- #36
Stingray
Science Advisor
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Rogue, please look up what a multipole expansion is. I agree that the force between two objects in Newtonian gravity (or electrostatics) is not given by the inverse square law between their center's of mass. This is hardly a great conspiracy, though. It is just the monopole approximation to the infinite series that actually describes the forces between two extended bodies. It is an approximation valid only when the distance between the center's of mass is much greater than the radii of either object. Here's a quick lesson:
Hurkyl gave the exact expression for the force, but I'm going to put it into a more convenient form. The potential due to body A is equal to (using slightly different notation)
[tex]
V_{A}(x) = G \int \frac{\rho_{A}(x')}{|x - x'|} d^{3} x'
[/tex]
Then the force on body B is
[tex]
F_{AB} = - \int d^{3} x \rho_{B}(x) \nabla V_{A}(x)
[/tex]
Looking at the potential in some more detail, we can use the identity
[tex]
\frac{1}{|x-x'|} = 4 \pi \sum_{l,m} \frac{1}{2 l +1} \left( \frac{ |x|_{<}^{l} }{ |x|_{>}^{l+1} } \right) Y_{lm}^{*} ( \theta, \phi ) Y_{lm} (\theta',\phi'),
[/tex]
where the [tex]Y_{lm}[/tex]'s are spherical harmonics, and [tex]|x|_{<}[/tex] is the smaller of [tex]|x|[/tex] and [tex]|x'|[/tex].
For simplicity, center the coordinate system on body A's center of mass. Also assume that every point in body B is at a greater radius than any point in body A. Then
[tex]
V_{A}(x) = 4\pi G \sum_{l,m} \frac{ M_{lm} }{2 l +1} \frac{ Y^{*}_{lm} (\theta, \phi) }{ |x|^{l+1} }
[/tex]
where [tex]M_{lm}[/tex] are the multipole moments of body A:
[tex]
M_{lm} := \int d^{3} x' \rho_{A}(x') |x'|^{l} Y_{lm} (\theta', \phi')
[/tex]
In spherical symmetry, the only moment that survives is the monopole, and we find that the potential is given by the standard point mass formula. In general, though, this does not occur. I'll leave it as an exercise to figure out what the full force equation is, why the point mass (monopole) approximation is usually very good.
By the way, a theorem is not something of "intermediate" certainty. It is something which is proven with complete rigor from a set of axioms (whether the axioms are useful ones or not is irrelevant). If something is a theorem, it has a precise proof.
Hurkyl gave the exact expression for the force, but I'm going to put it into a more convenient form. The potential due to body A is equal to (using slightly different notation)
[tex]
V_{A}(x) = G \int \frac{\rho_{A}(x')}{|x - x'|} d^{3} x'
[/tex]
Then the force on body B is
[tex]
F_{AB} = - \int d^{3} x \rho_{B}(x) \nabla V_{A}(x)
[/tex]
Looking at the potential in some more detail, we can use the identity
[tex]
\frac{1}{|x-x'|} = 4 \pi \sum_{l,m} \frac{1}{2 l +1} \left( \frac{ |x|_{<}^{l} }{ |x|_{>}^{l+1} } \right) Y_{lm}^{*} ( \theta, \phi ) Y_{lm} (\theta',\phi'),
[/tex]
where the [tex]Y_{lm}[/tex]'s are spherical harmonics, and [tex]|x|_{<}[/tex] is the smaller of [tex]|x|[/tex] and [tex]|x'|[/tex].
For simplicity, center the coordinate system on body A's center of mass. Also assume that every point in body B is at a greater radius than any point in body A. Then
[tex]
V_{A}(x) = 4\pi G \sum_{l,m} \frac{ M_{lm} }{2 l +1} \frac{ Y^{*}_{lm} (\theta, \phi) }{ |x|^{l+1} }
[/tex]
where [tex]M_{lm}[/tex] are the multipole moments of body A:
[tex]
M_{lm} := \int d^{3} x' \rho_{A}(x') |x'|^{l} Y_{lm} (\theta', \phi')
[/tex]
In spherical symmetry, the only moment that survives is the monopole, and we find that the potential is given by the standard point mass formula. In general, though, this does not occur. I'll leave it as an exercise to figure out what the full force equation is, why the point mass (monopole) approximation is usually very good.
By the way, a theorem is not something of "intermediate" certainty. It is something which is proven with complete rigor from a set of axioms (whether the axioms are useful ones or not is irrelevant). If something is a theorem, it has a precise proof.
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