Tuesday, 15 September 2015

Why does gravity decrease as we go down into the Earth?




We all know that gravity decreases as the distance between the two increases. Hence


$$ F = G \frac{Mm}{r^2}. $$


Hence the acceleration due to gravity $$ g =\frac{F}{m}= G \frac{M}{r^2} $$


increases as $r$ decreases. Then why does it decrease as we go deep into the earth?



Answer



It's actually not entirely true that the strength of the Earth's gravitational field decreases as a function of depth. It is true for certain regions in the Earth, but it's untrue for others because of the non-trivial dependence of the Earth's density on depth.


To see what's going on, assume that the Earth is a sphere whose density is spherically symmetric.


Now consider a mass $m$ at some radius $r$ from the center of the Earth. Using Newton's Law of Gravitation, one can show that that given the spherical symmetry, the gravitational attraction on $m$ of all mass with radii greater than $r$ exert no net force on it. It follows that only the mass with radii less than or equal to $r$ contribute to the gravitational force on $m$, which, by the Law of Gravitation is \begin{align} F(r) = G\frac{M(r)m}{r^2} \end{align} where $M(r)$ is the mass of stuff at radii less than or equal to $r$. Notice, then, that $F(r)$ will be an increasing function of $r$ (and will decrease as $r\to 0$), provided $M(r)/r^2$ is an increasing function of $r$.


Now, If the Earth were uniformly dense with density $\rho_0$, then the mass within a radius $r$ would be \begin{align} M(r) = \frac{4}{3}\pi r^3 \rho_0 \end{align} namely just the density times the volume of a sphere of radius $r$, and in this case the strength of the gravitational field as a function of radius would be \begin{align} g(r) = \frac{F(r)}{m} = G\frac{1}{r^2}\frac{4}{3}\pi r^3\rho_0 = \left(\frac{4}{3}\pi g\rho_0\right) r \end{align} So in this case, it would be true that the strength of the gravitational field would decrease with increasing depth.


However, the Earth's density is not constant, and instead has some non-trivial dependence on $r$. See, for example, this PREM table. You can clearly see that there are regions in the Earth where the strength of the gravitational field increases because of the varying density.



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