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Newton's Shell Theorem Question

  1. Aug 27, 2011 #1
    Hello everyone! I'm a math major taking a university physics course, and I have a question about the proof that perfectly spherical objects have equivalent gravitational force as a point mass located in the center.

    The proof in question can be seen in one form at:

    It's been a while since I did this kind of calculus (solids of rotation come to mind), but I'm having trouble accepting the step 1 assumption:

    "Applying Newton's Universal Law of Gravitation, the sum of the forces due to mass elements in the shaded band is

    [itex]dF = \frac{Gm \;dM}{s^2}[/itex]"

    It feels like there is a misuse of the infinitesimal here... the ring cut from the shell has width [itex]d\theta[/itex], therefore not all the mass is exactly [itex]s[/itex] units from [itex]m[/itex] (though it will approach this scenario as [itex]d\theta \to 0[/itex]). Consider the integration by cylinders method where the width of the cylindrical shell [itex]dx[/itex] plays the role in determining volume). It seems that we would need to involve [itex]d\theta[/itex] here in that assumption. Also, since the sphere is only infinitesimally thin, wouldn't we need to account for whatever thickness it did have in our equation? Sadly, after thorough research, every variation of the proof (there are surprisingly few references) makes similar assumptions, usually in the form of:

    "by symmetry... this is the sum".

    I can understand the cancellation of non-perpendicular vector components, but the suggestion that the sum of all the force can be determined so easily is confusing me. Obviously with that assumption, the rest of the math is straight-forward (though the substitutions were quite clever).

    Can anyone help? I am quite possibly over-thinking this (the conclusion seems reasonable) but I tend to prefer a level of rigor that I haven't found in my searching.

  2. jcsd
  3. Aug 27, 2011 #2


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    The reason the force must be towards the centre of the spherical object is this:
    Imagine the spherical object is centred at the origin, and the z-axis points towards the small test mass.
    To get the total force on the small test mass, integrate:
    [tex] \int - \frac{mG \rho}{s^2} \hat{s} dV [/tex]
    Over the volume of the sphere of the object. (Where [itex]\vec{s} \ [/itex] is the vector from the infinitesimal mass element you are integrating to the test mass). (And [itex]\rho[/itex] is the density which I'm taking to be constant).

    If we say the position of the infinitesimal mass element is [itex]x \hat{i} + y \hat{j} + z \hat{k} [/itex] and that the position of the test mass is [itex]z_t \hat{k} [/itex] then [itex] \vec{s} = x \hat{i} + y \hat{j} + (z_t-z) \hat{k}[/itex].

    Therefore, the x-component of the force is given by:
    [tex] \int \ - \frac{mG \rho}{s^3} x dV [/tex]
    You can then break this into two hemispheres (one in the positive x region and one in negative x region):
    [tex] {\int_{-x}}^0 \ - \frac{mG \rho}{s^3} x dV + {\int_0}^x \ - \frac{mG \rho}{s^3} x dV [/tex]
    Clearly, the two integrals cancel, so the total force in the x direction is zero (and similar can be done for y), so the only force is towards the centre of the spherical object.
  4. Aug 27, 2011 #3
    Wow, it really has been a long time since I've done this kind of work. First of all, this does seem like a good method, though it doesn't really solve my concerns with Newton's method (which should be correct)... perhaps Wikipedia overlooked something? I really didn't feel like trying to get an English translation of the Principia to see how he did it himself :-).

    Next: Would not [itex]\vec{s} = -x\hat{i} -y\hat{j} + (z_t - z)\hat{k}[/itex] since [itex]\vec{z_t} = <0,0,z_t>[/itex]? Ultimately this wouldn't hurt your conclusion about the [itex]x, y[/itex] components. However, this was the trivial part of the proof by symmetry...
    I would say the key here is whether or not the integral with respect to the [itex]z[/itex] component is [itex]<0,0,-1>\frac{\rho \cdot V}{z_t^2}[/itex]. Not quite sure how I would setup that integral though... unless I've forgotten something (I can't remember the details on vector integration), won't this be rather complicated since it is not volume which ranges from [itex]-x \to x[/itex]. We would need to either have to get it in terms of [itex]dx[/itex] or relate the whole equation to volume to actually solve it...
    Last edited: Aug 27, 2011
  5. Aug 28, 2011 #4


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    Well, wikipedia soon make a change to their equation to give:
    [tex]dF_r = \frac{GmdM}{s^2} cos(\phi) [/tex]
    Which, believe it or not, is the same as the z-component of my equation. They say in passing that the x and y components of the total force vanish due to symmetry, but I have explicitly shown you why they vanish.

    So Wikipedia's method is correct, but they don't go into detail on why the force is only towards the centre of the spherically symmetric mass.

    Yes, you're right, I should have written [itex] \vec{s} = -x \hat{i} -y \hat{j} + (z_t - z) \hat{k} [/itex] and you're also right that this error doesn't hurt my conclusion that the x and y components are zero.

    About the integral - think of it like this: for the upper hemisphere, we can write x = f(y,z) and for the lower hemisphere, x = -f(y,z). So now it is a volume which ranges from -x to x, its just that the x is not a constant.
  6. Aug 28, 2011 #5
    I appreciate your being thorough, but again what you previously showed was indeed the trivial part of my question - from the picture alone and basic understanding of physics/vectors it is clear that the non-axial components cancel. My concern is the jump to the conclusion that the sum of all gravitation forces in the ring shaped slice is identical to the gravitation of a point particle with equivalent mass. If this isn't what's happening, then perhaps their terminology has me confused - indeed, I don't know that I care for their use of the 'd' when indicating small quantities when it should be more accurately representing an increment (perhaps this is standard for physics applications though).

