Talk:Pioneer anomaly/Archive for 2006

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Pioneers Anomaly 2 - another alternative idea

Latest comment: 18 years ago7 comments2 people in discussion

FYI, for illustration only, since new article on "Pioneers Anomaly" is scheduled for delete, as of December 2005, as such a page already exists:

ANOMALOUS ACCELERATION OF PIONEERS 10 AND 11 may be modeled on a gravitational delta effect, per the Equivalence Principle

By I. D. Alexander December 16, 2005

The Pioneers Anomaly (1), as measured by John D. Anderson, M.M. Nieto, S.G. Turyshev, P.A. Laing, et al, shows a constant rate of acceleration towards the Sun at $-a=~8E-8cm/s^{2}$ . This can be interpreted, conversely, as a gravitational delta effect, where Newton's G constant is not so, but grows at the same steady rate as the anomalous acceleration towards the Sun. The rate of growth of delta G of ~1G per 1AU (here shown 'hypothetically as if') translates into the anomalous rate of acceleration of a (hypothetical) delta $-(G/AU)=6.4E-11m^{2}/kg.s^{2}$ , as a function of a variable 'phantom' Newton's G for our solar system.

The falsifiable test of Newton's G in the outer solar system has been proposed by the group working with the European Space Agency (3). At present, assuming a universal constant Newton's G, all calculations for distant mass and density of astronomical bodies had been determined using this constant. If there should be a 'phantom' G phenomenon at work in space, away from Earth's known 1G, it may not have been detectable until the Pioneers revealed their anomalous acceleration. This anomalous acceleration is made more poignant in that both probes, Pioneer 10 and 11, are moving out of the solar system in opposite directions, both showing the same constant rate acceleration. Both probes are being pulled back towards the Sun by the same rate. Though this measured rate at present remains unexplained with know physics, it is compelling enough to be researched further. Another anecdotal evidence (possibly supported by the constant delta G growth rate for our solar system), is the atmosphere of Saturn's largest moon, Titan. Titan's atmosphere, according to Nature (2), is ten times the 'thickness' of Earth's atmosphere. Both are largely nitrogen in composition, though residing on bodies of vastly disproportionate mass and temperature. It may be a function of the gravitational 'constant' G being greater at Saturn's orbital region than on Earth. The fact that the Cassini-Huygens mission to Saturn, and successful landing the Huygens probe on Titan, shows that whether G is a constant, as now believed, or a variable, as may be evidenced by Pioneers, is not relevant to our space missions. However, it may prove relevant to understanding planetary atmospheric density, as evidenced by Titan's atmosphere, for example. All this, however, is a subject for future study.

Translating Pioneers's Anomalous acceleration into a constantly growing G with distance from the Sun, it could be illustrated (hypothetically) as follows:

delta G'/G = 1, where G' represents 1G per 1AU, at a constant rate.

delta G'/G = 1 divided by 1AU in meters (1.5E+11 m) = delta 0.6667E-11 (per $m/s^{2}$ for acceleration)

which translate in centimeters to: delta $-a=~66.67E-15cm/s^{2}$

Where did the $cm/s^{2}$ come from? It's not at all clear how you are implementing your variable G idea, but regardless the number you arrived at is certainly not an acceleration. You can prove this to yourself by forgetting the actual numbers and simply working through your equation with just the units. You will find that the units you end up with are not $m/s^{2}$ . --Fringec 00:22, 24 January 2006 (UTC)

9 February, 2006:

Thanks Fringec, for pointing out the inconsistency in the $cm/s^{2}$ , in my above. In taking (delta)G'/G, which per reason G' increases 1G per 1AU, I used (delta)G'/G = 1, which still retains SI units $m^{3}kg^{-}1s^{-}2$ for Newton's G. To make the unit analysis work, it would have to imply that the (delta)G'/G relationship per AU invokes somehow that $m^{3}kg^{-}1s^{-}2$ be multiplied by $kg.s^{-}2$ (though I have no way to explain this) to make the units work out: $m^{3}kg/kg.s^{-}4$ . When this is divided by AU, which is in "m", then the result is $m^{2}s^{-}4$ , which taking square root is our familiar acceleration unit: $m/s^{2}$ . That said, I have no way to justify why Newton's G (hypothetically) growing at the rate of 1G per 1AU should invoke the needed $kg.s^{-}2$ units. --Ivan Alexander, 9 Feb. 2006.

