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Given a solid ball with density $\rho$ and radius $R$, we dig a "thin tunnel" through its center and place an object with mass $m$ at distance $r$ from the center.

Computing the gravitational force, how come can we ignore all the mass further away than $r$? All answers I found rely on Gauss' Law for gravity, which although I know I don't want to use. My main problem is setting up the explicit integral we need to calculate.


marked as duplicate by David Hammen, Floris, Kyle Kanos, Qmechanic Jul 12 '17 at 23:26

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You need only to establish zero field inside an infinitesimally thin homogeneous spherical shell. The same result follows immediately for shells of finite thickness, as these can be regarded as nests of infinitesimal shells! Note that the density can vary from shell to shell without spoiling the result.

I'd then divide the infinitesimal shell into 'hoops' following lines of latitude. You should be able to form an integral to give you the field strength at an arbitrary fixed point on the common axis of the hoops. It's clear from the symmetry of the hoop that components of field perpendicular to this axis will cancel.

Hope this will get you started. If you find it difficult, take comfort from the fact that Newton himself got stuck over this for some time…

I've just had a go at this myself, and got stuck trying to get the integral of the fields in terms of a single variable (though I'm sure it can be done). And (unlike Newton) I have at my disposal all the slick notation and methods of 'modern' algebra and calculus! But adding the potentials due to the hoops was much easier, partly because potential is a scalar. The total potential came out to be independent of the position of the 'fixed' point, meaning that the field strength is zero.

  • $\begingroup$ I thought about the thin spheres, but I didn't think about dividing them further into loops. Let's say that the mass density (per length) of the loop is $\sigma$. The field's size at angle $\theta$ from the line between the center and the object is $\frac{G \sigma (R d\theta)}{R^2+r^2-2Rr\cos\theta}$. Multiply by $\cos\theta$ for the component parallel and I get $\sigma GR\int_0^{2\pi}\frac{\cos\theta}{R^2+r^2-2Rr\cos\theta}d\theta$, which doesn't turn out to be $0$, and has a rather ugly antiderivative. Am I missing something? $\endgroup$ – Theorem Jul 13 '17 at 20:08
  • $\begingroup$ I'm sorry but I don't really understand what you're doing. Is $R$ the radius of the shell? In that case the radius of a hoop at polar angle $\theta$ from the sphere's centre is $R sin \theta$. You need to find the field strength at some point other than the centre. I found it at a point P a fixed distance $Z$ from the centre, along the diameter which is the common axis of the hoops (the z-axis if you like). I then found I wanted another angle, $phi$, at P, between the axis and any point on the loop… This is probably incomprehensible. I'll try and append my diagram if you're interested. $\endgroup$ – Philip Wood Jul 13 '17 at 21:26
  • $\begingroup$ Haha, I divided them into lines of longtitude appearantly. Still odd that it didn't work though. I think I understand your approach and the field of the loop at angle $\theta$ is $2\pi\sigma G R \frac{\sin \theta \cos \theta}{R^2+r^2-2Rr\cos\theta}$ which is also not nice to integrate... $\endgroup$ – Theorem Jul 14 '17 at 8:46
  • $\begingroup$ @Theorem The potential at the point P I talked about above, due to a spherical shell of thickness $R$ is $$\int_{r=R+z}^{R-z} \frac{GR d \theta 2 \pi R sin \theta dR}{r}$$ in which $r$ is the distance from the hoop circumference to point P. But if you've drawn the diagram you'll see that $$r^2=R^2 +Z^2 -2RZcos \theta$$ in which $R$ and $Z$ are constants. Differentiating: $$r dr =RZ sin \theta d \theta.$$ Eliminate $sin \theta d \theta$ and integrate wrt $r$. Good luck! $\endgroup$ – Philip Wood Jul 15 '17 at 21:53
  • $\begingroup$ Typo in last comment (top line): I should have said, "…of thickness d$R$…" $\endgroup$ – Philip Wood Jul 16 '17 at 8:01

Because the mass at a distance greater than r is pulling mass in different directions. Due to the spherical symmetry of the problem however all these force vectors cancel. If you consider a 2-dimensional plane through the center of earth, a point on the right side of your mass, with a distance greater than r, is pulling your mass to the right. There is a corresponding point on left doing exactly the same but pulling to the left. Hence all these outward forces cancel.

  • 1
    $\begingroup$ They only cancel because of the inverse square law. For other central forces, the cancellation only occurs at the centre. $\endgroup$ – badjohn Jul 12 '17 at 19:19
  • $\begingroup$ You proved they cancel in the axis orthogonal to the plane, I think. What about the field in the direction parallel to it? $\endgroup$ – Theorem Jul 13 '17 at 19:49
  • $\begingroup$ @Theorem, the circle (circular disk really) I referred to could be any circle with the center of earth as the center of circle. You can just rotate the spherical earth. All circles yield the same due to the spherical symmetry. $\endgroup$ – jmh Jul 14 '17 at 20:15

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