    To rephrase my concern:
    They say divide the shell into rings... but the shell does indeed have thickness (apparently ignored in their example) and the ring too will have depth corresponding to the horizontal component of Rdθ. With this in mind, the gravitational force across our infinitesimal volume is not equivalent... if volume were truly zero, then there could be no mass, and if there is volume, then across the the volume will vary. Intuitively, it would seem more accurate to represent this problem as a multiple integral across the entire volume with the force at each point.

    Is there something I've forgotten or missed that will put my mind at ease knowing that Newton was right in his assumption? (I should hope so since we have relied on the calculus he used for a long time) :-) Unfortunately, thanks to scheduling and class availability my Math studies have been out of order and my memory of these core principles are fading (Calculus -> Linear Algebra -> Proofs -> Diff. Eq -> Advanced Calculus (proofs) -> Numerical Analysis ... ugh...).

    As such, your explanation of how the integral can be evaluated is lost on me... it seems that the variable of integration must be with respect to the parameters of integration... Volume itself doesn't range from -x to x... I can see if we perhaps convert the bounds like we do with the cosine it might work, but other than that I'm lost :-)
  7. Aug 28, 2011 #6


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    Isn't there a simpler proof based in the superposition principle? You assume two concentric uniform spheres: one with positive gravitation and one with negative gravitation (the cavity). Then you just add the g-vector from both.

    Using the above method you can easily answer the flowing question:

    What is the g-field in a spherical cavity within a uniform spherical mass, if the cavity and mass are NOT concentric?
  8. Aug 28, 2011 #7


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    Jacob - Yeah, I was just trying to explain why the non-axial components cancel because I thought that was where you had a problem.

    About wikipedia's derivation - they do get the right result, but I don't really like the way they did it either. If I was doing the problem, I would have done it as a multiple integral across the entire volume with the force at each point (as you suggested).

    Doing it this way would get:
    [tex] F_z = {\int_0}^r \ {\int_0}^\pi \ {\int_0}^{2 \pi} \ - \frac{mG \rho}{s^2} (z_t - z) r^2 sin(\theta) d\phi \ d\theta \ dr [/tex]
    and we have [itex]s^2 = r^2 + {z_t}^2 - 2z_tz[/itex] and [itex]z=rcos(\theta)[/itex] therefore:
    [tex] F_z = -2 \pi mG \rho \ {\int_0}^r \ {\int_0}^\pi \frac{(z_t - rcos(\theta))r^2 sin(\theta)}{r^2 +{z_t}^2 - 2z_trcos(\theta)} \ d\theta \ dr [/tex]
    Note: zt is simply a constant equal to the distance from the centre of the spherical object to the test mass. This integral is a bit tricky, but I think it could be done with some patience.
    Last edited: Aug 28, 2011
  9. Aug 28, 2011 #8


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    A much easier way is to use the equation: [itex] \nabla \cdot \vec{g} = -4 \pi G \rho [/itex] and then integrate both sides over a spherical volume of some radius greater than the radius of the spherical mass, to give:
    [tex] \oint \vec{g} \cdot d\vec{A} = -4 \pi G M [/tex]
    The reason the right-hand side integrates to get M is because [itex]\rho[/itex] is only non-zero inside the volume of the actual spherical object, and outside [itex]\rho[/itex] is zero, so no matter how big our volume of integration is, the right-hand side is always the same.
    The reason for the left-hand side is due to the divergence theorem, so the area integral represents the area bounding the volume we have chosen to integrate. So it will be:
    [tex] {\int_0}^\pi \ {\int_0}^{2 \pi} \ \vec{g} \cdot \hat{r} \ r^2 sin(\theta) \ d\phi \ d\theta = -4 \pi G M [/tex]
    Now the trick is to assume that [itex]\vec{g} \ [/itex] depends only on the radius, and that it points opposite to the radial unit vector, so it goes outside the integral:
    [tex]-gr^2 \ {\int_0}^\pi \ {\int_0}^{2 \pi} \ sin(\theta) \ d\phi \ d\theta = -4 \pi G M [/tex]
    And this integral is easy, giving us: [itex] gr^2 = GM [/itex] and we know the direction of g is opposite to the radial unit vector, so we get:
    [tex] \vec{g} = - \frac{GM}{r^2} \hat{r} [/tex]
    And remember, r is any arbitrary distance from the centre of the spherical mass to some point outside the spherical mass. So this explains why the force is equivalent to when all the mass is concentrated at the centre.
  10. Aug 28, 2011 #9
    Thanks, those are indeed what I would have thought necessary to prove his results most thoroughly. The Wikipedia page *should* be just modelling Newton's own proof, though from a little looking on the university's library database, it appears that Newton's proof was likely significantly lengthier and more rigorous ... their conclusions are the same, but Wikipedia's version of the proof does seem to leave some doubt as to whether or not the integral itself accurately describes the situation.

    I am a touch frightened that I don't remember very much of this stuff even though it is my major, but it's coming back - this has been a much appreciated review. While I know *how* to integrate that, you're right that it looks rather tedious and I'm sure that I would have forgotten any "tricks" that may prove necessary. Interestingly, MATLAB didn't appreciate it either (it took like a minute for it to even compute this far lol):

    (angle represented theta since it's actually a reserved variable).

    So, I'm glad I'm not the only one who didn't care much for the proofs that be :-)

    As far as your second proof, this seems to be somewhat similar to what Gauss came up with correct? Certainly more elegant, though gradients and line integrals... hmmmm... I'm feeling rather dumb now :-)
  11. Aug 29, 2011 #10


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    haha, yeah the gauss version is a lot easier. Its surpising how quickly people forget maths if they don't keep using it. I had to learn a lot about integration of complex functions for my final year exams, but I can't remember the equations now.
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