mooeypoo Posted May 17, 2009 Posted May 17, 2009 Geistke, now you're being utterly rude. You're quoting people and putting the quote under different member names. Half of the quotes you attribute to me I did not say. This: Nobody says this is realized in nature. It is a mathematical realization that you get the same answer when you integrate over the volume, and thus is an allowable shortcut. A spherically symmetric mass behaves as if all the mass were concentrated at the center. Is swansont from post #49, which is later copy-pasted in its entirety again in a jumbled quote that seems to come from you (?). Sort out your quotes! Do not put words in other people's mouths and make sure you quote others' words as theirs and not as yours. You should also go over the rules of the forum, as well as the Speculation Policy. And for the last time: Stop using indent. This isn't a friendly request, it's a moderator's demand.
D H Posted May 17, 2009 Posted May 17, 2009 (edited) Yet another way to look at this problem is from the perspective of potential. Since gravitation is a conservative force, for any given mass distribution exists a scalar potential function [math]\phi(\boldsymbol x)[/math] such that the gravitational force one an object of mass M located at a point x is the additive inverse of the produce of the mass M and the gradient of the gravitational potential at that point: [math]\boldsymbol F = -M\,\nabla \phi(\boldsymbol x)[/math] What is this potential function? The gravitational potential at a point x due to a point mass of mass m located at a point ξ is [math]\phi_{\text{pt. mass}}(\boldsymbol x) = -\,\frac{Gm}{||\boldsymbol x - \boldsymbol \xi||} = -\,\frac{Gm}{l(\boldsymbol x,\boldsymbol \xi)} [/math] where [math]l(\boldsymbol x,\boldsymbol \xi)\equiv ||\boldsymbol x - \boldsymbol \xi||[/math] is the distance between the points x and ξ. The superposition principle applied to potential functions. Therefore, the gravitational potential for a mass distribution characterized by a density function [math]\rho(\boldsymbol x)[/math] over some volume V is [math]\phi(\boldsymbol x) = -\,\int_V \frac {G \rho(\boldsymbol \xi)}{||\boldsymbol x - \boldsymbol \xi||}\,d \boldsymbol \xi = -\,\int_V \frac {G \rho}{l}\,d \boldsymbol \xi = -\,\int_V \frac {G dm}{l} [/math] The problem at hand is to develop the gravitational potential for an infinitesimally thin spherical shell of mass, with the mass uniformly distributed over the shell. Denote the density (units = mass/length2) of the shell as ρ. Denote the radius of the shell as r and the distance between the shell and the point in question as R. We're working with a sphere, so using describing the points on the sphere in terms of spherical coordinates will take advantage of the spherical symmetry of the problem. Define [math]\hat{\boldsymbol z}[/math] as the unit vector from the center of the sphere toward the point at which the potential is to be determined, [math]\hat{\boldsymbol x}[/math] as a unit vector normal to [math]\hat{\boldsymbol z}[/math], [math]\hat{\boldsymbol y}[/math] = [math]\hat{\boldsymbol z}\times \hat{\boldsymbol x}[/math], [math]\theta[/math] as the angle between the [math]\hat{\boldsymbol z}[/math] axis and a point on the sphere, and [math]\phi[/math] as the angle between the [math]\hat{\boldsymbol x}[/math] axis and the project of a point on the sphere onto the x-y plane. With this notation, the Cartesian coordinates (x,y,z) of the point on the sphere with angular coordinates (θ,φ) are [math]\aligned x &= r\cos\theta\cos\phi \\ y &= r\cos\theta\sin\phi \\ z &= r\sin\theta \endaligned[/math] The distance between this point on the sphere and the point in question is, by the law of cosines, [math]l = \sqrt{R^2+r^2-2Rr\cos\theta}[/math] Note that the angle φ is not involved in