Debate:What is the exponent of r in Newtonian gravity?

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Do you think the exponent of r in Newtonian gravity can be slightly different than 2, such as 2.000001?


Actually, yes, I do think this is impossible. Very simply, because the power of 2 wasn't arrived at by experimentation or observation, but by pure hard logic. Additionally, in Newtonian gravity, it is 2 - that is part of what defines Newtonian gravity. If it wasn't 2, then we wouldn't be talking about Newtonian gravity anymore.

I just did this for fun, don't go jumping down my throat for anything! JacobB 19:40, 1 October 2009 (EDT)

You're open-minded. Congratulations! (I altered two of the questions since you responded.)
I do think it is quite possible that the power of 2 in Newtonian gravity is not precisely exact, and that a slight deviation from 2 (such as 2.0000001) is to be expected. The physical world is not a perfect in a mathematical sense.--Andy Schlafly 23:05, 1 October 2009 (EDT)
Hmmm, that's an interesting question. It seems to me that the power of 2 would be forced to be exact if we take two other facts for granted, which seem plausible:
  • The universe is a 3-dimensional Euclidean space
  • Gravitational fields are divergence-free away from a source. This physically means that there is no contribution to the gravitational field of a mass from any point that's not part of the mass.
This is basically a consequence of the divergence theorem (if I'm thinking straight). These are both basic tenets of the Newtonian picture. It's really not possible to do Newtonian mechanics where gravity is an inverse-2.00001 law: it would look to bodies as if gravitational force was being exerted by a body outside the mass in question! Similarly the fact that Coulomb's law is precisely an inverse-square law is a consequence of Gauss's law, which also follows from very basic assumptions. On the other hand, Newtonian physics isn't exact in the first place. I think I have read that gravity seems to be weaker than an inverse square law over extremely large distances (on the astronomical scale). This is because of some non-Newtonian, and perhaps not fully understood, physics. But there's no way to describe gravity in GR by something as simple as an inverse-square law anyway! --MarkGall 23:31, 1 October 2009 (EDT)

That's a great analysis, Mark. Even before the invention of calculus, Newton figured this with basic geometry, on the assumption that the force of gravity caused by a point mass would be the same in any direction, that is, at any other point at some distance from the mass. This logically means the force would be spread out equally over a sphere at a given radius r. Then the strength of the force at a point at radius r would be some "inherent" strength, divided by the area of the sphere. The surface area of a sphere is 4*pi*r^2. That is where the exponent 2 comes from. That formula for the area is geometric; the exponent couldn't possibly be anything other than 2. ON THE OTHER HAND, this is all in Newtonian gravity. In GR, a derivation of the exponents works similarly, but the possibility of them being slightly off 2 is equally remote, that is, 0, but even then, GR is not an ultimate formulation of gravity. Who knows how the world really works. JacobB 23:53, 1 October 2009 (EDT)

