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We know that the value of g in the center of earth is zero(0). But applying the classical formula for calculating g: We find g=(GM)/0 which is undefined. So how can this be zero. I don't understand.

Qmechanic
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Theoretical
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1 Answers1

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The acceleration due to gravity is only $g = G\, M / r^2$ when you are outside a sphere of mass $M$ (and at a distance $r$ from the center). When you are inside the sphere, that formula doesn't apply. Instead, you use the same formula with $M$ replaced by the amount of mass inside your radius.

If you think about the point at the precise center of the earth, just think about all the little pieces of earth that are pulling on that point. For each piece, there is an exactly equal piece on the opposite side from the center that is pulling the exact same amount, but in the opposite direction. So adding up the contributions from all those little pieces, you get a total of no acceleration.

You would expect this result in general because the acceleration has to accelerate you in some direction. Everywhere other than the center of the sphere, there's a natural choice for the direction in which you would be accelerated: towards the center. But at the center, there's no choice of direction that respects the spherical symmetry of the situation. The only acceleration that doesn't break the spherical symmetry is zero acceleration.

It turns out that the general formula for acceleration that works even inside the sphere of mass doesn't use the total mass $M$, but the portion of the mass that's at a radius less than $r$. This is called the Shell Theorem. It even works in general relativity, where it's called Birkhoff's Theorem

Of course, all of this is an idealization, as Martin points out in the comments. First of all, you have other objects around the Earth that alter the acceleration — the Moon, the Sun, the other planets, etc. Of course, even if you ignore all of them, the Earth itself is not spherically symmetric or even rotationally symmetric, so the "center" might not even have zero acceleration (depending on how you define center).

Mike
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    Although in practice it doesn't work for Earth because the Earth-moon system CoG is not in the center of the Earth. – Martin Beckett Feb 08 '18 at 03:09
  • @MartinBeckett Okay, yes. That's certainly true. Plus even the Earth alone is not spherically symmetric, etc. I'll add the clarification that this is an idealization to the answer. – Mike Feb 08 '18 at 03:14
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    Just to add, we should also look at how we got a formula before we start using it. The assumptions and approximations that were used are really important to know. You would have figured out the answer yourself if you go back and go through the proof again. – LostCause Feb 08 '18 at 03:15
  • Spherical symmetric isn't required, so long as it's rotationally symmetric. Earth's equatorial bulge cancels out - it's just that annoying 2nd body! – Martin Beckett Feb 08 '18 at 04:17
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    It’s not rotationally symmetric either. – Mike Feb 08 '18 at 04:56
  • @Mike Can you give me the formula of calculating g when a mass is underground in Earth and its derivation? – Theoretical Feb 08 '18 at 09:13
  • It's just what I said: $g=G m/r^2$, where $m$ is the portion of the mass that's at a radius less than $r$. The derivation is simplest with Gauss's Law: $\oint_{\partial V} \mathbf{g}\cdot d\mathbf{A} = -4\pi G \int_V \rho, dV$. Taking the volume of integration $V$ to be a sphere of radius $r$ centered on the mass, and noting the spherical symmetry, this simplifies to $-g, 4\pi r^2 = -4\pi G m$. Rearrange that, and you get $g = Gm/r^2$. – Mike Feb 08 '18 at 16:04
  • @Mike I am not familiar with integration very much.So an easier proof without integration will help me understand better. – Theoretical Feb 08 '18 at 16:34
  • Newton's proof isn't as convincing, but it doesn't explicitly use integrals. – Mike Feb 08 '18 at 16:43