Of course you can't justify the needed units. The method you are using to determine what you are trying to determine doesn't make any sense! The fact that the units don't balance should be a sanity check that is telling you that you are doing something wrong.

You're trying to compute the difference in acceleration predicted by standard Newtonian physics and some variable G concept. Clearly the way to do this is to first calculate the acceleration predicted by Newtonian physics. Naturally this is done all the time so you can do it too. Next you would perform the same calculation factoring in this variable gravitational constant. This will be more interesting to hash out, but perhaps you could look to existing variable G theories to see how they do it.

The difference between the two results is the value you are looking for.

The point is that there is a mathematically sound way to do what you're trying to do. Granted, it would require a lot of research and work to perform these calculations with any kind of accuracy, but you ought to be able to get in the ballpark just modelling a simple Sun-Pioneer system, with Pioneer travelling straight out of the solar system. In any case, what you've done here is just not correct.--Fringec 22:35, 21 February 2006 (UTC)

Admittedly, Fringec, it does appear peculiar to apply kg/s^2 to Newton's G, as I did here, even to me. But upon further reflection, it may not be that odd. If "per second per second" is rate of change, then m/s^2 means rate of acceleration "in terms of distance"; likewise, kg/s^2 would mean rate of acceleration "in terms of mass". This could be applicable to the Pioneer Anomaly if the two are contingent upon a commensurate increase in the "inertial mass" per the Equivalence Principle, if G is "accelerating" in terms of kilograms with distance from the Sun: the end result is a commensurate "acceleration" in terms of distance (towards the Sun), if this is true. For this reason, ESA's search for gravitational anomalies in the outer solar system, which may reveal a G anomaly, may be more important than we suspect. And if it is found that G is variable in a continuous way, i.e., ~1G per 1AU, then there may be a match between rate of change of inertial mass and rate of change of distance traveled. However, this would then call for a new physics, if so.-- Ivan Alexander, 23 February 2006.

I think we need to back up a little bit. Acceleration is defined as "the rate of change of velocity of an object with respect to time". Velocity is "the rate and direction of motion". It is from these definitions that the units are derived. Units are not just chosen arbitrarily. Nothing you've written above changes the fact that you are ultimately trying to calculate an acceleration which is measured in m/s², but the result of your method is not acceleration. Put another way, you're just performing various mathematical operations and then changing the units to whatever you like when you're done.

There's nothing magical about what you are trying to do. You're just trying to determine the difference between two accelerations. This is very straightforward.

Any model of gravity needs to be able to calculate the rate of acceleration of an object due to the effects of gravity on that object. The trick in this case is figuring out how to calculate the acceleration on Pioneer under a variable G model. If you're determined to model the Pioneer Anomaly as the result of a variable gravitational constant this is where you need to start. Once this is accomplished you need only calculate the acceleration predicted by Newtonian physics and compare it to the acceleration predicted by the variable G model. The difference between these two results will have the correct units, and is the value you are looking for.

If none of this is getting through, consider this: How was the value you are trying to duplicate, (8.74 ± 1.33) × 10⁻¹⁰ m/s², arrived at in the first place? It was calculated by subtracting the observed acceleration of Pioneer from the acceleration predicted by Newtonian physics. If this variable G model is correct, you should arrive at the same number, using the same method, merely by replacing the observed acceleration with the acceleration predicted by the variable G model.

So why not approach the problem correctly and try to calculate the acceleration on Pioneer using the variable G model? --Fringec 19:18, 24 February 2006 (UTC)

Per Equivalence, where change in G is reflected in comparably-equivalent inertial mass, taking the square root of this delta -a, it becomes:

$-a=~8.165E-7.5cm/s^{2}$

This is largely irrelevant because of your first mistake, but you should be aware that when taking the square root you need to consider the units as well. The square root of $cm/s^{2}$ is not $cm/s^{2},$ . Additionally, it's unclear why you decided to take the square root of your previous result in the first place, but that's another story.

