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Precession of Mercury’s Orbit |
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| Mar31-11, 11:11 AM | #18 |
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Precession of Mercury’s Orbit
I read it here > http://milesmathis.com/merc.html
Now, if we subtract Mercury's precession from the Earth's, we achieve the apparent precession of Mercury as seen from the Earth. This is the number we want. This gives us a difference of ΔP = .8 arcsec/yr Or 80 arc secs per century. Therefore we will see Mercury precess about 80 arc seconds per century, due to curvature of the field alone. This is almost double Einstein's 43, which is enough to disprove his math and postulates. It also means that we will have to re-figure the perturbation total. The number 528 from above cannot be correct, as I said, since that is a Newtonian number, not a Relativistic number. |
| Mar31-11, 01:32 PM | #19 |
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1.) Is it because of higher speed by perihelion relative how strong the curvature of space is ? 2.) Is it because of inertia, - increasing resistance against motion the further a body approach c ? - I mean when accelerating 100 km/h + 100km/h is not 200 km/h but 199,999...etc..km/h, 3.) or how can the cause / result of it be described. |
| Mar31-11, 03:06 PM | #20 |
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Recognitions:
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The simplest model derives a general-relativistic 1/r³ correction to the usual 1/r potential which causes the angular deficit.
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| Mar31-11, 04:29 PM | #21 |
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![]() ![]() From: http://www.physics.ucla.edu/demoweb/...spacetime.html However, I was told (and a quick estimation seemed to confirm it) that the purely spatial curvature causes just a small part of the 43 arc second relativistic contribution. While most of it is due to gravitational time dilation (time curvature) and effective potentials: http://www.bun.kyoto-u.ac.jp/~suchii/eff.potent.html |
| Apr2-11, 11:01 PM | #22 |
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157. How to Build a Nucleus without the Strong Force. With simple logic and diagrams. 129. A Recalculation of the Roche Limit. Showing that current math is wrong, and how to find a different kind of limit. 175. The Extinction of π. Here I show that the true value of π, defined as the ratio of circumference to diameter, is 4. ![]() So, I think that site is only good for humor. |
| Apr3-11, 02:01 AM | #23 |
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![]() I understand that space is curving (Fig.1+2+3+4+5+6+7). But the confusion is with references to Fig 8 + 9. Fig 8 - “Similar Mercury executes its Newtonian ellipse in local space” Fig 9. ---but the global curvature causes the ellipse to precess. I understand; that: Fig 8.; “ Newtonian space is without any curvature” and -- Fig.9 is with curvature. So what the illustration only say is; That the curvature causes the ellipse to precess, - not WHY. ![]() The essence from this link is so far I understand Fig 2 > “precession because of extra dwell time at inner part. According to classic understanding I don’t think there should be any extra “dwell time” – WHY should that happen due to general relativity? I still do not understand the cause of that anomaly, (and do not have a mathematical background to do so). Is there someone what can explain the cause in simple words? I simply cannot find any description in this thread, that explain the cause of that precision anomaly. * * * I have considered another option; - we know it requires more and more energy to get a diminishing increase in speed. So each time when Mercury accelerate towards perihelion, the planet need more potential energy to be able to reach that speed it should according to classis orbit mechanics, to overcome the increasing inertia against acceleration (towards perihelion). But where must that extra energy come from? As I see it this must mean that Mercury will lose speed, and hence approach the Sun. (Circling inwards to the Sun, in the same way as Illustration 3 above; “Captured and then plunged”. - But this does not fit with reality, - why not ? . What prevent that from happen? Maybe this question do not have anything with the precession anomaly to do, But I would anyway appreciate to get some suggestion why Mercury due to Inertia (by acceleration) not is losing speed / energy. - It must be a fact that it will not reach that speed it "should" according to classis orbit mechanics. |
| Apr3-11, 07:26 AM | #24 |
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Mentor
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From the perspective of the Earth, Mercury's observed apsidal precession is 5600 arcseconds/century if the observations are expressed in mean of date coordinates or 574 arcseconds/century if the observations are expressed in J2000 or ICRF coordinates. The difference between these two values results from the 5026 arcseconds/century general precession of the equinox. In either case, the observed precession is not 43 arcseconds/century. Let's call the observed precession 574 arcseconds/century. Newtonian mechanics explains most but not all of this observed precession: The outer planets cause Mercury's orbit to precess by 531 arcseconds/century -- which is off by 43 arcseconds/century. That 531 arcseconds/century is a calculated result; all that can be observed is the 574 arcseconds/century value. |
| Apr3-11, 08:29 AM | #25 |
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I tried this. The ellipse on my piece of paper didn't precess. What's the trick? |
| Apr3-11, 09:05 AM | #26 |
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| Apr3-11, 11:37 AM | #27 |
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I’m not stating this as an insult. I know no one in this forum invented that exercise. But it tells me nothing about relativity. I am a little better at Paper Mache though. |
| Apr3-11, 11:56 AM | #28 |
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| Apr3-11, 12:41 PM | #29 |
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| Apr3-11, 01:05 PM | #30 |
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Note that in reality the trajectory caused by the time curvature is different but they want to show only the effect of purely spatial curvature on the precession. Note that in reality the spatial geometry is not cone, and the curvature is everywhere, but the cone approximation is good to illustrate the principle. |
| Apr3-11, 02:44 PM | #31 |
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If I draw a vertical ellipse on the upper part of a disc symetric to the y axis and make a cutout in the lower part of the disc symetric to the y axis and close the disc into a cone symetric to the y axis my ellipse is still vertical and symetric to the y axis. I don't get any precession. |
| Apr3-11, 02:55 PM | #32 |
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| Apr3-11, 04:22 PM | #33 |
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But it's easy to show mathematically. Just consider the angles between your original ellipse and your cuts. On the tip side they total less than 180°, so when you weld the cut lines, you don't get a smooth path. |
| Apr3-11, 05:23 PM | #34 |
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Here's a very exaggerated version (not to scale) with 120° precession to get the point across.
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