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 [itex]M[/itex] and a particle of mass [itex]m[/itex] located at a distance [itex]r[/itex] from the center.
The shell has been divided into narrow rings.[itex]R[/itex] has been assumed to be the radius of the shell with thickness [itex]t[/itex] ([itex]t<<R[/itex]). The ring at angle [itex]\theta[/itex] which subtends angle [itex]d\theta[/itex] has circumference [itex]2\pi R\sin\theta[/itex].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 [itex]\alpha[/itex] be the angle between the force vector and the line of centers, [itex]dF=\frac{Gm\rho dV}{r'^2}\cos\alpha [/itex] where [itex]r'[/itex] is the distance of each part of the ring from [itex]m[/itex].
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.
The shell has been divided into narrow rings.[itex]R[/itex] has been assumed to be the radius of the shell with thickness [itex]t[/itex] ([itex]t<<R[/itex]). The ring at angle [itex]\theta[/itex] which subtends angle [itex]d\theta[/itex] has circumference [itex]2\pi R\sin\theta[/itex].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 [itex]\alpha[/itex] be the angle between the force vector and the line of centers, [itex]dF=\frac{Gm\rho dV}{r'^2}\cos\alpha [/itex] where [itex]r'[/itex] is the distance of each part of the ring from [itex]m[/itex].
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.