Ah, that's a better way to picture it, no need for calculus. Thanks for sharing. If only I were as smart as Newton! --MarkGall 23:58, 1 October 2009 (EDT)
This question is a very interesting one, but I think you'll agree after pondering it further that there is no logical obstacle to observing a gravitational force that proportional to the inverse of, say, 2.0000001. Math is fine for what it does, but mathematical elegance does not dictate physical laws.
Invariably math/physics types initially insist that it must be precisely 2 in order to make the integrals work nicely. But physical observation is what counts, and there is no logical obstacle to mathematical inelegance. Thanks for discussing this.--Andy Schlafly 23:56, 1 October 2009 (EDT)
Mark and Jacob, I encourage you to ponder this question (#12) further. I do think that an insistence on mathematical elegance as dictating physical phenomenon cannot withstand scrutiny.
God and logic define nature, not mathematical elegance. We have to observe first, and then come up with the best math to describe the observations. There is no logical problem to 2.0000001 here. Indeed, I find it implausible to expect that exponent to be precisely 2 for our calculating convenience.--Andy Schlafly 00:22, 2 October 2009 (EDT)
I am afraid we will have to agree to disagree on this one. Is it possible that the exponent is slightly more or less than 2? Of course it is, although the difference would be amazingly small for us not to have noticed it by now. Is it likely? Well as I've said, we'll have to agree to disagree. JacobB 00:29, 2 October 2009 (EDT)
I think there was a physics paper published about a hundred years ago indicating that an observed anomaly in orbits could be explained by an exponential like 2.000001. But mathematical physicists instinctively oppose the suggestion, just as we've seen here, and moved to relativity theories rather than explore this further. After all, who would expect to win a Nobel Prize for explaining anomalies in orbit with such an inelegant approach!--Andy Schlafly 00:33, 2 October 2009 (EDT)
If I understand correctly, it is claimed that relativity accounts for the red shift of light from deep-space objects. Would a variation in the exponent do the same? (Tapering off of gravitational wells at a distance would mean that photons would lose energy approaching us, thus drop in frequency) DouglasA 00:41, 2 October 2009 (EDT)
I don't mean to argue that it's a priori impossible that the exponent isn't exactly 2. My point is that the strength of gravity is something that's closely tied to the geometry of space, it's not just an arbitrary constant. In Newtonian mechanics, space is flat, and there's no escaping an inverse square law. If the exponent weren't exactly 2, this would just mean that space isn't quite flat. That certainly doesn't imply that the theory is wrong: relativity too is just another prediction of how space is curved! DouglasA, I'm not a physicist by any means, but that sounds plausible to me. The effect in either case would really be the same phenomenon: deviations from a flat Newtonian universe.
Mr. Schlafly, I don't know about the paper you cite, but I believe there was another paper just a couple years ago arguing that redshift observations imply that gravity isn't an inverse-square law on very large scales. I'll see if I can't dig it up. --MarkGall 01:01, 2 October 2009 (EDT)
I'd like to learn more about the red shift phenomenon in order to answer the question above, and welcome information here with an open mind. I'll do some looking myself too.
In reply to Mark, I agree that there is a close connection between gravity and the geometry of space. But isn't 2 and 2.0000001 close enough to satisfy that requirement? There is probably greater imprecision in simply modeling space with mathematics, don't you think?--Andy Schlafly 15:33, 2 October 2009 (EDT)

I know I said I wouldn't get re-involved with this debate, but let me just make a few points. Re: the red shift - yes it's true a massive object is slightly red-shifted due to the energy light loses to its gravitational field, but this effect is no stronger for distant objects than near ones, and the effect is extremely minuscule for all but the strongest gravitational fields. The red shift of distant cosmic object is a result of their incredible speed away from us, or at least their speed away from us at the time the light was emitted.

Second, 2 and 2.000001 would not be close enough to "match up" the geometry of space with the mathematics of geometry, here is why. The equation for the surface area for a perfect sphere isn't an approximation, or a guess, it's exact. Therefore, the imaginary spheres of different radii we imagine around a point mass would, if made manifest, actually have area4 \pi r^2 \ , and nothing else, just as the circumference of a circle is precisely 2\pi r \ . On this, there can be no debate - it's math, not physics.

So lets move to physics now. Newtons formula for the acceleration caused on a test particle a distance r from a point mass was a_g = G \frac{m_1}{r^2} \ . (Marking the mass of the test particle m2 gives, by F = ma the more familiar F= G \frac{m_1 m_2}{r^2} \ ) Basically, what his formula for acceleration is saying is that no matter how far you are from a point mass, there is a constant effect due to that point mass' gravity, but spread out over the area of the sphere which has the radius that is the test particles distance from the point mass. Since these spheres grow larger and larger as the test particle moves away, the acceleration grows smaller.

