the density $\rho$ is constant throughout the earth, we can produce an
analytical expression for the gravity vector (don't try to integrate above
equation somehow -- it leads to elliptic integrals; a simpler way is to
- notice that $-\Delta\varphi(\mathbf x) = 4\pi G \rho
+ notice that $-\Delta\varphi(\mathbf x) = -4\pi G \rho
\chi_{\text{earth}}(\mathbf x)$ and solving this
partial differential equation in all of ${\mathbb R}^3$ exploiting the
radial symmetry):
x)=\rho(\|\mathbf x\|)=\rho(r)$. In that case, one would get
@f[
\varphi(r)
- = 4\pi G \int_0^r \frac 1{s^2} \int_0^s t^2 \rho(t) \; ds \; dt.
+ = 4\pi G \int_0^r \frac 1{s^2} \int_0^s t^2 \rho(t) \; dt \; ds.
@f]
vector that varies with space and time, and does not always point straight
down.
- In order to not make the situation more complicated than necessary, we'll
- here just go with the constant density model above.
+ In order to not make the situation more complicated than necessary, we could
+ use the approximation that at the inner boundary of the mantle,
+ gravity is $10.7 \frac{\text{m}}{\text{s}^2}$ and at the outer
+ boundary it is $9.81 \frac{\text{m}}{\text{s}^2}$, in each case
+ pointing radially inward, and that in between gravity varies
+ linearly with the radial distance from the earth center. That said, it isn't
+ that hard to actually be slightly more realistic and assume (as we do below)
+ that the earth mantle has constant density. In that case, the equation above
+ can be integrated and we get an expression for $\|\mathbf{g}\|$ where we
+ can fit constants to match the gravity at the top and bottom of the earth
+ mantle to obtain
+ @f[
+ \|\mathbf{g}\|
+ = 1.245\cdot 10^{-6} r + 7.714\cdot 10^{13}\frac 1{r^2}.
+ @f]
<li>The density of the earth mantle varies spatially, but not by very
much. $\rho_{\text{ref}}=3300 \frac{\text{kg}}{\text{m}^3}$ is a relatively good average