this distance function. The gravitational potential at the point in question is [math]\aligned \phi &= -\,\int_S \frac {G dm}{l} \\ &= - \int_0^{\pi} \int_0^{2\pi} \frac{G \rho}{\sqrt{R^2+r^2-2Rr\cos\theta}} r^2\sin\theta\,d\phi d\theta \\ & = -G 2\pi \rho r \int_0^{\pi} \frac {r\sin\theta d\theta}{\sqrt{R^2+r^2-2Rr\cos\theta}} \endaligned[/math] To evaluate the above, make the substitution [math]l^2 = R^2+r^2-2Rr\cos\theta[/math]. With this substitution, [math]Rr\sin\theta\,d\theta = l\, dl[/math]. The numerator in the [math]\phi = -G 2\pi \rho r \int_{l(\theta=0)}^{l(\theta=\pi)} \frac {l\,dl}{lR} = -\,\frac{G 2\pi \rho r}{R}(l|_{\theta=\pi}-l|_{\theta=0})[/math] The integration limits, [math]l|_{\theta=0}[/math] and [math]l_{\theta=\pi}[/math], are simply [math]|R-r|[/math] and [math]|R+r|=R+r[/math], respectively. For points outside the sphere, R>r and thus [math]|R-r|=R-r[/math]. For points inside the sphere, R<r, and thus [math]|R-r|=r-R[/math]. Thus for points outside the sphere, [math]\phi = -\,\frac{G 2\pi \rho r}{R}((R+r)-(R-r)) = -\,\frac{G 4\pi \rho r^2}{R} = -\frac {G M_s}{R}[/math] where [math]M_s=4\pi \rho r^2[/math] is the mass of the sphere. This is exactly the same as the potential due to a point mass, and thus a spherical shell looks exactly like a point mass for points outside the shell. For points inside the shell, [math]\phi = -\,\frac{G 2\pi \rho r}{R}((R+r)-(r-R)) = -\,\frac{G 4\pi \rho r}{r} = -\frac {G M_s}{r}[/math] This is a constant. As the gradient of a constant function is zero, a spherical shell results in zero gravitational acceleration for all points inside the shell. This proves the shell theorem. I could end with that. However, geistkie (geistkiesel) has asked about hemispheres as well. Using the same coordinate system as before, let S be the half of the spherical shell with positive z coordinates. Without going through the math, the gravitational potential for this hemisphere for points located on the z axis is [math]\phi(0,0,z) = -\,\frac{G 2\pi \rho r}{|z|}(\sqrt{z^2+r^2}-|z-r|))[/math] Note that this expression applies only to points on the z axis. I'll leave it up to you, geistkie, to take the gradient of this expression (Hint: compute [math]\partial \phi/\partial z[/math]). I made a claim in an earlier post that for any point mass located inside the spherical shell, the gravitational forces resulting from any partitioning of the shell into two measurable parts will be equal and opposite. Suppose you partition the shell into two complementary sets, [math]S[/math] and [math]S_C[/math]. After Herculean series of calculations, you manage to find [math]\phi_S(x,y,z)[/math]. Since the sets are complementary, the potential function for the complement of [math]S[/math] must be, by virtue of the definition of the Lebesgue integral, [math]\phi_{S_C}(x,y,z) = \phi_{\text{shell}} - \phi_S(x,y,z)[/math] For points inside the sphere, the potential for the entire sphere is just a constant. Thus for points inside the sphere, [math]\phi_{S_C}(x,y,z) = -\frac {G M_s}{r} - \phi_S(x,y,z)[/math] The gravitational forces on a test point mass of mass m located at (x,y,z), with [math]x^2+y^2+z^2<r^2[/math] are thus [math]\aligned \boldsymbol a_S &= -m\,\nabla \phi_{S}(x,y,z) \\ \boldsymbol a_{S_C} &= -m\,\nabla \phi_{S_C}(x,y,z) \\ &= -m \nabla\left(-\frac {G M_s}{r} - \phi_S(x,y,z)\right) \\ &= m \nabla \phi_{S}(x,y,z) \\ &= - \boldsymbol a_S \endaligned[/math] Edited May 17, 2009 by D H 1
swansont Posted May 17, 2009 Posted May 17, 2009 As geistkie has been banned, I am closing this thread. Anyone who has legitimate questions about the shell theorem/Gauss's law should start a new thread on it, so that the odor of these leavings does not pollute the discussion.
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