I've heard few explanations, but none in good detail that make it clear to those with less mathematical knowledge on the equations of complex physics.

I would like to think that since you are on the surface of the Earth, for example, everything, at least from my introspection, seems to be "pulled" towards the center of the celestial body itself.

If that is true, and everything, including me, is being pulled towards the center, and our (human, for example) bodies use skeletal muscles to maintain supportive structure to fight the gravity all the time (when awake), how can there be no gravity within the very center of the Earth?

It just seems hard to believe that something attracts you to it with energy, yet it has no energy in its core (gravity) to do this. Where and why does gravity stop at the center, and how does that make good, detailed sense for an average to relatively understand?

Also, taking in to accounts with centers of mass, wouldn't the Earth's core be the "center of mass" in this sense?

I mean if the energy or force "gluing" us to the Earth is not caused by the Earth itself, how come we are "pulled" towards Earth in particular?

  • 3
    $\begingroup$ If you're at the center of a planet, what direction are you going to be pulled in? $\endgroup$ – DumpsterDoofus Oct 19 '13 at 19:02
  • $\begingroup$ Good question. However, that doesn't clarify why there exists no energy of gravity within it. If Earth is what "glues" us to it, when and why does the energy that does so end within the mantel far enough inwards? Basically, why should gravity end inside the Earth, existing solely on the surface as the most powerful? $\endgroup$ – Child Fondalen Oct 19 '13 at 19:03
  • $\begingroup$ related: physics.stackexchange.com/q/81324 $\endgroup$ – Ben Crowell Oct 19 '13 at 19:06
  • 2
    $\begingroup$ You seem to be mixing up energy with force. Force is a vector. At the center, symmetry causes the force vectors to cancel out. $\endgroup$ – Ben Crowell Oct 19 '13 at 19:07
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/18446/2451 and links therein. $\endgroup$ – Qmechanic Oct 20 '13 at 2:45

You dont even have to go the core of the earth to ponder the question. Dig a hole 6ft deep and go stand in it. You now weigh less in this hole than you did while outside of it. That is because all the earth's mass to your front, back, left and right are now pulling you in those directions...no longer DOWN towards earth's center. If you dug deeper there now would be a portion of the earth's mass pulling you UP. Note the decreasing amount of potential energy you have due to earth as you go deeper.

If you dug all the way to the core there is an equal amount of mass in every direction around you, so you can't be pulled anywhere. Analagously you are a marble at the bottom of a bowl. So why is there no energy at the bottom of the bowl? Because you were already given all the gravitational potential energy the earth could give you while you made your way down. You squandered it as heat in your footsteps going down...


You can think of a planet (for example) as some type of stable configuration of matter, under gravity.

One little piece of matter attracted another, then the two attracted some more, etc... None of the pieces cared about where (the direction in space) a new piece of matter came from and, after a while, they built themselves up into something ball-like.

So, even though the planet looks like a ball with a center, it was really the individual pieces of matter that made it all happen. If you pretended that you were a little piece of matter on the surface of the ball, then it's just you and this large ball of little pieces of matter, each pulling you. Mathematically, it looks like you are being pulled to the center.

If you were a little piece of matter in the center, then it's the same kind of thing. All the little pieces of matter in the ball around you are pulling you. But, since they are arranged in a ball (a thing that looks the same, no matter how you rotate it), you are pulled the same all around you. Mathematically, it looks like there is no net pull, so you don't go anywhere.

Edit: Yes, the center of mass would be at the center of a perfect ball planet. If the ball was a little less perfect, the center of mass might be a little off from where it would be in the perfect case.


Well there is plenty of gravity at the center of a (the) planet; there just isn't any significant net gravity. Imagine a very thin perfectly spherical shell of matter. Take ANY point inside that shell (try one about halfway from the center, and draw two straight lines at a small angle through that point to the shell. Imagine those are the sides of a small angle cone. You can easily see that the two pseudo triangles created, are similar triangles, so the triangle side that is a short arc of the shell is proportional to the other sides of the triangle, or the mean (s) of those two (long) sides. So the small elliptical caps on the surface have areas proportional to the square of the mean distance from your point, and so are their masses. So m / s^2 is the same for both of those sections so the net gravity force at that point is zero. A little thought, and you will see that it doesn't matter how small, or how large are the cone angles, or where the point is.

"There is NO NET GRAVITATIONAL FORCE inside a spherical shell.

So when you go down inside a planet, the entire mass of the planet OUTSIDE your position produces zero net gravity; but you are being pulled in all directions. Only the sphere inside your radius produces a net measurable gravity.

This is exactly comparable to the fact that there is no net electric field inside a charged spherical conducting shell.

  • $\begingroup$ It's actually an interesting question. If you are being pulled every which way, could you be torn apart by the opposing forces ?? Well, you see, there is no gravitational shield, so gravity can pull right through any material; so those cancelling forces pulling every which way are doing so AT EVERY SINGLE POINT, so no part of you is being pulled in any direction; so you aren't even aware of the effect of all the forces acting on you. $\endgroup$ – user26165 Oct 21 '13 at 1:11

Not the answer you're looking for? Browse other questions tagged or ask your own question.