Doc Al said:
I don't think so. Taking a simplified model of a spherically symmetric earth, as long as the symmetry is maintained it doesn't matter if all the mass were contained in a thin shell: the gravitational field outside the shell will be the same.
But what I'm saying is that model in unphysical. For a hollow earth, its simply wrong.
Here, I'll work through the math:
For a spherical shell of radius R, width \Delta r, and uniform density \rho the total mass M is:
<br />
M = \int\int\int \rho r^2 \sin \theta dr d\theta d\phi
<br />
\qquad = \int^{R+\Delta r}_{R} \int^{2\pi}_0 \int^{2 \pi}_0 \rho r^2 \sin \theta dr d\theta d\phi
so we can stipulate that the mass of some volume slice in the shell is given by dm as:
<br />
dm = \rho r^2 \sin \theta dr d\theta d\phi
The gravitational field generate by this mass slice is given by:
<br />
g = G \frac{dm}{s^2} \vec{\epsilon}
where s is the distance to the mass slice, and \vec{\epsilon} is the unit vector point towards the mass slice. It can be show geometrically that when calculating the gravtiational field at the point P(R, 0,0) (surface of the sphere)
<br />
s = 2 r \sin \frac{\theta}{2}
The unit vector point from P to the mass slice \vec{\epsilon} is:
<br />
\vec{\epsilon} = \left [ \begin{array}{l} 1 \\ \theta \\ \phi \end{array} \right ]
Now, because this is a sphere, and as such is rotationally symmeticraly we can assume that
<br />
g_\theta = 0
<br />
g_\phi = 0
so we get:
<br />
g_r = \int \int \int G \frac{\rho r^2 \sin \theta dr d\theta d\phi}{4 r^2 \sin^2 \frac{\theta}{2}} dr d\theta d\phi
Integrating only over a single hemisphere, and then multiply by two (to eliminate negative r) gives
<br />
\qquad = \frac{G \rho}{4} \int_{R}^{R+\Delta r} \int_0^{\pi} \int_0^{2\pi} \frac{\sin \theta}{\sin^2 \frac{\theta}{2}} d\phi d\theta dr
<br />
\qquad = \frac{G \rho}{8} \int_{R}^{R+\Delta r} \int_0^{\pi} \int_0^{2\pi} \sin \theta - \frac{\sin 2 \theta}{2} d\phi d\theta dr
(check your trig identities if you're not sure how I got that. i used the double angle identity)
<br />
\qquad = \frac{ G \rho \pi \Delta r}{2}<br />
Now, to get a number for that, let's say our Earth has a \Delta r of 0.01 percent the normal radius of the Earth (this derivation requires that \Delta r \ll R). This gives:
<br />
G = 6.67 \times 10^-11
<br />
\Delta r = 63800 \text{m}
<br />
\rho = 575641 \frac{kg}{m^3}
gives g = 3.847 \frac{m}{s^2}.
Well that is odd. it came out smaller. Now this is going to bug me even more. I don't suppose someone could check this for me to make sure I didn't screw something up along the way? interesting result though, if its right.
