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I need the explanation and proof of it. Thanks for your answer.

2007-01-24 00:02:06 · 9 answers · asked by Anonymous in Science & Mathematics Physics

9 answers

I don't have any proof but if you were in the middle of the Earth, the pull from gravity would be equal from all directions making a net zero gravity.

2007-01-24 00:06:09 · answer #1 · answered by scruffy 5 · 0 0

All mass in the universe is gravitationally attracted to all other mass in the universe. "Gravity" exists everywhere.

That said, the effects of gravity can be canceled out. If gravity pulls you one way, something else can pull the other way and cause an effective "weightlessness" as if there was no gravity.

To analyze this question easier, let us neglect all the other celestial bodies in the universe (like the sun, other stars, planets,), and only deal with the Earth.
In this case, if you Earth were a uniform, hollow, spherical shell, then ANYWHERE inside, the net gravitational force would be zero...in effect there would be no "gravity" / gravitational force.
The math which goes into proving this fact is beyond the scope of this post, but if you were anywhere inside a uniform spherical shell...there would be no net gravitational attract because of the symmetry of the system.
Think of it like this...If you were in the exact center of the sphere, I think it is pretty easy to conceptualize the fact that all the Earth's mass would be pulling you equally in all directions because you are in the center, thus no gravity. But it is also true if you are not in the center, but rather closer to one 'side' of the sphere than the other. If you are closer to the shell, you are that much closer to some mass which will pull on you...but the amount of mass which pulls on you is less than if you were in the center, the other mass is pulling the opposite direction so the net effect is zero.
You might recognize this effect is similar to that of electric fields inside charged conducting spherical shells if you have studied Gauss's law.

If we then take the other celestial bodies into consideration again, then there would be a net gravitational force…the same force which keeps you and the Earth in orbit around the sun and the rest of the galaxy.

http://en.wikipedia.org/wiki/Shell_theorem
http://www.kineticbooks.com/physics/trialpse/13_Gravity%20and%20Orbits/06/sp.html

The second link, above, offers an excellent mathematical derivation / proof of this concept...but in order to see the page you first have to click on the "continue on" link if you don’t want to install their software (which is not required).

2007-01-24 01:55:10 · answer #2 · answered by mrjeffy321 7 · 0 0

You can prove by integration that the net gravitational field at any point inside a hollow sphere is zero. Assume that a sphere is infinitesimally thin, where R is the radius of the sphere. Then integrate the contribution of gravity at your point over a series of infinitesimally narrow circular bands that make up the body of the sphere, where the location of interest lies on a diameter of the sphere, which is always true. Choose bands that lie perpendicular to the chosen diameter. You will find that the net gravitation is zero.

2007-01-24 00:06:32 · answer #3 · answered by DavidK93 7 · 0 0

the gravitation or attraction between two masses depend on the distance of the masses. In the case of sphere, the distance is calculated from the center of the sphere. Even if the sphere is hollow. So , the gravitation would be more intense in the inner of the earth and surely not zero

2007-01-24 00:51:52 · answer #4 · answered by maussy 7 · 0 1

Ummm...
I'm writing this from what i know so it might be wrong.

All particles in the universe attract each other with a force that is directly proportional to the product(multiplication) of their masses and inversely proportional to the square of the distance between them.

So you can say that since the earth wieghs thousands of times more than you, you are attracted towards it and not the other way round, so if the earth has no mass, then the force (F) would be zero, as:

F = (mass of earth) * (your wieght) / (Square of distance between you)

Therefore, there will be no gravitation.

And... The formula above was formulated by Isaac Newton ,I think. So that must be proof enough.

2007-01-24 00:13:55 · answer #5 · answered by Kaku 1 · 0 1

There would be zero gravity right in the center because the pull from all directions would be equal. If you were a little bit off of the center, even the little mass of the crust would have an effect and eventually pull you "up" to the underside of the crust.

2007-01-24 01:38:45 · answer #6 · answered by Anonymous · 0 1

It is because of the pull of gravity from all directions canceling out and resulting in a zero g environment.

2007-01-24 19:18:29 · answer #7 · answered by Qyn 5 · 0 0

Gravitational force is a consequence of mass. A hollow earth still has mass, and so, gravity.

2007-01-24 00:06:59 · answer #8 · answered by Jerry P 6 · 0 2

It is virtually impossible for the earth to be hollow beneath the techtonic plates since they basically float on top of molten materials. The earths crust would have to be one complete un moving shell to support this idea, and we know thats not true!

2007-01-24 00:09:15 · answer #9 · answered by Anonymous · 0 1

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