Earth and You: Physics

[Aphex_Twin]

Would say, the Earth-Sun system be constantly losing energy due to the mutual gravity tug? Or would such an effect have to be greater than the "fine constant" (quantum discontinuity) in order to happend?

And Off Topic, something I've been wondering. Why is the Moon getting further away from the Earth? For all intents and purposes it should do the opposite.

 

[betazed]

Would say, the Earth-Sun system be constantly losing energy due to the mutual gravity tug? Or would such an effect have to be greater than the "fine constant" (quantum discontinuity) in order to happend?

And Off Topic, something I've been wondering. Why is the Moon getting further away from the Earth? For all intents and purposes it should do the opposite.

had the earth-sun system or the earth-moon system be a simple two body system then yes over time they would be constantly losing energy to todal forces and orbit each other more closely till they coalesce into one body.

However, our solar system is a chaotic system. So it is not so simple anymore. For example, the moon-earth system is being acted on by the sun and the other planets which affect the system in unpredictbale (in the long term) ways. That is the reason why the moon may be moving away from earth.

 

[Aphex_Twin]

Ok, so the solar system taken as a whole should be losing energy, even if some parts of it gain it. But how influent is the said discontinuity?

 

[smalltalk]

Chairmen Meow, microbe
v=sqrt(2GM/d)
...
actually it's a bit more complicated than that, but I'll stay out of calculus for this..

I doubt this can be done without calculus.

Your solution is only true if earth were a point mass.

Of course it isn't, so we'd have to calculate the residual gravitational pull while falling down. I would assume to get 9,81 m/s/s on the surface, and 0 m/s/s at the core. Then we'd have to integrate the stuff.



Bonus question:

say, Earth is hollow. Would we have zero gravity everywhere in the interior, or would we feel a slight pull to the "wall" when approaching it?

 

[betazed]

say, Earth is hollow. Would we have zero gravity everywhere in the interior, or would we feel a slight pull to the "wall" when approaching it?

What, two days and no one answered this?

If Earth were a perfect sphere and hollow then there would be zero gravity everywhere, but knowing that earth is slightly flattened at the poles there will always be a slight gravitational pull towards the center of the earth (opposite from the wall).

 

[Pirate]

Why would there be zero gravity everywhere? The center of a hollow sphere is still the only place where the gravity would be balanced, right? Here's a sketch of how I see it. The horizontal dashed line is the plane of your center of gravity perpendicular to the center of the earth. The attraction of the mass below that line dwarfs the attraction of the mass above that line. (Any component of the vectors that is parallel to that line are cancelled out.) So you would still be pulled to the center, would you not? The only reason I can think of that would make gravity zero everywhere is if the proximity of the small mass "above" you perfectly balances the magnitude of the mass that lies far away 'below' you. <edit- image size>

 

[microbe]

I doubt this can be done without calculus.

Your solution is only true if earth were a point mass.

Of course it isn't, so we'd have to calculate the residual gravitational pull while falling down. I would assume to get 9,81 m/s/s on the surface, and 0 m/s/s at the core. Then we'd have to integrate the stuff.

I got the impression from the original question that this was the intended simplification. Not in real life, as you will hit the hard ground and die.

I kinda remember that inside earth the force to its core is inverse proportional to the distance to the core, not the square of the distance. In that case my answer might be wrong. Heck, I used to remember all these in high school.

Bonus question:

say, Earth is hollow. Would we have zero gravity everywhere in the interior, or would we feel a slight pull to the "wall" when approaching it?

My memory tells me zero gravity, but I could be wrong.

 

[betazed]

@pirate: The result that inside a spherical shell the gravitational pull of the shell is everywhere 0 is actually pretty simple to prove. It is a standard 3D integral best done in spherical coordinates. But qualitatively you can understand it by the following.

Take the man on your left picture. The mass above the dashed line is pulling him up. The mass below the dashed line is pulling him down. The mass above the line is closer and the mass below the line is further away; but there is more mass below the line than above the line. All this comes together to make sure that the pull from the mass above the line exactly balances for the pull for the mass below the line giving a net zero pull of gravity.

