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Journey to the center of gravity

  1. Aug 27, 2010 #1
    Suspend all disbelief and imagine digging to the center of the Earth, there you dig out a small spherical chamber, you sit in the middle, do you float or are you depatched violently to the wall around you instantaneously.
     
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  3. Aug 27, 2010 #2
    If your chamber resisted the high pressures and temperatures without collapsing and somehow protects you from being burned too, I guess you could expect to float, not sure about it though.
     
  4. Aug 27, 2010 #3

    CompuChip

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    Exactly in the center of the earth you would indeed be weightless and just sit there (neglecting the influence of gravitational forces of objects other than the earth).
    If you didn't sit exactly in the center but just a little outside it, you would go into oscillating motion. (In fact, it is a popular mechanics assignment to have students calculate that if you just dug the tunnel and jumped in, you would bounce from one end of the earth to the other in exactly 43 minutes; neglecting rotational effects which would make you bump into the walls).
     
  5. Aug 27, 2010 #4
    What puzzled me was this: I thought the mass of the Earth created a gravitational pull which was spread over its surface and thereby diluted but within the spherical chamber the surface area is relatively tiny and so the gravitational pull would be greatly concentrated. But I suppose I must be wrong.
     
  6. Aug 27, 2010 #5

    Doc Al

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    Think of each piece of the earth's mass as exerting its own gravitational pull. On the surface, the resulting pull from all the earth's mass is the usual force acting towards the center of the earth. But at the earth's center itself, the gravitational pull of the earth's mass from one side exactly counters the pull from the earth's mass from the other side--the net effect is that everything cancels out and there is zero gravitational field at the center.
     
  7. Aug 27, 2010 #6
    Would the same hold true for a black hole? (theoretically, of course.)
     
  8. Aug 27, 2010 #7

    CompuChip

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    In fact, if you are at a distance r from the center of the earth, the gravitational pull you feel is just that of all the mass which is within distance r of the center. If you think of the earth as a homogeneous sphere of radius R with total mass M, then the effective mass in the gravitational formulas is [itex]\alpha M[/itex], where
    [tex]\alpha = \frac{4/3 \pi r^3}{4/3 \pi R^3}[/tex]
    is the fraction of mass that lies within "your" spherical shell.
     
  9. Aug 27, 2010 #8

    CompuChip

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    A black hole is a different story altogether. To understand that, classical mechanics won't do and we will really need to look at the general relativistic formulas.
    Also, when you go to the "center" (i.e. the singularity) of a black hole, you don't simply go through it and keep going until you are pulled back again, thereby creating an oscillatory motion like in the earth.

    In fact, we don't really know what happens at the 'center' of a black hole, probably it's just death and destruction. The question is somewhat useless though: apart from the practical aspects (deadly radiation, huuuuuuge gravity pulling you apart like spaghetti, probably extreme temperatures), an outside observer can only track you until some distance away from this center (the event horizon). Once you get inside the event horizon, there is no way that you can send any information to the outside observer, so in a sense whatever happens from there on is irrelevant (in my view).
     
  10. Aug 31, 2010 #9
    Uh, the neutral spot would not be at the geometric centre of Earth as you must allow for Lunar and Solar gravity and their tides. If you were fussy about it, you could factor in Venus' pull, too...
     
  11. Aug 31, 2010 #10

    CompuChip

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    Ah yes, I mentioned that the earth should be homogeneous, but of course the universe should be otherwise empty as well :)
    Don't we physicists live in a perfect world?
    Thanks for pointing it out.
     
  12. Sep 2, 2010 #11
    In fact, at the Center of Earth you would not be weightless at all, but would be pulled along the walls of your spherical cave (if that's the shape it was). The reason is that you must consider the Earth/Moon gravitational system, and the balance point is somewhere beneath the surface of the earth, very close to the center, but not the exact center, and it moves as the moon revolves around the earth. This is the reason that you do in fact weigh less (not less mass of course) when the moon is directly overhead versus on the other side of the earth from you. This center of mass point between the earth and the moon is what you'd be attracted to as the moon revolved around the earth and the earth rotated. So if you were at the exact center of the earth, you could very easily tell what direction the moon was from your location as you'd be drawn to that side of your little spherical cave.
     
  13. Sep 4, 2010 #12

    DrGreg

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    First, if we ignore the effects of the Moon, Sun, etc, the effective gravity will be zero throughout the spherical cavity, not just at the very centre. So if you are motionless anywhere in the cavity you will remain motionless; if you are moving you will continue to move unless you hit the wall or get slowed down by air friction.

    When we add the Moon, then the centre of the Earth will now orbit around the centre-of-gravity of the Earth-Moon system. But objects in orbit are still "weightless". An object at the centre of the Earth would be gravitationally attracted towards the Moon, but that force would be exactly the centripetal force needed to maintain orbit, so the object would actually remain motionless relative to Earth. However the cancellation would not be perfect at other points within the cavity, so in this case some objects not at the centre would start to drift.

    (I've phrased this answer in terms of Newtonian gravity. The explanation in general relativity would use different words but come to the same conclusion.)
     
  14. Sep 6, 2010 #13
    If you were in a chamber at the center of the Earth's mass, I would think that while you would "fall" toward the moon, the Earth would as well. Wouldn't the Earth's and your acceleration pretty much cancel out?
     
  15. Sep 6, 2010 #14

    DrGreg

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    Yes. (Which is why I said you'd remain motionless relative to the Earth.)
     
  16. Sep 12, 2010 #15
    If you were at the center of the Earth you would not "feel" any effect from the Sun or the Moon. The Earth and Moon rotate about each other, the point they rotate around is called the barycenter. That is a point that the center of masses rotate around. Since you are at that center of mass you and the moon are both rotating around that point.

    If you have had some calculus you should be able to see that the forces balance out. One way of calculating gravity is to use the "shell method". That is, you separate the Earth into a series of shells and add up the force of gravity from each shell. If you are inside a shell all of the forces balance out. So if you are one thousand miles deep into the Earth you can ignore all of the mass that is less than one thousand miles deep all around the Earth. And the force of gravity can be calculated from the mass below you. One interesting side point is that gravity does not go down as you go deeper into the Earth as most would think. It actually goes up until you hit the mantle core boundary. That is because the core is much denser than the mantle and its gravitational effects fall off, as do all objects, the further away you get from it. If the Earth only consisted of its core the acceleration due to gravity on its surface would be roughly 10.8 m/sec^2.
     
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