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Efficiently calculating the magnitude of gravitational force

by Richard Nash
Tags: gravitation force
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Richard Nash
Dec9-12, 10:31 PM
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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.
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Dec10-12, 10:06 AM
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You can use Gauß' law and symmetry. But if you want to calculate it via an integral, I don't think there is an easier way.
Andrew Mason
Dec10-12, 10:11 AM
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Quote Quote by Richard Nash View Post

Question: I would like to avoid these calculations and I was wondering if there exists a better solution.
Welcome to PF!

You certainly need to use calculus. When one does this kind of calculus problem one has to use as much symmetry as possible. It seems that is achieved by dividing the sphere into rings perpendicular to the axis through the centre of the sphere and the point mass, calculating the gravity from a ring and integrating from one end to the other. If that is what they are doing, that is as simple as it gets.


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