine the person is in the center. The crust exerts the same amount of force in each direction, so the net force is zero. Easy, right?

But what if we move the guy a little off to the side? Well, now he's closer to one side, so that side is going to have a greater force. But there is also less mass that is closer to him and more mass that is farther away. In fact, there's exactly enough more than it still cancels!

The mathematical formalism is as follows:

\int_S \boldsymbol{g}\cdot d\boldsymbol{a} = 4\pi G M_\textrm{enclosed}
where S is the Gaussian surface to which I referred, i.e. the inner surface of the shell and g is the gravitational acceleration.

You will notice that the mass enclosed is 0 (nothing inside the shell, right?), so the right hand side of the equation will be 0. It's easy to see (or maybe not so easy, for many) that the only way the left hand side can be zero (since the surface is quite large) is if the gravitational acceleration is zero. Consequently, no acceleration due to gravity inside the shell.

As for outside the shell, the full mass of the Earth is enclosed, so the acceleration due to gravity is as we know it, ~9.8m/s, despite the fact that all the mass in in a shell. That being said, you can't pass through the ground, which is all that I was mentioning.

cookiemonster
 
So with cancelling out gravitational forces would you be, in effect, weightless and suspended?
 
Yup. Just like the shell weren't there.

cookiemonster
 
Originally posted by cookiemonster
Yup. Just like the shell weren't there.

cookiemonster

Hmmm...

That may make an even more interesting scenario than my original thought.
I am picturing the astronauts in space giving themselves a gentle push and careening across the ship to the other side.

Hey, what if it weren't a perfect sphere?
What if the Eastern crust was twice as thick?
Would you be attraced to the Eastern side and be able to walk on the surface like the little people in my really bad picture, then as you walk more West you get lighter and lighter, until...
Until what?
Would you just not be able to go further?
Would you become weightless, until you push yourself back into the gravity side?
Would you get very light until you had no traction, then just "fall" back to the gravity side?
 
Well, once you get away from a perfectly spherical shell and a radial density (i.e. one that has spherical symmetry), all bets are off. I imagine you could make some weird fields, but I'd hate to be the one to find an expression for them.

However, if you made one hemisphere more massive than the other, then there would be a gravitational attraction in the general direction of the more massive hemisphere. The precise attraction depends precisely on the characteristics of the shell, but I think you get the idea. You'd probably be able to walk on the surface, but it'd probably be more like walking in a bowl than anything else. Keep walking up the side and you're just going to fall on your ass.

cookiemonster
 
Thank!
This is great!
I would add you to the list of acknowledgments on my book, but people would just think I was talking about the one that lives on Sesame Street.
Unless... You aren't him, are you?
 
  • #10
No, just an admirer!

Actually, I just like cookies.

cookiemonster
 
  • #11
Thanks for the great info.

I wonder if I can hire someone to do a Java model that will take input to represent different scenarios and play with that.

You know anyone that would be interested in that?
 
  • #12
Actually, there's a physics CAD program that could probably do it. I, er, used to have it, but it's seemed to have disappeared.

It's called Interactive Physics or Physics Interactive or something like that. Maybe somebody else has it or something like it.

cookiemonster
 
  • #13
Does it make an interactive model http://www.arseiam.com/fx/52.htm
 
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  • #14
Well, you set up situations and let them run.

It certainly wasn't as pretty as that, last I checked, but it could pull what that program does off (I think?). I'm not so sure it can handle three dimensions (more important than you might think), so that might be a problem.

I've only ever used it once, and that was my high school physics class, where we shoot imaginary circles at each other. I did play around a bit with the gravity and stuff of it all, and it seemed to work pretty well.

Hopefully somebody who has more experience with programs like these will pipe in and bring this man some good solid knowledge!

cookiemonster
 
  • #15
I would imagine that if the planet were rotating sufficiently fast the occupants on the inner surface should feel an equivalent of gravity (such as inside a gravitron).
However, that would be limited to the equator, and decrease as you travel away from the equator, correct?
What if the rotation was not just on a single plane?
Let's say the rotation was not linear, and the planet's rotation plane constantly and quickly precessed? (would that be the right word there?)
Is there a name for this type of non-linear rotation?
What would be the effects of it?
 
  • #16
Hey Raven!

That last link is very amusing. Thanks!
 
  • #17
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