# Does gravity increase the closer to the core you get?

Or does the mantle and crust above you counteract the increase at one point and it actually decreases?

The below figure, taken from Wikipedia shows a model of the free fall acceleration, i.e., 'gravity'. The left-most point corresponds to the center of the Earth; then further right at $$6.3\cdot1000$$ km you are at the Earth's surface; and then further out you move into space. You can follow the blue line for PREM to get an idea of the average (expected) gravity. As you see, the gravity actually increases slightly within the Earth (reaching a maximum at the core-mantle boundary), but tapers down within the core.

To make this kind of calculations, you must think of the Earth like an onion: made up of many concentric spheres. Whenever you move a bit deeper into the Earth, you strip off all the layers you've crossed. As you get closer to the center of the Earth, there are fewer and fewer layers, and eventually, there's nothing left at the center! The reason why gravity goes up ever so slightly within the Earth is that you get close to the much denser core material. If the density of the Earth were constant (per the green 'constant density' line), the gravity would just decrease linearly.

See the other answer and the discussion below for some more details on the math and procedures required to make these calculations.

• How well-validated is this PREM model? In other words, how certain are we about the accuracy of the dark blue line? – gerrit Feb 6 at 9:14
• @gerrit PREM stands for en.wikipedia.org/wiki/Preliminary_reference_Earth_model , it is the most commonly used geophysical model of the Earth. Of course, it's simplified (it's 1-D spherically symmetric), but it matches all known Earth properties very well. See the introduction (or the whole document) here: geo.uu.nl/~berg/geodynamics/lecturenotes.pdf for more details. – Erik Feb 6 at 10:39
• "Whenever you move a bit deeper into the Earth, you strip off all the layers you've crossed." This implies that if you are anywhere inside of a hollow ball, the effect of gravity on you from the ball is always zero, irregardless of the thickness of the material that the ball is made of. Is this true? Would there be zero gravity from the ball, even if you were touching the inside of it? – John Pankowicz Feb 6 at 17:09
• @JohnPankowicz Yep! I think it's a pretty cool result! See: Shell Theorem – Mark Omo Feb 6 at 17:24
• @NuclearWang I'm not sure "enjoy" would be the right word here - unless you take pleasure from being crushed inside a molten ball of nickel-iron alloy... – Darrel Hoffman Feb 6 at 19:49

This is a companion to the other answer. The other answer gives values for what the acceleration actually is based on knowledge about the composition of the planet. In this answer I want to introduce some of the mathematical tools that allow you to reason about how gravity varies in and around planets. David Hammen's answer, which I think he wrote overlapping with the second part of this one, also has versions of some of the maths in the second part of this one, and is very worth reading.

## An idealised (and incorrect) Earth

First of all consider an idealised Earth: in this model Earth let's assume that the planet is:

• spherical;
• not spinning;
• has constant density throughout.

The first two of these are reasonably good approximations. the third is not a good approximation: the density of the Earth in fact varies significantly with depth.

For this simplified model we can use a famous result, proved by Newton, known as the shell theorem. This says two things:

1. a spherically-symmetric mass distribution can be treated gravitationally outside it as if it was the same mass concentrated at a point at the centre of the sphere;
2. a spherically-symmetric shell of mass (so a hollow sphere) has no gravitational influence on any body inside the shell.

From this it immediately follows that, at any point inside our model Earth, you can ignore all the mass outside that point and then just consider the gravitational influence of the mass inside the point, and we can also consider that mass to be concentrated at the centre of the model Earth.

So, well, let the density of the planet be $$\rho$$. Then the mass within a radius $$r$$ is $$m(r) = 4\pi r^3 \rho/3$$, for $$r\le R_E$$, where $$R_E$$ is the radius of the Earth. And using Newton's law of gravitation, we get the acceleration due to gravity, $$g(r)$$ to be

\begin{aligned} g(r) &= \frac{Gm(r)}{r^2}\\ &=\frac{4\pi G\rho r}{3}\quad r \le R_E \end{aligned}

Which shows you that $$g$$ reduces linearly as you approach the centre of the Earth in this model.

## A more realistic model Earth

A nice thing is that you can use most of the above even if $$\rho$$ is not constant: so long as $$\rho = \rho(r)$$, only, then you can still use the shell theorem to compute $$g(r)$$, but you now have to do some integral over $$r$$ to compute $$m(r)$$. So you can relax the third requirement above, at the cost of a little more work. I believe that, in practice, $$\rho$$ does depend only on $$r$$ to a fairly good approximation, so this model will let us compute how gravity actually does vary.