For math nerds here is a much more elegant proof that I found (that does not involve the somewhat messy integral - I hate integrals!) It involved writing Newtons law as L(phi) = c d , where phi is the potential, L is the Laplacian, d is the density, and c is some constant (Gravitational constant times something - which we do not care about). Understanding phi is a harmonic function (because of radial symmetry) and d = 0 inside the sphere. use the weak maximum principle of harmonic function, which says that a (smooth) harmonic function does not have local minima and maxima. Hence the potential inside the sphere must be constant (since it is constant on any shell inside the hollow sphere). Since Force =grad(phi) , we have F=0, as required.

 

[Pirate]

That's what I meant when I said

The only reason I can think of that would make gravity zero everywhere is if the proximity of the small mass "above" you perfectly balances the magnitude of the mass that lies far away 'below' you.

-that the mass above you is small but close, and the mass below you is great but far away. I just didn't think that the balance would be perfect. That's really interesting.

<edit>Well, considering that latest revelation, consider this. If instead of the moon there was a second earth, you would obviously be weightless at the midpoint between the two bodies (equal masses, equal distances, everything is cancelled out). Based on the weightlessness at everypoint inside a hollow earth, does it stand to reason that you would be weightless at every point on an imaginary line connecting the centers of mass of the two bodies? You could be standing on the surface of earth#1 and would be completely weightless as long as earth#2 were directly overhead, regardless of how far away it is?

 

[microbe]

You could be standing on the surface of earth#1 and would be completely weightless as long as earth#2 were directly overhead, regardless of how far away it is?

No you cannot. The zero gravity results from the particular environment. You have to do the math to see if it balances out or not. In the hollow sphere case, yes; in the two-earth case, not.

 

[Pirate]

That's what I figured, but Betazed will please forgive me if I don't know the nature of harmonic functions. I proposed the 2 earth system as the opposite extreme because I wondered a what geometries the zero gravity breaks down. The two earth model can be simplified as two microscopic black holes and represents a one dimensional system - zero gravity at all interior points doesn't work. I can see now that it wouldn't work for a 2-dimensional hollow hoop. And probably wouldn't for a 4-dimensional hypershpere, either, right? But there is something intrinsic in gravity that a 3-dimensional sphere is perfectly balanced this way. What is it?

Sorry, I got out of engineering because I hated the number crunching. I'm more of an artistic designer so I can visualize these problems very well, but I just don't like doing the math...

 

[betazed]

That's what I figured, but Betazed will please forgive me if I don't know the nature of harmonic functions. I proposed the 2 earth system as the opposite extreme because I wondered a what geometries the zero gravity breaks down. The two earth model can be simplified as two microscopic black holes and represents a one dimensional system - zero gravity at all interior points doesn't work. I can see now that it wouldn't work for a 2-dimensional hollow hoop. And probably wouldn't for a 4-dimensional hypershpere, either, right?

Any such two body system will have points called L points (Lagrange points) where the gravitational pull of the two bodies will balance. In the static two body system there will be 5 such points. The earth-moon system, the earth-sun system all have L points and we use them regularly for placing satellites.

But there is something intrinsic in gravity that a 3-dimensional sphere is perfectly balanced this way. What is it?

Radial symmetry and inverse square law.

you can see the same thing happening in electromagnetism. Radial symmetry and inverse square law tells us that the electromagnetic potential inside a charged spherical shell is contant and hence the electric field and force is 0.

 

[Aphex_Twin]

Betazed, aren't there 5 Lagrange points (two stable, three unstable)?

 

[betazed]

Betazed, aren't there 5 Lagrange points (two stable, three unstable)?

Of course. I am losing it.

 

[Pirate]

you can see the same thing happening in electromagnetism. Radial symmetry and inverse square law tells us that the electromagnetic potential inside a charged spherical shell is contant and hence the electric field and force is 0.

But Faraday cages don't need to be perfectly spherical to have a zero field inside, don't they?

 

[betazed]

But Faraday cages don't need to be perfectly spherical to have a zero field inside, don't they?

No they do not. But Faraday cages do not depend on radial symmetry to block fields. They depend on the conductivity of the cage. When you put a charge on the cage, an electric field does form inside the cage. However that electric field then forces the charge on the cage to allocate itself in such a way that entire cage becomes a equipotential surface.