Also note that the square root of $6.67E-14$ is $2.58E-7$ , so the magnitude of your result is incorrect as well as the units.--Fringec 00:22, 24 January 2006 (UTC)

My mistake. Looking at it again, $8.165E-7.5$ equals $2.58E-7$ , so the square root is in fact correct, it's just an odd way of expressing it. Of course, the magnitude still doesn't match the anomalous acceleration and, more importantly, the comments about the units being incorrect still stand.--Fringec 19:30, 30 January 2006 (UTC)

This is a result within one half order of magnitude of the measured Pioneers Anomaly, $-a=~8E-8cm/s^{2}$

One should remain very much aware that such a constant rate of change for Newton's G in the outer solar system had never been demonstrated, and other explanations for the Pioneers Anomaly, searching for systemic reasons, have not as yet yielded verifiable results. Therefore, neither should this hypothetical 'phantom G' explanation be taken as a definitive reason for the Anomaly. It is shown here merely as a representation of how this anomalous acceleration may be numerically explained using a constant rate of change for Newton's G, where this 'phantom' G grows at a comparable rate of 1G per 1AU - shown here for illustration purposes only.

As pointed out, the value you have calculated is not an acceleration. Units are critically important. You have a lot more work to do to model the Pioneer Anomaly using a varying gravitational constant..--Fringec 00:22, 24 January 2006 (UTC)

References:

(1)"Study of the anomalous acceleration of Pioneer 10 and 11" by Anderson et al. http://arxiv.org/abs/gr-qc/0104064

(2) Fig. 1: Comparisons between atmospheres of Titan and Earth. http://www.nature.com/nature/journal/v438/n7069/fig_tab/438756a_F1.html

(3) "The Pioneer Anomaly: The Data, its Meaning, and a Future Test", by Nieto, Turyshev, and Anderson http://arxiv.org/pdf/gr-qc/0411077v2

(This alternate idea cannot be included in Wiki, so list it here merely as a curiosity on the Discussion page. Any comments welcome.)

Original research

This WP talk page is not really the right place to have a conversation on the type of original research being conducted above. Please find another forum for having this conversation. linas 17:31, 26 February 2006 (UTC)

ERRATUM?

Latest comment: 18 years ago1 comment1 person in discussion

RE text in the Pioneer Anomaly text, it says: "The effect is seen in radio Doppler and ranging data, yielding information on the velocity and distance of the spacecraft. When all known forces acting on the spacecraft are taken into consideration, a very small but unexplained force remains. It causes a constant sunwards acceleration of The effect is seen in radio Doppler and ranging data, yielding information on the velocity and distance of the spacecraft. When all known forces acting on the spacecraft are taken into consideration, a very small but unexplained force remains. It causes a constant sunwards acceleration of (8.74 ± 1.33) × 10−10 m/s2 for both spacecraftfor both spacecraft."

Just a note, but shouldn't $-a=8.74x10-8cm/s2$ read as $-a=8.74x10-6m/s2$ when converted from centimeters to meters? The text in the article on Pioneer Anomaly reads as "It causes a constant sunwards acceleration of (8.74 ± 1.33) × 10−10 m/s2 for both spacecraft," which looks wrong. After all, if the acceleration is taken over a greater distance, shouldn't it be greater? The way the article now reads, it appears the conversion from centimeters to meters makes it smaller.

Ivan D. Alexander

Hi, -- the short answer is "no", that is not how one converts units. An acceleration of 1 m/sec^2 is equal to an accelearation of 100 cm/sec^2. linas 01:02, 18 January 2006 (UTC)

Hi linas, sorry for my very slow response, as my neurons fire as fast as those of a jellyfish, at times.

I could plainly see how velocity in meters is smaller, per second, than centimeters. The source of my confusion was that acceleration, as a rate of change per second per second, was larger over the distance of a meter than a centimeter, and still larger over the distance of a kilometer, which is why I had it reversed from the obvious answer. This ends my end of this discussion, Thanks. Ivan Alexander, 15 April, 2006

The Planetary Society Project

The Planetary Society is currently (2006) sponsoring a project to analyze thirty years of Pioneer data in an attempt to shed light on the anomaly.

Joe Macke 06:14, 17 Jan 2006 (UTC)