Why would we weigh less at earth's center?

  1. I don't understand why we would weigh less at earth's center(as opposed to sea level). Looking around the net, I have seen different approaches to this problem, each with different solutions.

    According to Newton's universal law of gravitation:

    F= Gm1m2/r2

    If this is the case, then as r approaches nearly zero as the object gets closer to the center of the earth, the denominator gets smaller and smaller, making the quotient approach infinity.

    On the other hand, I've heard the explanation that all the mass around you cancels each other out at the center of the earth.

    This explanation also seems to make sense, but both explanations contradict each other..

    Could someone please point out where I rationalized improperly
     
  2. jcsd
  3. Drakkith

    Staff: Mentor

    The law of gravitation only applies when you can consider the two objects as "point sources". For most situations, this approximation works well and will give good results. However, it does not work well when you burrow beneath the surface of the Earth, as you are now surrounded by matter on all sides and cannot use the formula anymore. At the center of the Earth the gravity from all the matter around you cancels out and you end up with zero net pull, meaning you don't weigh anything.
     
  4. Okay. I am still slightly confused. What are "point sources", and how come the law of universal gravitation doesn't apply here?
     
  5. Drakkith

    Staff: Mentor

    A point source when you approximate an object's mass as being concentrated at a single point instead of being spread out. We can use the law of gravitation to find the force between the Earth and Moon if we consider both objects to have all of their mass concentrated at a single point at the center of each object. Then the r2 in the equation becomes the distance between these points. If you don't approximate the Earth and the Moon as point sources, you cannot use the equation because you no longer have a single mass and radius.
     
  6. phinds

    phinds 9,155
    Gold Member

    A "point source" is exactly what the phrase sounds like ... a source that can be TREATED as though it were a point. So when calculating the gravitational force between the Earth and the moon, both are treated as point sources.

    You find the center of mass of a body and consider it as a point source at that point but with the mass of the whole thing. It is a convenient mathematical fiction.

    When you are IMMERSED in a source, you can't treat it as a point source because it is all around you.
     
  7. sophiecentaur

    sophiecentaur 13,916
    Science Advisor
    Gold Member

    They don't contradict each other, actually. Your formula only works when you are outside a body (and when it can be treated as the equivalent of a single mass, all concentrated in one point. So, if the Earth were a totally uniform sphere, then the formula would be totally accurate because you can treat it as a single mass right at the centre. You actually need to be at some distance from the Earth for that assumption to work because the various parts have different densities and it's not symmetrical, either.
    As you go beneath the surface, each spherical 'shell' that lies outside the level you're on has no net effect because the attractions of all the individual parts of the shell cancel each other out and it's only the stuff 'beneath your feet' that gives you a weight force. When you get to the centre, there is nothing beneath your feet ( the effective mass becomes nothing) so you will 'weigh' nothing. Of course, as you leave the centre, there will be a small restoring force, pulling you 'downwards.
     
  8. Thanks for clearing this up everyone.
     
  9. A.T.

    A.T. 6,189
    Gold Member

  10. Let's say m1 is the mass of the earth and m2 is your mass.

    Per the shell theorem, any mass above your current depth cancels out. So, as R goes to 0 m1 also goes to 0 so at R=0 you have 0/0. In order to figure out exactly what that means you have to express m1 in terms of R and then take the limit as R goes to 0.

    As it turns out in this case F goes to zero, but this is a prime example of a problem that is easy to solve with limits. If you are unfamiliar with them you should make a point not to remain so.
     
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