Gravitys effect on individual atoms.

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Ginjar

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I am not sure of this but i was just wondering if it is possible. If a person is pulled past the event horizon of a black hole, the differences of the gravitational pull on opposite ends of the person is so great that the person would be pulled to look like "spaghetti." However what about the individual atoms that make up that person? Wouldn't gravity have a different effect on different parts of the atom? Gravity would pull certain electrons more than others, and therefore essentially pull the atom apart. If this is true than wouldn't a black hole have a different effect on more complex atoms, such as a helium atom being effected in a different way than an atom of iron?
 
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ramparts

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Ginjar":2rsua6px said:
I am not sure of this but i was just wondering if it is possible. If a person is pulled past the event horizon of a black hole, the differences of the gravitational pull on opposite ends of the person is so great that the person would be pulled to look like "spaghetti." However what about the individual atoms that make up that person? Wouldn't gravity have a different effect on different parts of the atom? Gravity would pull certain electrons more than others, and therefore essentially pull the atom apart. If this is true than wouldn't a black hole have a different effect on more complex atoms, such as a helium atom being effected in a different way than an atom of iron?

You have the conceptual framework right, but the forces that hold atoms together are extremely strong - far more so than gravity. It would take unimaginably strong gravitational forces (as you wouldn't even find throughout most of a black hole) to actually rip an atom apart.
 
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dangineer

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According to the Standard Model of physics, gravity is actually the weakest of the four fundamental forces. Although, on the molecular side of things, I would think that long molecules, such as hydrocarbons and polymers, would encounter gravitational potentials high enough to break their molecular bonds (essentially their tensile strength). This would also depend on what sort of bonds are there (covalent, hydrogen, etc.) Although these are entirely dependent on the electromagnetic force (which is comparatively stronger than the gravitational force), I wouldn't think that it would be impossible gor gravity to break the weaker bonds.

A calculation to figure this out could probably done: you would need to know the bonding strangth of the particular molecule and then see if there's a distance from the singularity that has a high enough force gradient along the length of the molecule to break that bond, if anyone wants to try that out.
 
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origin

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Ginjar":3sn0e45b said:
I am not sure of this but i was just wondering if it is possible. If a person is pulled past the event horizon of a black hole, the differences of the gravitational pull on opposite ends of the person is so great that the person would be pulled to look like "spaghetti." However what about the individual atoms that make up that person? Wouldn't gravity have a different effect on different parts of the atom? Gravity would pull certain electrons more than others, and therefore essentially pull the atom apart. If this is true than wouldn't a black hole have a different effect on more complex atoms, such as a helium atom being effected in a different way than an atom of iron?

Great question! I think this boils down to the following: is the graviational difference near the event horizon over the radius of an atom greater than the electrostatic attraction between the nucleus and the electrons. This calcuation can be a complete nightmare (for me at least) because there is no way to discount the warpage of space near the event horizon which means you would need to use general relativity. But you could use newtonian gravity for an first approximation.

The amount of energy needed to pull an electron from the atom can be found by the energy of a photon that causes the photo electric effect.

The difference in the gravity over the radius of the atom can easily be found using the 1/r^2 since the black hole would be a point source. The baseline gravitational force and distance from the blackhole would be calculated at the event horizon with the Schwarzschild radius. Again this is completely disregarding general relativity and assuming that the Schwarzschild radius is a straight line radius (which will throw in significant errors).

I don't have time to calculate this since I am at work and suppose to be working! :oops:
 
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ramparts

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origin":a1f23d1o said:
Ginjar":a1f23d1o said:
I am not sure of this but i was just wondering if it is possible. If a person is pulled past the event horizon of a black hole, the differences of the gravitational pull on opposite ends of the person is so great that the person would be pulled to look like "spaghetti." However what about the individual atoms that make up that person? Wouldn't gravity have a different effect on different parts of the atom? Gravity would pull certain electrons more than others, and therefore essentially pull the atom apart. If this is true than wouldn't a black hole have a different effect on more complex atoms, such as a helium atom being effected in a different way than an atom of iron?

Great question! I think this boils down to the following: is the graviational difference near the event horizon over the radius of an atom greater than the electrostatic attraction between the nucleus and the electrons. This calcuation can be a complete nightmare (for me at least) because there is no way to discount the warpage of space near the event horizon which means you would need to use general relativity. But you could use newtonian gravity for an first approximation.

The amount of energy needed to pull an electron from the atom can be found by the energy of a photon that causes the photo electric effect.

The difference in the gravity over the radius of the atom can easily be found using the 1/r^2 since the black hole would be a point source. The baseline gravitational force and distance from the blackhole would be calculated at the event horizon with the Schwarzschild radius. Again this is completely disregarding general relativity and assuming that the Schwarzschild radius is a straight line radius (which will throw in significant errors).

I don't have time to calculate this since I am at work and suppose to be working! :oops:

It will, and moreover, my guess is that by the time the forces are more or less equivalent, you're already very close to the center (and well within the event horizon), at which point quantum effects will come into play as well. Suffice it to say, the exact problem is non-trivial ;)
 
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