I (and we all) know that acceleration due to gravity $g\propto\frac{1}{r^2}$.Now my question is can I use this for depth.If not,why?If we can use it for depth or not struck me when I was trying to prove that acceleration due to gravity at center of earth is zero (hence weightlessness).If I use the above formula that the value of $g$ tends to infinity and not zero(I put radius equal to 0).Please help me with the limitations of the above formula and how to prove that $g$ is zero at the center of the earth where radius is zero.

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    $\begingroup$ Inside uniform spherical bodies, only the mass present inside the spherical shell you are present at contributes to the gravitational force. The gravitational force due to a uniform spherical shell is zero inside it. Read: en.wikipedia.org/wiki/Shell_theorem $\endgroup$ Commented Aug 27, 2015 at 13:24
  • $\begingroup$ Related: physics.stackexchange.com/q/18446/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Aug 28, 2015 at 1:43

2 Answers 2


Limitations of $g \propto \frac{1}{r^2}$:

The relationship

$g(r) = G \frac{M}{r^2} \rightarrow g \propto \frac{1}{r^2}$

(where $g$ is the acceleration due to gravity, $G$ is the universal gravitational constant, and $r$ is the distance between the massive object and the accelerating object) is just fine in Newtonian physics—it doesn't need to be fixed. Its only limitations are in the context of general relativity, wherein gravity is much more complicated. (Well, that and the fact that it assumes the massive object is infinitely small, which is the crux of your question!)

Can we use depth as a coordinate?

Let's define depth $d$ as

$d = R-r$

where $R$ is the radius of the spherical object. (Hence $d=R$ at $r=0$ and $d$ is negative when $r>R$.) Equivalently, $r=R-d$. Now, let's plug this into the formula above:

$g(d) = G \frac{M}{\left(R-d\right)^2}$

Even here, we can see that if $d\rightarrow R$ (e.g. at the center where $r=0$), we're getting infinite gravity. This isn't right. This is because we used a formula that assumed that the massive point is infinitely small, and we're interested in the gravitation at the center of an extended object (i.e. an object that has volume). If all of the Earth's mass were concentrated at the center, and $R$ took on any finite value, that formula would be valid. Since it isn't, we'll need a new approach....

The Real Answer

An extended object can be thought of as being made of an infinite number of smaller point objects, sort of like as in this diagram:

voxel sphere

Each of those little masses $m_i$ is going to exert its own little force which will cause a little gravitational acceleration $g_i = G \frac{m_i}{r_i^2}$ (where $r_i$ is the distance from the little mass $m_i$ to the thing being accelerated).

One can show that if you have a spherical shell of uniform density, then the force of gravity inside that shell is zero, by summing up the forces from all of the little point masses in that shell. The force of gravity outside that shell is equal to the force of gravity if the mass of the shell were contained at a point in the center of the sphere. This is called the Shell Theorem, and is described in more detail here.

As a result, if you're in the middle of a ball with constant density $\rho$, you should think of the ball as a bunch of concentric spherical shells. All the shells above you are not contributing a net force on you. All the spherical shells below you are contributing as if all of their mass was contained at the center. Effectively, the force of gravity at radius $r$ from the center can be calculated via $g(r) = G \frac{M(r)}{r^2}$, where $M(r)$ is the mass of all the layers beneath you (i.e. a ball of mass $M(r) = \frac{4}{3} \pi r^3 \rho$).

Thus, the net force of gravity a distance $r<R$ from the center of the ball is

$g(r) = G \frac{M(r)}{r^2} = G \frac{\frac{4}{3} \pi r^3 \rho}{r^2} = \frac{4}{3}\pi G \rho r$

If you don't know the density of the ball offhand, but you know its total mass $M$ and its radius $R$, then its density is

$\rho = \frac{\text{mass}}{\text{volume}} = \frac{M}{\frac{4}{3} \pi R^3} = \frac{3 M}{4 \pi R^3}$

We can plug this into our equation for the gravity at any radius $r<R$:

$g(r) = \frac{4}{3}\pi G \rho r = \frac{4}{3}\pi G \left(\frac{3 M}{4 \pi R^3}\right) r = G M \frac{r}{R^3}$

Note that the $\frac{r}{R^3}$ term looks almost like the $\frac{1}{r^2}$ term. That means the units work out.