Now, moving to physics, let's ignore all theories of relativity, just for a moment, since the question was, after all, asked about Newtonian gravity. If the force of gravity a_g \propto \frac{1}{r^{2+x}}, x>0 \ where x is some tiny positive number, say, 0.00001, then there are either two possibilities: you have the part of this which is due to the spreading out of the effect over the spheres, and another component,  G \frac{m}{r^x} , which would mean the gravitational constant is not constant at all but a function of distance, OR, the surface areas of the spheres different than expected.

The former of these possibilities is what must be true if the exponent is indeed 2.0001, because the latter implies that space is negatively curved far away from the point mass, where the increased exponent would result in larger-than-Euclidean surface areas, and positively curved close in to the point mass, where the increased exponent would result in smaller surface areas - this would mean that the formula would work only for this one point mass, and no others in the universe; that this point mass occupied a place of privilege, the only part of the universe where space would be positively curved. Keep in mind this isn't the ST curvature of GR, which causes the effect we call gravity, and is itself caused by mass; this would be a curvature of 3-space independent of the mass distribution and having no properties that might be interpreted as a force.

But the former possibility is somewhat possible, that G is in fact G(r) - it's just that if this was the case, it would be far better described by saying that "the gravitational constant is a function of distance" rather than "the exponent is slightly different than 2." JacobB 17:42, 2 October 2009 (EDT)

That is a fascinating analysis that I want to digest further. But my initial reaction is that the obstacle to acceptance of an exponent of 2.000001 is due to an insistence that the overall gravity field remains precisely equal over the surface of the sphere. The basic assumption of mathematics driving physics remains as the underlying fallacy. The physical world is not necessarily mathematically pure.--Andy Schlafly 20:25, 2 October 2009 (EDT)
Upon further reflection on Jacob's superb contribution, I realized that the above argument that the exponent of r must be precisely 2 is defective. It cannot be justified by the surface area of the sphere because Newtonian gravitation is not a wave theory. There is no meaningful connection between the increasing surface area of a sphere as a function of r and gravitation as a function of r.--Andy Schlafly 23:44, 2 October 2009 (EDT)


I just wanted to point out that this may not be a very good measure of open mindedness. 2 and 2.00001 are arbitrary numbers, yes, but that isn't why there is an inverse square relation ship. The square of the radius in this case comes from a derivation which is a direct result of geometry. That is to say, it isn't that the variable R was saddled with some arbitrarily accurate exponent, it specifically means that the equation holds that the value of R must be multiplied by itself. If I may use a thread of logic common at conservapedia, 1+1=2, not 2.0001. Likewise, r*r=r2 not r2.00001

Open mindedness is all well and good, but I feel this question overreaches a bit, considering the fact that this proportionality holds definitively that the exponent is exactly 2.

