How did Newton determine the average density of Earth is twice the density of the surface rocks? [closed]

Question

I've read that Newton determined the average density of Earth is twice the density of the surface rocks, but I can't find his computation anywhere. I presume he used the differential calculus in some way. This is the correct result.

The density of granite is 2.765 g/cc, so the density of Earth should be about 5,530 kg/m^3.

https://www.nature.com/articles/ngeo954

For a spherically symmetric planet ( concentric shells of density that vary with radius), Gauss's Law says the gravity you feel at $r_0$ depends only on the mass at $r<r_0$. The force on you depends only on the Mass enclosed by the Gaussian surface.

So if you go down a 100m mine shaft, the gravity from the top 100m of the entire planet has no affect on you.

Any mass, $M$, below you acts as if it were concentrated at the center, with force per unit mass:

$$ g = G\frac M {r_0^2} $$

So for a spherical Earth, the gravity at the surface is the same if all the mass is at the center.

The acceleration due to gravity at the surface is:

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

If we reduce $r$, there is less mass $M$ pulling on us, but it is closer, so it pulls harder. 100 meters down g is essentially g at the surface.

At this point we use the quotient rule to find $\frac {dg}{dr}$, and set it equal to zero, to find a critical value.

$$\frac{dg}{dr}=\frac{G}{r^2}\left(\frac{dM}{dr}-\frac{2M} r \right)=0$$

or

$$\frac{dM}{dr} = \frac{2M} r $$

The LHS is the surface area times the surface rock density:

$$\frac{dM}{dr} = 4\pi r^2\rho(r)$$

For the RHS, the mass is the volume times the average density of Earth:

$$ M =\frac {4\pi} 3 r^3 \bar{\rho}$$

Therefore

$$4\pi r^2\rho(r) = \frac {4\pi} 3 r^3 \bar{\rho}\frac 2 r$$

Therefore

$$\rho(r) =\frac 2 3 \bar{\rho}$$

This isn't

$$2 \rho(r) = \bar{\rho}$$

Taking the result at face value, the gravitational constant is

$G=\frac{g}{2765 2πR_{earth}}$

So

$G=8.85408 \text { X } 10^{-11} \frac{m^3}{kgs^2}$

I noticed that this is almost the permittivity of free space $\epsilon_0 = 8.854187817 \text{ X } 10^{-12} \frac{C^2}{Nm^2}$.

If you define the permeability of free space as $\mu_0 = 4\pi \text{ X }10^{-8} \frac{N}{A^2}$ instead of $4\pi \text{ X }10^{-7}$, then you get the result that

$$G=\epsilon_0$$

Forgive the units.

Could someone be playing games with the physical constants for some reason unknown to me? Gauss's law is pretty solid.

Answer 1

I do not find any error in your calculations , it seems they are okay and gives lower bound estimate of Earth density $3/2\rho_{granite}\approx4\rho_{water}$. Btw, nice analysis.

However, according to this citation of Newton "principia" book :

Since, therefore, the common matter of our earth on the surface thereof is about twice as heavy as water, and a little lower, in mines, is found about three, or four, or even five times more heavy, it is probable that the quantity of the whole matter of the earth may be five or six times greater than if it consisted all of water

Newton was guessing Earth mean density by extrapolating density increase from surface water to those in mines. Your estimates are off from a Newton guess lower bound by $20\%$ error. So I would not say that your shot would be too far from that one of Newton.