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This one has had me stumped for years. Imagine being in a very sturdy air-conditioned spherical pod right at the centre of the Earth. Now, you have equal amounts of mass pulling on you from every side, so the gravitational attraction should equal out, and you should be weightless, just floating around in the pod. Could that be right? But if weightlessness rules at the Earth's centre, then how does that affect the enormous pressure that's supposed to be down there.

And how does this weightlessness tie in with Einstein's General Theory of Relativity, which says that mass curves space, which causes the gravity effect? Does that mean all that mass unwinds the curvature at the centre of massive objects like the Earth.

2007-03-27 17:09:48 · 16 answers · asked by Anonymous in Science & Mathematics Astronomy & Space

16 answers

You would be weightless, but retain your usual mass. Consider it in terms of atoms or molecules, the ones at the surface are attracted to the centre of mass (centre of gravity), with the maximum force, because all the other mass is warping/deforming spacetime (exerting gravitational attraction) in the one direction. Underneath that is the next one, with a huge attractive force in one direction, and a miniscule one in the opposite direction. Next layer down, there is a big force in one direction, and a small one in the other. All those forces are cumulative, increasing the pressure, but reducing the weight, until the centre is reached, where pressure is at a maximum, but weight is at a minimum (zero). The curvature of spacetime is at a maximum at the surface, and approaches zero at the centre of a massive object (it is being attracted to the nearest large concentration of mass, so never reaches zero). It may help, if you consider the behaviour of either light, or an object moving at a given speed, at that place. If neither are deflected noticeably, spacetime curvature (gravitational force of attraction) is virtually zero.

2007-03-27 23:56:16 · answer #1 · answered by CLICKHEREx 5 · 0 0

You're trying to have it two incompatible ways, technically weightless but under no pressure. You would feel weightless because mass is pulling you equally in all directions. But it's also pulling everything else OUTSIDE of it. The mass to your left pulls the mass on your right, the mass "above" pulls the mass "below", etc. Everything else is being pulled in your direction, which is why your little sphere would not exist, no matter how magically it was constructed. Gravity is not "cancelled out", it's just everywhere at once, which is a very different thing than no gravity.

2007-03-28 02:55:07 · answer #2 · answered by skepsis 7 · 0 0

Yes,

Yes. You would be weightless.

Literally suspended with equal gravitation pull from all directions at once.

Gravity is a form of energy. It manifests itself in wave form. Just as two wave of equal force cancel each other so would the gravitational waves at the planets core.

With certain conditions.

We are assuming the planet is a perfect sphere and that no outside gravitational influences affect your place in your air conditioned room suspended at Earths center.

Lets clarify a few things to make this hypothesis valid.

1. Earth would have to be perfectly symmetrical. It would need to be absolutely perfectly round because if it wasn't the enormous gravitational pulls would be off balance.

Just for arguments sake lets say your were only 2 atoms off balance.

Does not seem like much right? And really, it's not. But we are dealing with a planetary body that is roughly 25,000 miles around at the center. The gravitational pull of all the elements of the planet are needed to calculate the exact force that is going to exert the exact pull to keep you centered in your air conditioned room or container. If you were even one electron's mass off center you would begin to oscillate. A vibration that would quickly accelerate until the forces of gravity literally pulled you apart atom by atom.

Still, your question did not require much hypnosis about the nature of physics and relative forces so let me answer your question based just on the info gathered from the query.

You are floating weightless at the center of the planet Earth. You were transported to a special room that is perfectly centered.

Your body has to be equal too the forces acting on it.
So would your mass. Densities would be re-arranged so that your mass would be equal in each direction. Just as those same forces shape planets and suns into circular shapes so would your body now be shaped.

No worries. the pressure of all the planets mass would squeeze all the atoms of your body to the smallest area possible but slightly larger. We can't compress your atoms too much or you would burn.

You are weightless. And the size of a period divided by ten thousand.