Lastly, to calculate this in terms of depth $d = R-r$:

$g(d) = G M \frac{R-d}{R^3}$

Thus, at the center of the ball, when either $r\rightarrow 0$ or $d \rightarrow R$, we can see that the net gravitational acceleration is zero:

$g(d=R) = G M \frac{R-R}{R^3} = 0$

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    $\begingroup$ Having derived that equation $g(r) = \frac{4}{3} \pi G \rho r$, one can have some fun. Imagine you're on a planet with constant density. If you were to increase the radius of that planet (but kept the density constant, always increasing the mass), the gravity at the surface of that planet would increase proportionately to its radius. Of course, this keeping the density (approximately) constant with radius usually doesn't happen, except in the cases of neutron stars and planetesimal accretion. $\endgroup$
    – jvriesem
    Commented Aug 27, 2015 at 14:53
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    $\begingroup$ @jvriesem-What do you mean by -'This is because we used a formula that assumed that the massive point is infinitely small'.Why is this assumed? $\endgroup$
    – Harmonic
    Commented Aug 28, 2015 at 16:44
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    $\begingroup$ Good question. This is complicated, and what I said wasn't strictly true. The law is best stated in its differential form: $dg(r) = G \frac{dM}{r^2}$, which works for an infinitesimal amount of mass $dM$ a distance $r$ away. This doesn't have limitations in Newtonian physics: it's true. The $g(r) = G \frac{M}{r^2}$ law is only valid for certain situations: 1) when all of the mass $M$ is concentrated at a single point a distance $r$ away (this is what I meant by that phrase), or 2) for special arrangements of mass, such that they add up just perfectly to give that law. $\endgroup$
    – jvriesem
    Commented Aug 28, 2015 at 18:35
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    $\begingroup$ @soham: To illustrate the 2nd point: you could imagine a large, thin disk of steel that has a mass $M$. In that case, if a test particle were a distance $r$ perpendicularly away from its center, the particle's acceleration would not follow that law, but would be different. The only special arrangement of mass (that I know of) that gives that $g(r)=G \frac{M}{r^2}$ law is a ball with total mass $M$ and spherically symmetric mass density--and only then if $r<R_{sph}$ (that is, you're outside the ball). This is a consequence of the Shell Theorem. (I could explain this more, if you'd like.) $\endgroup$
    – jvriesem
    Commented Aug 28, 2015 at 18:41
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    $\begingroup$ The vector and differential form of our favorite physical relationship is $d\vec{g}(\vec{r}) = -G \frac{dM(\vec{r})}{r^2} \hat{r}$. This is valid for an infinitesimal point-mass $dM(\vec{r})$ located at $\vec{r}$. For a distribution of mass (not a point-mass), we have to integrate. First, $dM(\vec{r})=\rho(\vec{r})dV$ (where $\rho(\vec{r})$ is the mass density at $\vec{r}$ and $dV$ is the differential volume element size ($dx dy dz$ in Cartesian coordinates). Thus we get $d\vec{g}(\vec{r}) = -G \frac{\rho(\vec{r})}{r^2} \hat{r} dV$. $\endgroup$
    – jvriesem
    Commented Sep 1, 2015 at 17:19

If you have a spherical body of radius $R$ with mass $M$, the gravitational field at any point at a radial distance $r$ is given by: $$\phi=\frac{GM(r)}{r^2}$$ where $M(r)$ is the mass enclosed inside a spherical shell of radius $r$. This is the only mass that matters in this case. (because of the Shell Theorem: https://en.wikipedia.org/wiki/Shell_theorem)

$M(r)$ is: $$\int_V\rho \ dV $$ where $\rho$ is the density and $V$ is the volume of the spherical region. (This integral is automatically zero for regions where there is no mass present, so this would evaluate to $M$ in any region which completely encloses the sphere.)

For the interior of a spherical mass distribution (with uniform density), $$M(r)=\frac{Mr^3}{R^3}$$ (This simply is $\rho V(r)$ in this case, because we've assumed the mass density to be constant.)

and hence the field expression in the interior of the body is: $$\phi=\frac{GMr}{R^3}$$ which is zero for $r=0$.

This is true for all mass distributions with spherical symmetry. (i.e those that only vary with $r$)

In a quite simpler way, you could arrive at this from symmetry. If there is some gravitation field at the center of a uniform mass distribution, which way would it move your test mass?

  • $\begingroup$ +1 for the symmetry argument! That kind of thinking should be encouraged. $\endgroup$
    – jvriesem
    Commented Aug 28, 2015 at 18:23
  • $\begingroup$ @Hritik Narayan-"If there is some gravitation field at the center of a uniform mass distribution, which way would it move your test mass?"....Is it towards the center? $\endgroup$
    – Harmonic
    Commented Aug 29, 2015 at 13:24
  • $\begingroup$ The mass distribution is symmetric (spherically), and hence the field has to be too. If the field is towards a certain direction in the center, it implies an asymmetry. (Why should the field be directed towards a certain direction if the mass distribution around it is symmetric?). From this we can conclude that the field has to be zero. $\endgroup$ Commented Aug 29, 2015 at 13:31

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