From 5edcf3cae9e70c00b786b81fffac525969420692 Mon Sep 17 00:00:00 2001 From: bangerth Date: Thu, 2 Dec 2010 05:00:34 +0000 Subject: [PATCH] A better gravity model. git-svn-id: https://svn.dealii.org/trunk@22897 0785d39b-7218-0410-832d-ea1e28bc413d --- deal.II/examples/step-32/doc/intro.dox | 21 +++++++++++++++++---- 1 file changed, 17 insertions(+), 4 deletions(-) diff --git a/deal.II/examples/step-32/doc/intro.dox b/deal.II/examples/step-32/doc/intro.dox index 7ef5f97651..a1d56a227a 100644 --- a/deal.II/examples/step-32/doc/intro.dox +++ b/deal.II/examples/step-32/doc/intro.dox @@ -780,7 +780,7 @@ the following quantities: 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): @@ -818,7 +818,7 @@ the following quantities: 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] @@ -836,8 +836,21 @@ the following quantities: 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]
  • 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 -- 2.39.5