The Pioneer Anomaly |
The Pioneer Anomaly |
Aug 16 2005, 04:27 PM
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Rover Driver Group: Members Posts: 1015 Joined: 4-March 04 Member No.: 47 |
http://www.planetary.org/news/2005/pioneer_anomaly_faq.html
The planetary society may be checking it out... QUOTE The Planetary Society has committed to raise the funds to preserve the priceless Pioneer data from destruction.
After years of analysis, but without a final conclusion, NASA, astonishingly, gave up trying to solve the "Pioneer Anomaly" and provided no funds to analyze the data. The Pioneer data exists on a few hundred ancient 7- and 9-track magnetic tapes, which can only be read on "antique" outdated computers. The agency is going to scrap, literally demolish, the only computers able to access and process that data in the next few months! |
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Jun 27 2007, 06:49 PM
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Member Group: Members Posts: 723 Joined: 13-June 04 Member No.: 82 |
Well... (thinking about it) ...I would assume that the gravitational force exerted on the orbiting body always remains the same as it would be under 'standard' physics, for the reason that you had mentioned. But in addition to the the 'standard' effect that varies with radial distance, approaching zero as radial distance R approaches infinity, and inversely proportional to the square of the radial distance:
1) f1 = G1*m1*m2*R^-2 Where f1 is the 'actual' gravitational force exerted on the body, and G1 is Newton's gravitational constant. A second-order effect of that force on the object's trajectory would vary with radial velocity (distance over time), reaching zero as radial velocity reaches zero, and linearly proportional to (the negative of) the radial velocity: 2) f2 = -Ga*m1*m2*v Where f2 is the pseudo-force exerted on the body, v is the radial velocity and Ga is a second gravitational constant, possibly equal to G1. f2 would be much smaller than f1 at non-relativistic velocities, reaching zero as v reaches zero, but I would not be surprised if it approached f1 as v approaches c, so that Ga equals G1. Combining the two: 3) f1+f2 = (G1*m1*m2*R^-2) - (Ga*m1*m2*v) The 'Pioneer effect' can be viewed as a modification of the well-known Newtonian force f1: 4) f2 = f1*R^2*v*G2 where G2 = Ga/G1 Resulting in: 5) f1+f2 = (G1*m1*m2*R^-2)*(1 - R^2*v) if G1 = Ga, as I think it could be. So, again assuming that Ga = G1, the change in the second-order 'force' must be proportional to v, with the inertial mass equal to the first object's rest mass times (1-v) in natural units (c = 1). This is of course the opposite effect as that described in special relativity, in that with special relativity, the inertial mass equals the rest mass times the square root of (one over (1-v)) Bill |
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Jun 27 2007, 09:56 PM
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#3
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Senior Member Group: Members Posts: 3516 Joined: 4-November 05 From: North Wales Member No.: 542 |
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Lo-Fi Version | Time is now: 5th June 2024 - 03:47 PM |
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