2007-03-28 02:49:50 · answer #3 · answered by whastheheck 2 · 1 0

As another posted indicated, in the real universe, you can never be truly weightless because there are millions of other objects (stars, planets, galaxies, the sun) pulling at you. But let's suppose you created a universe with just two objects in it - a big sphere (the earth) and another object (you) suspended safely at the precise center with all of your mass concentrated in a perfect uniform sphere, and with a perfectly even distribution of matter in the (Earth-like) sphere around you. In that situation I believe you would feel no gravitation pull, because all the forces would be balanced out. But that does not negate the idea of gravity curving space-time. The situation you describe is kind of like a ball-bearing sitting at the bottom of a valley between two hills, where it is not going to roll in any direction because it is in a stable situation and perfectly balanced - but it's still at the bottom of a valley.

2007-03-28 02:06:19 · answer #4 · answered by Anonymous · 0 0

Thats very interesting... I never thought about this. (there was another problem... make a hole through the center of the earth, then drop a ball... what will happen? - ans: ball will oscillate till it comes to rest at the center)

anyway, man isnt that perfectly symmetrical, so you wouldnt have equal mass pulling in from all sides... so that is answered...

but i have no answer if you replace man with a sphere or symmetrical body..

You feel weightless in orbit because centrifugal force (of rotation around the earth) is balanced by gravity...

in this case gravity should itself cancel out gravity...

but there is still a problem... you wouldnt be 'floating around' as you said... if you (your centre of mass, that is) budge even a bit from the centre, then there will be unbalanced forces and you will instantly be dragged back to the centre. so there will only be floating... no floating around;

2007-03-28 02:13:15 · answer #5 · answered by Vishnu Unnikrishnan 1 · 0 0

Heavy thinking, dude!

Well, why don't you propose it for a science experiment on the International Space Station?

One way of thinking, aside form the small problem of getting there, would be the problem of everything else wanting to get there too.

I think it 'makes sense' that the gravity would even out if there was an even amount of mass all around you.
Then you have to think about little things like attraction forces of individual molecules, atoms, and subatomic particles.

An experiment in miniature in zero gravity could give you an answer, but you'd need a fairly big hunk of stuff to develop enough gravity force to test it, unless you cheated with magnetism, which is a lot stronger than gravity.

2007-03-28 00:23:41 · answer #6 · answered by Big Bruce 6 · 0 0

I am going on a limb here but what the heck. Sadly, you are never weightless. Yes you feel that you have no weight due to earth's mass. The explanation is given by the previous answers. You do however have to consider the effect of the Sun, that you still orbit trapped in the center of the earth. Then your weight is equal to W2 = G*M(you)*Msun/ (earth-sun-distance)^2. Compared to your weight on the surface of the earth due to earth alone (W1) that is equal to W2/W1 = Msun/Mearth * (earth radius/earth-sun-distance)^2 = (2 x 10^30)/(6 × 10^24) * (6378.1/ 150,000,000)^2 = 6 x 10^(-4). So if you were 200 lbs, 6x10^{-4} x 200 = 12x10^{-2} = 0.12 lbs of it is due to the sun's gravity. If now you move to the center of the earth, you are only left with 0.12 lbs ~ 2 oz.

So you do always have a weight. Therefore, your last point is not necessary.

2007-03-28 01:39:24 · answer #7 · answered by John Doe 2 · 2 0

Yes. You would have equal amounts of mass pulling you in all directions. These gravitational pulls would cancel each out, effectively leaving you weightless.

Of course, the entire mass of the earth would be gravitating towards your location, and this would put you under huge pressure, but that's another issue.

2007-03-28 00:36:25 · answer #8 · answered by Bramblyspam 7 · 3 0

Yes you would be weightless. The pressure comes from the weight of all that stuff that isn't at the center of the Earth pushing on the center. It has nothing to do with relativity though.

2007-03-28 00:16:45 · answer #9 · answered by campbelp2002 7 · 3 0

You would float with the mass and gravity pulling equally from all sides.

2007-03-28 07:03:11 · answer #10 · answered by Billy Butthead 7 · 0 0

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