Re: Where on, or in, the Earth would you weigh the most.

Very nice question! The answer is not as obvious as we usually think, because the density of the earth is not constant throughout its volume. If we do take the density as constant then we may calculate that the acceleration of gravity will be maximum at the surface of the earth. In reality, because the density is not constant, the maximum acceleration occurs at a depth of about 2000 kilometers below the surface!!!!

I will start the calculations by referring to a law that is called the "Gauss Law of Gravitation" (see any freshman-level college physics textbook. My favorite book, because I was in undergraduate school in the late 1960's, is Halliday and Resnick's physics text). This law states that, for a spherically symmetric distribution of mass, the gravitational effect is equivalent to the total mass of the shell residing at the center of the sphere!! One of the effects is that if one goes below the surface of the earth, the spherical shell above has no effect on the net gravitational acceleration. Again, this assumes a spherically symmetric distribution.

So we will assume that the distribution of mass of the earth is spherically symmetric, which is a reasonable assumption. This is not precisely true because the earth is just not entirely smooth in its distribution of mass. But any differences are not important to this discussion. The main thing that is important is the radial distribution of the mass of the earth. That is, the earth is more dense as we go toward the center.

I consulted a book called "Smithsonian Physical Tables" (published by the Smithsonian Institution Press. My copy is 9th edition, 1969). I found a table of the density of the earth versus depth, which is based on calculations such as the average density, the precession of the earth, the flattening of the earth, and the known elastic behavior of the earth. Anyway, I put the table data into a spreadsheet, plotted it, and then fitted a sixth-order polynomial to the data (that's the highest-order polynomial that my version of Excel supports). Using that polynomial to represent the density versus depth, I did the following calculations, using the spreadsheet because it's easy.

Here is the table I started with:

And here are the coefficients of the sixth-order fit:

The law of gravitation (not Gauss's law) states that the gravitational force between two masses is
F = GmM/r²
where G is the universal gravitational constant, m is one of the masses, M is the other mass, and r is the distance separating the centers of mass of the objects. We can "remove" the test object (the one that we want to experience the acceleration, and usually the smaller object) by dividing its mass through in the equation, and we obtain an acceleration:
F/m = a = GM/r²
We get into the nuts and bolts of the details when we want to compute the mass, M, and for that we use the polynomial from above. We can represent a thin spherical shell of mass as dM (which means "a little itty bitty thin shell of M"), and, assuming that the earth is spherically symmetric then the density will be constant over a little itty bitty thickness of the earth, dr. So dM = r dV where r is the density and dV is the volume of the thin shell. So now we have
a = G/r² S (r dV)
(I am using the symbol S to represent an integral)
where the integral is from zero to r.
Now, the area of the surface of a sphere of radius r is 4pr² so for a thin shell of thickness dr the volume is
dV = 4pr² dr
Putting this together we have
a = G/r² S (r 4pr² dr) = 4pG/r² S (rr² dr)
It is tempting to cancel out the r dimension in this equation!! But that is the wrong thing to do!! If we were to cancel out the r² terms we would be off by a factor of 3 in the final calculation. This is because the r outside the integral is the final radius at which we are calculating the acceleration, but inside the integral the variable r is a function of r and needs the r² term to be multiplied times the polynomial.

Let's first use the assumption that the earth's density is constant. If we do that then r can be taken outside the integral and we obtain
a = 4pGr/r² S (r² dr)
and when we do the integral we obtain
a = 4pGr/r² (r³ / 3) = 4pGrr/3
Put all the numbers in, using the accepted average density of the earth as 5.522 gm per cubic centimeter, and we get a = 983 cm per second per second at the surface of the earth. As you probably know, the measured value for a is about 981 cm per second per second, so we aren't far off. Also, note that the function for a is a steadily increasing monotonic function in r, so the maximum value is at the surface of the earth. Once you leave the surface the acceleration declines by r^-2 because there is no additional mass to contribute and the distance between the centers of the masses increases.

Now, to re-do this using the non-constant density of the earth, the variable r is replaced by the sixth-order polynomial, which is a function in r. Doing the integral causes each term of the polynomial to increase by one in its order of r and to be divided by that final order of r. (I know this is a bit technical, and you can ignore it if you want, but I wanted to provide this level of detail in case any other reader of this answer wants to work through the problem!)

Finally, using the coefficients of the polynomial and the proper order of the terms of r from the integrand, the terms of the equation are computed in the spreadsheet and added up, and multiplied by all the constants (such as 4 and G and p). Amazingly, the total mass of the earth comes out to be about 6.0x10²⁷ g, which is only about 0.3% high, and the acceleration due to gravity at the surface comes out to be about 985 cm per second per second, which is only about 0.4% high!!! That's amazing since we started out with a table of the estimate of the variation of the density of the earth at different depths! A plot of the acceleration of gravity at various depths using the computed data shows that the maximum acceleration of about 1052 cm per second per second occurs at a depth of about 2000 kilometers. That acceleration is about 6.8% higher than at the surface.

Here's a table summarizing the acceleration information. (Notice that I have changed the units of the acceleration to m/sec² instead of cm/sec². I did this to have the same units as the table that follows in the next section below.)

After I finished the work above, I found in the Halliday & Resnick textbook a chapter problem that gives some apparently useful information. Problem 35 of Chapter 16 gives the following table of gravitational acceleration versus depth. It is apparently real data from the 1960's.

The overall structure of the data that this table represent is very very close to what I calculated above. Any differences are likely due to the rounding of the corners of the 6th-order curve that was found for the original table of information. That is, I think the density of the earth changes more rapidly than what my 6th-order polynomial allows. According to the table from Halliday & Resnick the maximum acceleration is at a depth of about 2900 kilometers, which isn't far from the 2000 kilometer estimate that my original calculations found! Also, Eder Molina (who is a PhD geophysicist in Sao Paulo Brazil at the Dept. of Geophysics, Institute of Astronomy and Geophysics, and one of our answering scientists) reviewed my calculations (at my request!) and found in his library a graph of both the density and gravity versus depth, and his information is very close to what I found in Halliday & Resnick.

Email me at madlink@mindspring.com and I will email you a copy of the spreadsheet if you would like to look at the details.

John Link, MadSci Physicist