Thanks for your comment, but your remarks serve to illustrate how this question exposes close-mindedness. Gravity is a physical phenomenon. It's observed, not derived. Many people are trained (by atheists) to think that physics must be a certain way, rather than observing with an open mind what it actually is. Those who insist that physics must conform to mathematical expectations are ... unjustifiably closed-minded.
Open your mind and you'll never want to go back.--Andy Schlafly 14:13, 14 February 2009 (EST)
Just to clarify, as the above user mentioned, the law of gravity is an inverse square law for a specific geometric reason. If you envision a gravitational field as a series of concentric spheres radiating outward from a body, you will see that as the surface areas of the spheres become progressively larger, the gravitational force must be spread over a greater area, and will therefore be weaker. The fact that the exponent is 2 comes from the fact that the surface area of a sphere varies with the square of the radius, not with the 2.00001th power of the radius.
I agree with you that observations do not always agree perfectly with theoretical models, but this is because the universe does not behave in an ideal mathematical way. Other forces (such as the gravitational fields from other planets) require that correction factors be introduced to adjust the result given by a physical equation. This does not mean the equation is wrong, only that it isn't complex enough to account for everything that is at work. Hope this helps. --Economist 18:29, 14 February 2009 (EST)
It does help ... illustrate your closed-mindedness. The "law of gravity" is a physical observation, not a postulate of Euclid. Only the closed-minded would insist that gravity cannot be inversely proportional to r2.00001. Open your mind and you'll agree. Don't and you won't.--Andy Schlafly
The essay specifically says "Newtonian gravity", not "the law of gravity" (whatever that might be). In Newtonian gravity the force absolutely has to be inverse square; as was pointed out above it's a direct and simple consequence of the fact that the area of a sphere is proportional to radius squared. The same applies for other classical field theories, such as classical electromagnetism (eg, the Coulomb attraction between two point charges is inverse square). But "Newtonian gravity" is not the same thing as "the way gravity really works" - we have General Relativity after all, and doubtlessly other successful theories of gravity will be developed in future. BrianW 07:13, 15 February 2009 (EST)
Your comment misses the point. The phrase "Newtonian gravity" illustrates that we're talking about the law of gravity as being in rough proportion to one over distance squared. If you insist that it must be precisely r-squared, as you seem to, then your mind is inexplicably closed. Open it, and your only regret will be that you didn't open it sooner.--Andy Schlafly 08:48, 15 February 2009 (EST)
No Andy, you're the only one missing anything here. The phrase "Newtonian gravity" means something very specific, which is a theoretical framework which leads directly to the consequence that force is inversely proportional to distance squared. Newton didn't just randomly pick 2 as the power because he felt like it; it is an immediate consequence of developing a classical field theory in a 3D space. By definition, if you had a law of gravity that isn't inverse-square (and you will note that I never claimed such a thing was impossible) then it isn't Newtonian gravity. BrianW 09:00, 15 February 2009 (EST)
Brian, it's a fool's errand to argue with a closed-minded person, and I'm not interested in nitpicky semantics to justify it. Are you willing to admit that gravity, an observed phenomenon, could vary with the inverse of r2.00001? Euclidian geometry does not define gravity, and it's close-minded for anyone to imply that it does.--Andy Schlafly 09:11, 15 February 2009 (EST)
Yes, I agree it is entirely possible that the way gravity works in our universe is something other than an exact inverse-square law. I was only trying to point out that Newtonian gravity is a specific physical theory and also part of a specific theoretical framework. I don't see how using technical definitions in a precise and well-defined way is nit-picking: "gravity" and "Newtonian gravity" are not the same thing. BrianW 09:14, 15 February 2009 (EST)
Brian, you haven't admitted that "gravity, an observed phenomenon, could vary with the inverse of r2.00001."
I'm not going to waste any more time debating with a closed mind. Please contribute to the encyclopedia, but I'm not optimistic about your ability to add insights. Godspeed.--Andy Schlafly 09:55, 15 February 2009 (EST)
I never attempted to claim that mathematics "defines" physics, only that mathematical models are developed to describe physical behavior. Once again, I agree with you that real-world observations necessitate refinements to these models, but this does not invalidate the models themselves. I fail to see why this makes me "close-minded." We must understand gravity fairly well if we can land rovers on Mars! --Economist 23:56, 15 February 2009 (EST)
Mr Schlafly, if I may I would like to explain a simulation I ran over a decade ago. Back in high school teachers taught about gravity. I was quick to note that the exponent 2 seemed quite arbitrary and it was the first thing I tried to disprove when I got home. So I took my laptop and ran a couple of simulations. I plotted the paths of several planets around the sun and used different formulas. First with factor 3, then 2.5, then 2.1, 2.01, 2.001, you get the point. I kept going closer to 2. The result was inviariably the same. Planets started to crash into the sun in a single orbit. The patterns were highly irregular and always involved moving through the sun. Only with exponent 2 flat the orbits remained elliptical as observed. I hope this clarifies the discussion. It's open minded to question the formula, it's open minded to test the formula and it's open minded to accept the formula once the tests come in affirmative. Marnick 13:41, 25 February 2010 (EST)
The orbits crash ... when? The orbits are not going to be significantly different over reasonable periods of time for an exponent that only slightly differs from 2.--Andy Schlafly 13:54, 25 February 2010 (EST)
At first glance that is the most logical deduction. However I challenge you to run the simulation with any commercial or free physics or mathematic software suite. I'm sure you can give this as a homework assignment to one of your students. The orbits change in very significant ways, and as I said, they are going to move through the sun (not near, exactly through).Marnick 13:56, 25 February 2010 (EST)
My statement is not merely at "first glance ... the most logical deduction." It is the only logical deduction. It is true with a mathematical certainty, which is why the denial of it is so illustrative of a closed mind.--Andy Schlafly 14:40, 25 February 2010 (EST)
You accuse ME of having a closed mind because I actually looked for evidence? The evidence might not be what you want it to be, but that doesn't make it less true. 2.00000001 will crash the planets into the sun, that's a mathematical and physical fact. I dare you to test it with a simulation, that would really show how open minded you are. Also admitting you're wrong is part of being open minded. I'm correct. Run the tests and admit it. You're so proud of your open mind, this shouldn't be a difficult task, now would it? Really mr Schlafly, run the simulation. As I said, I doubted my teacher so with my open mind I immediatly did some tests. And they confirmed what I learned. Sometimes logical deductions are simply wrong.Marnick 09:16, 26 February 2010 (EST)
Marnick, it's not necessary or helpful to run a simulation to test a logical truth. One doesn't "look for evidence" that "2+2=4", and your claim that the planets "will crash" into the sun (within a reasonable time) for any value of the exponent slightly different from 2 is mathematically fallacious. For starters, it implies an absurd discontinuity in the limit as the exponent approaches 2. Open your mind, please, and let logic in.--Andy Schlafly 10:39, 26 February 2010 (EST)
I'd like to add a bit of my own view on this. The world as we know it is strictly consequent in every way. When you change numbers, you change what happens.It is actually a fact that planets will crash into the sun when the variable is changed. For instance, if Newton's number was 3, planets would follow a path shaped like an 8, the sun being right in the middle of the 8. When you make in 2.5 , the shape will become more complex. It doesn't matter how close you get to 2; the sun will be right on the path of your planet. Though when it's 2, the path will be elliptical. I understand that this can be hard to understand / believe, but it's not a matter of "the closer it gets to 2, the closer it gets to being an ellipse." It'll only look even slightly like an ellipse when the number is exactly 2. MalP 08:41, 3 March 2010 (EST)
OK, I'm going to run a simulation. I'll be back in a day or two, perhaps with pictures. SamHB 23:09, 3 March 2010 (EST)