In the case where $$\rho = \rho(r)$$ we can use the shell theorem to write $$g(R)$$ at some radius $$R$$ (I'm using $$R$$ not because I want to integrate over $$r$$).

\begin{aligned} g(R) &= \frac{G m(R)}{R^2}\\ &= \frac{G}{R^2}\int_0^R 4\pi r^2 \rho(r)\,dr \end{aligned}

Where the integral is the mass of the planet to radius $$R$$. It's clear from this that $$g$$ is constant if $$\rho(r) = \rho_0/r$$: a planet with a density which goes down as $$1/r$$ has gravitational acceleration which is constant throughout its volume. Note that this planet has infinite density at its centre: since that's never true for planets – we're not dealing with black holes here, and if we were we could not use Newtonian gravitation – this means that for all real planets, $$g \to 0$$ as $$r \to 0$$.

More generally we can differentiate the above to get an expression for $$dg/dR$$:

$$\frac{dg}{dR} = 4\pi G\rho(R) - \frac{2G}{R^2}\int_0^R 4\pi R^2\rho(r)\,dr$$

Defining $$\bar{\rho}(R)$$ as the mean density of the planet to radius $$R$$, then the integral in the second term is just $$4\pi R^3\bar{\rho}(R)/3$$, so we get

$$\frac{dg}{dR} = 4\pi G\left(\rho(R) - \frac{2}{3}\bar{\rho}(R)\right)$$

This tells you how $$g$$ behaves with depth, and in particular whether it goes up or down with depth for a general spherically-symmetric mass distribution, if you know the radius, the density at that radius, and the mean density inside that radius (or equivalently the radius, the density, and the mass inside that radius).

So, if we can measure how the density of Earth varies with depth, and that it varies only with depth to a good approximation, then this model can be used to compute how $$g$$ varies with depth, to a good approximation.

• This cow is a bit too spherical. It would be better to show the circumstances under which gravitational acceleration increases or decreases with increasing distance from the center, assuming a spherical mass distribution. A uniform mass distribution does not make sense for an object the size of a planet (even a very small planet). – David Hammen Feb 7 at 4:39
• @David Hammen I do not understand your comment. On one hand, you say "it would be better....assuming a spherical mass distribution". Then you say "A uniform mass distribution does not make sense...". Which is the difference? – George Menoutis Feb 7 at 8:08
• @GeorgeMenoutis The first one means that the density depends on the radius. Since the density of a planet is far from uniform, the linear decrease isn't a useful approximation, but the mathematics of spherically symmetric planet isn't that much harder and it would be much closer to actual truth. – JiK Feb 7 at 12:56
• In spite of any objections or quibbles, this question and both these answers are great, all taken together, because they show how different models (point mass at center, uniform density, modeling density variations) get you different, and more and more accurate results. – David Conrad Feb 7 at 15:48
• @SemidiurnalSimon: In the current version it is more explicit that the initial idealisation is wrong and has a second section which deals with general spherically-symmetric mass distributions. Finally it also now explicitly says that it's there to provide references the mathematical tools you need, which it should always have said. – tfb Feb 12 at 11:20

I'll make the same assumption made in the Preliminary Reference Earth Model, which is that density inside the Earth depends only on radial distance from the center of the Earth. This means that Newton's shell theorem applies: The gravitational acceleration at some point inside the Earth depends only on the material closer to the center of the Earth than the point in question. Denoting $$M(r)$$ as the mass of all of the material at a radial distance less than $$r$$, then the gravitational acceleration at some point inside the Earth is $$g(r) = \frac{G M(r)}{r^2}\tag{1}$$ Differentiating with respect to $$r$$ results in $$\frac{dg(r)}{dr} = \frac G{r^3}\left(r\frac{dM(r)}{dr} - 2M(r)\right)\tag{2}$$ One way to calculate $$M(r)$$ is as a volume integral:$$M(r) = \int_0^r 4 \pi x^2 \rho(x) dx$$ where $$\rho(r)$$ is the local density at a radial distance $$r$$ from the center of the Earth. Differentiating with respect to $$r$$ results in $$\frac{dM(r)}{dr} = 4 \pi r^2 \rho(r)\tag{3}$$ Another way to express $$M(r)$$ is via the average density $$\bar\rho(r)$$ of all of the material at a radial distance less than $$r$$:$$\bar \rho(r) \equiv \frac{M(r)}{V(r)} = \frac {M(r)}{\frac43 \pi r^3}$$ or $$M(r) = \frac43 \pi r^3 \bar \rho(r)\tag{4}$$ Applying equations (3) and (4) to (2) yields $$\frac{dg(r)}{dr} = 4 \pi G\left(\rho(r) - \frac23\bar\rho(r)\right)\tag{5}$$ In other words, gravitational acceleration increases with increasing distance from the center of a planet in regions where the local density is more than two thirds of the average density of all of the stuff at a lesser distance, but decreases with increasing distance in regions where the local density is less than this limit. In the case of a hypothetical uniform density planet, the local density is equal to the average density and thus gravitational acceleration increases from the center up to the surface.