Proof like Shell theorem for non-spherical objects?

In summary, the article discusses the failure of two lunar orbiting satellites, which were released during the Apollo era.
  • #1
Zula110100100
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Is there one out there? Do we have any reason to believe we can treat other objects like point masses as well?

I ask because if you consider line-world, and there was a 4m segment with uniform density 3kg/m located with it's left end at (3), the center of mass would be at (5), and I am located at (0), the acceleration from gravity is:

[tex]A_{g} = \frac{Gm}{r^2{}}[/tex]
[tex]A_{g} = \frac{12G}{25} = .48G[/tex]

Now I cut the object in half, so I have two 2m segments, each still a uniform 3kg/m, and it hasn't moved, but the center of mass for the nearer segment is now at (4), and the other is at (6):

[tex]A_{g} = \frac{Gm_{1}}{r_{1}^2{}} + \frac{Gm_{2}}{r_{2}^2{}}[/tex]
[tex]A_{g} = \frac{6G}{16} + \frac{6G}{36} = .375G + .167G = .542G[/tex]

So then I tried to do it with an integral(not 100% sure I am doing it right) and this is what I got

[tex]\Delta A_{g} = \frac{G\rho \Delta x}{x^{2}}[/tex]
[tex]A_{g} = G\rho \int \frac{1}{x^{2}}dx[/tex]
[tex]A_{g} = G\rho \int^{7}_{3} \frac{1}{x^{2}}dx[/tex]
[tex]A_{g} = 3G(-\frac{1}{7}+\frac{1}{3}) = 3G(-\frac{3}{21}+\frac{7}{21}) = 3G(\frac{4}{21}) = \frac{12G}{21}\equiv .571G[/tex]

That's almost a fifth more than the original estimation, so one could assume if point mass fails in 1d, it could fail in some higher dimensional configuration. Unless there is an error above, or some reason why it works in 3d and not 1d...Anyone know this?
 
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  • #2
In reading that now I realize that the inverse-square law is for 3D so I imagine that makes the difference here. But the question on proofs for non-spherical objects remains
 
  • #3
Proof of what? I assume you want a proof that gravitational acceleration is still given by GM/r2 for non-spherical objects.

No such proof exists, because as you have found, it is not true. A similar issue arises in static electricity. Strictly speaking, Coulomb's law is only true for point charges. Newton's shell theorem also applies to this problem, so Coulomb's law is also valid for objects with a spherical charge distribution.

Just as electricity and magnetism are better described using spherical harmonics, so is gravitation. The non-spherical nature of an object leads to 1/r4 and higher-order terms. Note: electricity has 1/r3 terms that result from imbalances of negative and positive charge. There is no such thing as negative mass, so the first non-spherical terms in the case of gravitation are 1/r4. Another thing to note: These higher-order terms tend to zero much more quickly than does the 1/r2 term. As distances grow large, objects tend to look more and more like a point mass. At very large distances, the simple form of Newton's law is approximately correct.

The various space faring nations take advantage of the non-spherical nature of Earth's gravitational field. It is very handy for an Earth-observing satellite if the subsatellite spot on the sunlit side of the satellite's orbit is always near local noon to minimize shadows (or always near twilight to maximize shadows). If the Earth was spherical, this would require constant orbit adjustments to compensate for the Earth's orbit about the Sun. This constant compensation is not needed thanks to the non-spherical nature of the Earth. The Earth's equatorial bulge causes the line of nodes of a satellite's orbit to precess. Place a satellite at the right altitude and inclination and this nodal precession will make the satellite be in sync with the Sun. This is called a sun-synchronous orbit.

The non-spherical nature of the Earth also causes satellites in geosynchronous orbits to migrate toward either 75.3°E or 104.7°W longitude. Except for geostationary satellites placed at these longitudes, a geostationary satellite needs to periodically perform stationkeeping maneuvers to keep the satellite at the desired location.

One final example: Read this article, http://science.nasa.gov/science-news/science-at-nasa/2006/06nov_loworbit/, on the fate of a couple of lunar satellites released in the Apollo days.
 

1. What is the Shell theorem for non-spherical objects?

The Shell theorem states that for a spherically symmetric mass distribution, the gravitational force experienced by a point mass outside the distribution is the same as if all of the distribution's mass were concentrated at its center. This theorem can also be extended to non-spherical objects by using an equivalent spherical mass distribution.

2. How is the Shell theorem used in astronomy?

The Shell theorem is used in astronomy to approximate the gravitational force of non-spherical objects, such as planets or galaxies. By treating these objects as if their mass were concentrated at their center, calculations for orbits and trajectories can be simplified.

3. What are the limitations of the Shell theorem for non-spherical objects?

The Shell theorem is only an approximation and does not account for the actual distribution of mass within a non-spherical object. It also assumes that the object is spherically symmetric, which may not always be the case.

4. Can the Shell theorem be applied to all non-spherical objects?

No, the Shell theorem can only be applied to objects with a spherically symmetric mass distribution. Objects with irregular shapes and mass distributions, such as asteroids or comets, cannot be accurately modeled using this theorem.

5. How does the Shell theorem relate to Newton's law of gravitation?

The Shell theorem is derived from Newton's law of gravitation, which states that the gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The Shell theorem is a useful application of this law for non-spherical objects.

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