I'm quite skeptical of Marnik's claimed simulation. Crashing into the Sun in a single orbit is just not credible for an exponent not wildly different from 2. And "moving through the Sun" is incredible. It implies that he entered the Sun's own radius as one of the simulation parameters. In any case, the planet would have to lose nearly all of its angular momentum. (This is what makes Solar exploration spacecraft trajectories so difficult.)

Running simulations of gravity is not easy. I admit I haven't done this, so I suppose I could be accused of closed-mindedness on this issue. But I have done an analysis of the equations. The results are below. I agree with Andy that the only thing that will happen, for small perturbations in the exponent, is that the orbits will precess slowly.

The paper to which Andy refers near the top of this page was presumably the one by Simon Newcomb, who proposed an exponent other than 2 for the purpose of explaining the anomalous precession of the perihelion of Mercury. The value he gave was 2.0000001574. This was one of a number of proposed explanations for the precession, prior to general relativity. So, in the last couple of hours (when I should have been thinking about multivariable calculus!) I worked out the equations. Let the gravitational force be formulated as

F = \frac{G M m}{r^{2+k}}

Now if we let δ be the precession, as a fraction of an orbit, per orbit (that is, if δ is 1/360, the orbit will precess by 1 degree for each of the planet's own years) then (to first order):

\delta = \frac{k}{2}

This is (to first order) independent of the orbital eccentricity, and independent of the radius!

