54.5: Earth Density Model
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Suppose we have a uniform, spherical body (such as a planet) of radius R and mass M. What is the acceleration g due to gravity as a function of r for r both inside and outside the body (0 \leq r<\infty) ?
First, consider the case where we're inside the body (r \leq R). In this case, the acceleration due to gravity at r is g(r)=G m / r^{2}, where m is the total mass inside a sphere of radius r :
m=\frac{4}{3} \pi r^{3} \rho
where the (uniform) density \rho=M /\left(\frac{4}{3} \pi R^{3}\right). Thus
g(r)=\frac{G M}{R^{3}} r \quad(0 \leq r<R)
so inside the body, g \propto r.
Second, consider the case where we're outside the body (r>R). In this case, the total mass inside a sphere of radius r is M, and so
g(r)=\frac{G M}{r^{2}} \quad(r \geq R)
so that outside the body, g \propto 1 / r^{2}. The maximum value of g is at the surface, g=G M / R^{2} at r=R (Figure \PageIndex{1}).

However, planetary bodies are generally not uniform. For example, the Earth has a higher density closer to its core, and its density decreases closer to the surface. One density model of the Earth given by Dziewonski and Anderson { }^{1} is shown in Figure \PageIndex{2}. We can use this density model to compute a more realistic model of g(r) inside the Earth:

g(r)=\int_{0}^{r} \frac{G \rho(r)}{r^{2}} d V=\int_{0}^{r} \frac{G \rho(r)}{r^{2}}\left(4 \pi r^{2}\right) d r=4 \pi G \int_{0}^{r} \rho(r) d r
The result is plotted in Figure \PageIndex{3}. We see that in a more realistic model of the Earth's interior, the maximum value of the acceleration to to gravity g occurs just outside the outer core, where g=10.7 \mathrm{~m} / \mathrm{s}^{2}.

{ }^{1} Dziewonski, A.M., and Anderson, D.L., Preliminary Earth reference model. Physics of the Earth and Planetary Interiors, 25 (1981) 297-356.