# Gravity at the center of the earth

I’ve been having a very interesting exchange with an io9 “Ask a Physicist” reader named Pat about what would happen if you were to dig a hole through to the center of the earth. Would you get heavier or lighter as you descended? While this wasn’t really appropriate for “Ask a Physicist,” it did remind me of a classic quals problem, and since our own grad students are taking the quals in a few weeks, here’s my tribute to them.

And yes, I’m assuming the earth is a sphere, instead of an oblate spheroid. Not that that would make too much of a difference.

You calculate the gravitational acceleration at any height, r, using: $a(r)=frac{GM(r)}{r^2}$

Incidentally, I’m deeply sorry about the equations; that’s why this one is labeled “technical.”

(Note: The original version of this did not include LaTeX. I am indebted to twitter follower thrugl, who pointed me to an excellent wordpress plugin).

Anyway, back to the central thrust. For a uniform density earth (which it’s not), the solution is quite simple: $a(r)=gfrac{r}{R_odot}$

Which is kind of cool. You become weightless at the center, and since the gravitational force is proportional to the distance, you would bounce around to the other side of the earth and back with a period of $P=2pi sqrt{frac{R_odot}{g}}$
In case you’re curious, that’s about 1.4 hours, exactly the amount of time it would take to send a rocket around the earth that just barely skimmed the surface.

But Pat wanted to know if with the real earth there wasn’t any situation in which you might gain weight on the way down. This got me thinking about a much more general relationship. It’s fairly straightforward to show that: $frac{da(r)}{dr}=left(frac{3rho}{langle rho rangle}-2right) frac{a(r)}{r}$

Where $langle rho rangle$ is the average density interior to a particular radius. So for the gravity to increase inwards, we need: $frac{rho}{langle rhorangle} < frac{2}{3} [/latex] For the earth, the average density is about [latex]5.5 g/cm^3[/latex], about 5 times that of water, which means that if the surface has a lower density than about [latex]3.6 g/cm^3[/latex], you're in luck. What happens as you move through the crust? Well, as Pat points out, the density there is only about [latex]2.5 g/cm^3$, which more than satisfies our criterion. In fact, it looks like your weight wouldn’t start to actually decrease until you got near to the bottom of the Mantle or to the top of the outer core. The density there is about $5.6 g/cm^3$, which is still less than 2/3 of the average density of the core (closer to $10-13 g/cm^3$ ).

For what it’s worth, the additional gravity wouldn’t amount to all that much. Even in the outer core, you’d only weigh about 10% more than surface normal.

Fun question.

-Dave

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