For Mercury, δ is observed to be .077E-6 (43 arcseconds per (Earth) year.) This gives k=.154E-6, in good agreement with Newcomb's figure.

So why was Newcomb's theory discarded? Because it doesn't work in general. It was first found to be in error in analyzing the orbit of the Earth's moon. But it also gives the same precession for all the planets! We now know that the anomalous precession for the other planets (the δ value) is much less than the value for Mercury.

SamHB 20:35, 26 February 2010 (EST)

I have simulated the planetary orbits, with pictures. The orbital equations are not easy to simulate accurately. I used a "relaxation method" boundary-value solver, along the lines of chapter 17 of "Numerical Recipes" by Press, Teukolsky, Vetterling, and Flannery. Double precision (17 decimal place) arithmetic was used. The convergence criterion was 10^-10. The functions were interpolated with 2000 points.

Figure 00, exponent = 2.0
Figure 03, exponent = 2.03
Figure 2, exponent = 2.2
Figure 4, exponent = 2.4
Figure 436, exponent = 2.436

Figure 00 shows the familiar Kepler orbit with an exponent of exactly 2. It is an ellipse, of course. The parameters were set so that the eccentricity was about 2. (Way more than the planets in the solar system, by the way.)

Setting the exponent to 2.000000157, as Newcomb did, would give a graph visually indistinguishable from this.

Figure 03 shows the effect of setting the exponent to 2.03. The precession is clearly visible.

Figure 2 shows the effect of setting the exponent to 2.2. The precession has become so wild that it's not really following an ellipse.

Figure 4 shows the effect of setting the exponent to 2.4. The planet is looping around wildly, but still following a regular pattern.

Figure 436 shows the effect of setting the exponent to 2.436. At this value, the planet follows a double loop, sort of like the "figure 8" pattern referred to above. (For different values of angular momentum, the exponent required to get this shape would be different.)

In no case is the orbit not a regular repeating pattern, and in no case does the planet fall into the Sun. Simulations that show that are incorrect.

But this question is ill-formed as an aspect of "quantifying openmindedness". The question seems to be asking "Do you think is impossible that Simon Newcomb's 19th century formulation is superior to Isaac Newton's 17th century formulation? While Newcomb's formulation did finagle the exponent to get a correct precession value for one planet (Mercury), a 20th century formulation has superseded both. This gives extremely accurate explanations for all planetary precessions, as well as the correct (1.75 arc second) value for the bending of light beams close to the Sun, as well as the Shapiro effect, the Pound-Rebka experiment, the Hulse-Taylor pulsar orbit precession and decay, gravitational lensing, de-Sitter precession, Lense-Thirring frame-dragging precession, etc.

There is simply no reason to embrace Newcomb's formulation, except under a bizarre notion of what constitutes openmindedness. Asking whether one is openminded enough to accept the possibility that Newcomb's exponent fudging is correct is rather like asking:

  • Are you openminded enough to admit the possibility that Ptolemy's geocentric formulation of the heavens, complete with epicycles, as opposed to the Copernicus/Kepler model, is correct?
  • Are you openminded enough to admit the possibility that the phlogiston theory of combustion is correct?
  • Are you openminded enough to admit the possibility that Trofim Lysenko's genetic theories are correct?
  • Are you openminded enough to admit the possibility that the cosmological theories of Immanuel Velikovsky, Erich von Daniken, L. Ron Hubbard, Gene "time cube" Ray, etc., are correct?
  • Are you openminded enough to admit the possibility that the "Piltdown man" wasn't a hoax?

SamHB 23:20, 11 March 2010 (EST)

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