I am reading Kolenkow and Kleppner's Classical Mechanics and they have tried to calculate the gravitational force between a uniform thin spherical shell of mass $M$ and a particle of mass $m$ located at a distance $r$ from the center.
The shell has been divided into narrow rings.$R$ has been assumed to be the radius of the shell with thickness $t$ ($t<<R$). The ring at angle $\theta$ which subtends angle $d\theta$ has circumference $2\pi R\sin\theta$. The volume is $dV=2\pi R^2t\sin \theta d\theta$ and its mass is $pdV=2\pi R^2t\rho\sin\theta d\theta$. If $\alpha$ be the angle between the force vector and the line of centers, $$dF=\frac{Gm\rho dV}{r'^2}\cos\alpha$$ where $r'$ is the distance of each part of the ring from $m$.
Next, an integration has been carried out using $\cos\alpha=\frac{r-R\cos\theta}{r'}$ and $r'=\sqrt{r'^2+R^2-2\pi R\cos\theta}$.
Question: I would like to avoid these calculations and I was wondering if there exists a better solution.
Best Answer
This is basically a proof of the Shell theorem
Simple. First, you must know that gravitational forces superpose.
Ok. First, calculate the force at a distance $r$ from the origin for a sphere of mass $M$ and radius $R$. You ought to get the result that it is $\frac{GMm}{r}$ if $r\geq R$, 0 otherwise. This is a standard textbook formula.
Ok, now, pretend the shell is made up of a larger sphere with positive mass and a smaller sphere with "negative" mass that cancels out all the mass except a shell of thickness $t$ on the outside. If you feel uncomfortable doing this, convert it into an electrostatics problem, calculate the formula with a similar method, convert it back.
Knowing $M,R,t$, you can calculate the masses and radii of the component spheres. Both have equal magnitudes of density.
Now, calculate the force at $r$, and superpose the two, taking care to subtract the force of the inner "negative" sphere.
That's it. You have (ab)used the principle of superposition to calculate this force, which is the final answer (it will have three cases -- the gravitational field inside will be 0).
The harder you try to avoid integration, the more fun newtonian mechanocs becomes :)
I'm on mobile right now (hard to add math) if you want I'll put more math in the answer tomorrow.