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?

Please click here to view (http://www.geocities.com/talorweb/innerworld.htm)

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

Hi one_raven:

What would be the gravitational effects of living on the inside plane of a hollowed-out planet with a huge crust?

Funnily enough, there would be no gravity at all on the inside surface of your hollowed-out planet, provided it was spherical. Everything would just float around. This is a consequence of Newton's "Shell" theorem, which is easily derivable from Gauss's law.

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?

It doesn't matter how thick you made the crust - there'd still be no gravity inside.

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

The forces at A and B would be zero.
The force on a mass m at C would be equal to GMm/r², where M is the mass of the shell, r is the distance to point C from the centre, and G is Newton's gravitational constant.

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

Not in this case.

James R is right:

Funnily enough, there would be no gravity at all on the inside surface of your hollowed-out planet

This is because the gravitational force of the shell completely cancels out on the inside! There is no net gravitational field!

I was talking about this elsewhere and got into discussing what would happen if it weren't a perfect sphere, or of the planet's crust was not uniform.

Seems it's a pretty complex scenario.

I have seen some pretty impressive physics modeling in Java Like this whip cable, for example (http://www.arseiam.com/fx/52.htm).
Do you know how I might go about modeling this (or hiring someone to do it for me)?

just do the integral, it is pretty easy. you can look in any e+m book to understand the symmetry, it is synonmous(sp?)
to gravity.

While its true to say that you would feel no gravity in the center, effects such as time dilation as compared to the surface should still be present.
A clock inside the spherical homogeneous outer shell will run slower than one on the surface.

Beta: I would expect the clock on the surface of the hollow spherical shell to run slower than one at the center. Relativity physics suggests that you analyze what you would observe in a closed room with no observational evidence from outside.

A closed room on the surface of a hollow spherical shell, on the surface of a planet, or in room being accelerated are equivalent.

A closed room in free fall or unaccelerated with no gravitational forces acting are equivalent. A clock in this room would be slower than one with noticable gravitational or inertial forces. This is what would be observed in a room anywhere inside a concentric hollow spherical shell.

One Raven: The equations (differential, I think) of a uniform cable suspended at both ends is well known and can be found. Use some ingenuity in picking search keys. Solve the equations and plot the resulting curve. The code at the site you referenced looks damned efficient. The response to changing position of one end seems immediate.

If differential equations are being solved numerically, some special techniques might be in use. When solving differential equations numerically, some transendental functions can be computed using short cuts due to the small change in the independent variable at each time slice. For example, square root can be done with one Newton iteration and no check for convergence by using the last computed square root as the first approximation.

As posted by Dinosaur:
Beta: I would expect the clock on the surface of the hollow spherical shell to run slower than one at the center. Relativity physics suggests that you analyze what you would observe in a closed room with no observational evidence from outside.


Although it may seem counterintuitive, the clock on the surface does run faster than the clock in the center.
The potential at the center of the homogeneous sphere is not zero (which is what it is at infinity) - but it's flat (it's called an inflection point )
I can provide mathematical verification, but at this stage, regard the gravitation well as continuing all the way towards the center.
The center point in the example given can be regarded as the hollowed out area.

I don't think the planet would be rotating fast enough to where you would have to keep relativistic effects in mind.

The lack of gravity inside a perfect spherical shell is true for any inverse square law, and only inverse square laws. Otherwise the radial dependence does not cancel out of the integrals and the whole thing depends upon the shape chosen.

Beta: Where did you get the idea that the clock would run slower at the center of a holllow spherical shell?

It seems so contrary to what I know about relativity that I will not accept your view without more evidence. I do not consider myself particularly knowable about relativity, so I might accept a better explanation than what you have so far provided.

In every other situation with no measurable gravity, the clocks run faster than regions with measurable gravitational or inertial forces. What is different about the inside of a hollow spherical shell?

In free fall toward a large star, there are no measurable gravitational or inertial forces acting and relativity predicts that clocks run faster than at the surface of the Earth. Again, what is so different about the interior of the hollow sphere and free fall toward a large star?

>>Beta: Where did you get the idea that the clock would run slower at the center of a holllow spherical shell?

Dinosaur- The following is an extract from a post in another forum, which answers the question of clock on the surface and in the center. ( Note- the point of discussion was an idealized homogeneous spherical mass, which we called the earth for simplicity)
The relevant poster (Cusp) is an astronomer and particle physicist.

Take from-http://www2b.abc.net.au/science/k2/stn/newposts/920/topic920542.shtm

"If you have a sphere with a radius R and a uniform density, d, the potential at the surface is

phi = - G M / R (where M = (4/3) pi R^3 d)


Inside the sphere, the potential goes as;

phi = - G ( (2/3) pi d )[ R^2 - r^2] - GM/R

If you don't believe me, then from

F = -d/dr ( phi )

you get the force inside the sphere to be

F = - ( - G (4/3) pi d ( -r ) ) = - G M(
Let r = 0 and you see the potential at the centre of the earth is *deeper* than that at the surface!! It has to be - if you drop something from the surface of the earth to the centre, it passes through with positive kinetic energy - if the potential was zero at the centre, what ever you drop from the surface of the Earth out not get there (like you can't drop anything to infinity).

Now - the time dilation formula is

del_tr/ del_te = [ (1+2 phi_r / c^2 )/(1+2 phi_e/c^2 )]^0.5

where del_tr is the length of tick seen by the receiver, del_te is the length of tick as seen by emitter, phi_r is the potential at receiver, and phi_e is is the potential at the emitter.

Consider the case of the surface of the earth and a person at infinity -

del_tr(infinity) / del_te(surface) = sqrt( 1 / (1-2GMR/c^2) )

Which is greater than 1 so the person at infinity sees the clock at the surface of the earth tick slowly.

But

del_tr( centre ) / del_tr( surface ) is less than 1!!!! so the person at the centre sees the clock at the surface tick faster than his own local clock!"


>>From PhysMachine
I don't think the planet would be rotating fast enough to where you would have to keep relativistic effects in mind

I'm referring to Gravitation time dilation, not time dilation due to relativistic speeds.

Beta: He is talking about a solid planet, not a hollow spherical shell.

Beta: He is talking about a solid planet, not a hollow spherical shell.

With a hollow shell you get this situation:

A clock inside the shell would run slower than one sitting on the surface, and this time rate would be the same everywhere inside the hollow area. The time rate difference between surface and inside would depend on the mass, density and thickness of the shell.

>>Beta: He is talking about a solid planet, not a hollow spherical shell.

Yes and as Janus58 has correctly observed, the time rate is slower inside the hollow area and remains a constant anywhere within this hollow area.
The maths for the homogeneous sphere is applicable to this scenario- Just consider the hollowed area as the center and the time dilation is constant anywhere within this area.
The original question did not specify the size of the shell, so it could be vastly greater than the center hollow.

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?