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Mercury's precession 
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#1
Apr2510, 04:50 PM

P: 93

When calculation GR, the component due to other planets is removed. In that calculation outer planets are assumed to be rings of mass and the effect of that mass is seen to have an effect on mercury precession :
http://www.physics.princeton.edu/~mc..._47_531_79.pdf How does this square with the idea of the shell theorem that says you do not feel a gravitational force from the inside of a hollow sphere of mass : http://en.wikipedia.org/wiki/Shell_theorem Thanks 


#2
Apr2510, 06:02 PM

P: 1,568




#3
Apr2510, 08:18 PM

P: 93




#4
Apr2510, 10:25 PM

P: 1,568

Mercury's precession



#5
Apr2610, 07:27 AM

P: 100

http://farside.ph.utexas.edu/teaching/336k/Newton/node127.html Here is an application to the planets. http://farside.ph.utexas.edu/teachin...n/node128.html 


#6
Apr2610, 08:18 AM

Mentor
P: 15,055




#7
Apr2610, 12:28 PM

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P: 15,055

The force is not null except at the center of the ring. Take the gradient of the potential and you will see that this is the case.
Rings are used to prove the shell theorem, but the rings are carefully constructed so that the point in question lies on the axis of each ring. 


#8
Apr2610, 12:34 PM

Mentor
P: 6,219

http://www.physics.princeton.edu/~mc..._47_531_79.pdf the results in this reference and in http://farside.ph.utexas.edu/teachin...n/node127.html are the same. Again, you have jumped too quickly to a conclusion. 


#9
Apr2610, 01:15 PM

P: 1,568




#10
Apr2610, 04:13 PM

P: 3,967

Here is a simple visual demonstration that the gravity inside a ring is not null as inside a hollow sphere (in Newtonian terms):
In the above image the particle at point P inside the hollow sphere shell is at different distances from the shell but the shell mass contained in the circular intersection of the cone on the left is greater than the shell mass intersected on the right and this difference in intersected shell mass exactly compensates for the differences in distance. The regions outside the intersecting cones are identical and trivially cancel each other out, no matter what arbitary angle is chosen for the intersecting cones. See http://galileo.phys.virginia.edu/cla.../GravField.htm for the source of the diagram. If a section of the shell above and below point P (the green sections in the attached diagram) are removed, leaving an equatorial ring, then some of the compensating shell mass is removed (the light blue regions in the attached diagram) and the forces no longer cancel out. Anything in the plane of the ring, but not exactly at the centre, moves outwards as if there is a repulsive force at the centre of the ring. 


#11
Apr2610, 04:40 PM

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P: 15,055

In short, to reecho the claims of MIT students at the 1971 World Science Fiction Convention, "The Ringworld is unstable!"
It is also important to remember that Mercury's perihelion precession was known to be anomalously large well before Einstein. (The anomaly was first reported in 1859.) Those physicists were quite adept with calculation; they had to be because they did not have those electronic marvels that we rely on (possibly overly so) these days. 


#12
Apr2610, 05:24 PM

P: 1,568




#13
Apr2610, 05:41 PM

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P: 15,055

The proof of the shell theorem such as that in wikipedia constructs the rings so that the point in question is along each of the rings axes. That construction ensures radial symmetry.
There is no radial symmetry in this problem to take advantage of. Mercury is not located along the axis of the outer planet's orbits. It is more or less located on the orbital planes of those orbits. The projection of Mercury's location onto each of those orbital planes is not at the center of those orbits. 


#14
Apr2610, 05:46 PM

P: 1,568




#15
Apr2610, 05:52 PM

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P: 15,055

Try it. You're going to get something rather ugly. In fact, the integral that results does not have a simple solution. You will get what are called elliptical integrals.



#16
Apr2610, 06:05 PM

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P: 15,055

This isn't homework, so if you want I can give you a link to a solution. (You probably wouldn't believe me if I the words came from my fingers. Fortunately, the solution is out there on the 'net. Besides, I don't particularly want to typeset the LaTeX with my fingers when the solution is out there on the 'net.)
Or you can try it yourself. When/if you get stuck I'll be glad to help. 


#17
Apr2610, 06:21 PM

P: 1,568




#18
Apr2610, 06:23 PM

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P: 15,055

No, it does not. Show your work.



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