(Alpha) A new equation for escape velocity

Flawed attempt at showing his reasoning is flawed. Clearly, pi has no units. How can you add a unitless number to a number with units?

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You can always add a proportionality constant equal to one and with the right units (if you're really desperate).

 

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No I think Absane is right. I will revise my new proposal, then:

\(v_{esc} = \sqrt{\frac{2M}{r + \pi M}}\)

 

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It does not, because all things are not otherwise equal between GR’s equation and my equation. GR’s equation leads to GR being self-inconsistent, as I noted above. That invalidates GR’s equation. In a test of Occam’s razor, my equation can be compared only to another equation that is not known to be invalid, so that the condition “all things being equal” in Occam’s razor is satisfied.

Pulling an equation out of my ass is perfectly acceptable, according to the scientific method. But I didn’t do just that. I fit the equation to what is known. Namely, as I have shown, “the escape velocity must always be less than c” or else the theory of gravity will be self-inconsistent. And I fit the equation to observations. For example, had I created this equation instead:

v = sqrt(M / (r + M))

it wouldn’t be better because, although it predicts a velocity less than c above r=0 and is simpler than the one I gave above, it does not approximate GR’s equation for all of the experimental tests of it to date.

I created the simplest equation I could think of that fits within all the known parameters. That’s good science. How I created the equation doesn’t really matter.

Nobody knows that any theory’s predictive equation is right. Theories are not known to be valid.

The only reason to believe any theory’s predictive equation is faith.

No, I’m not telling you that. Conservation of energy is only one component of your derivation. How do you know that the predictive equations in your derivation are correct? You don’t.

Nobody can dictate how nature must work; that is what you are attempting to do. You cannot prove that nature works the way you think it does (i.e. you cannot prove that a theory is valid), regardless of any derivation you show. Nature can always trump you. And we don’t have the observational data to confirm GR’s equation except in relatively extremely weak gravity.

Consider as well that the derivation you gave is Newton’s. It is thought to yield the same equation as the general relativistic derivation only by happenstance. Andrewgray gave a general relativistic derivation; see the OP.

 

No it doesn't. A flawed derivation does not imply that its conclusion is false. It's perfectly possible for GR to be wrong and \(v = \sqrt{\frac{2M}{r}}\) to be the correct equation for escape velocity.

 

Occam’s razor does not refer to the derivations of explanations. It refers to only the explanations themselves. In this case the “explanation” is in the form of stand-alone equation. It need not be a complete theory. Ultimately all theories are based on ad-hoc patches, for nobody knows for sure how nature works.

 

I agree. That’s partly why I say that an equation’s derivation need not matter.

You are right that such is possible. I stand corrected. But I have shown that any theory of gravity that has that equation for escape velocity and in which SR applies locally everywhere will be self-inconsistent. The only way that that equation can be correct is if SR does not apply locally everywhere. Then things are still not otherwise equal between GR’s equation and my equation, or else you could say that my equation is far simpler because it doesn't require any exceptions for where SR applies locally.

 

Because they've been tested.

http://www.npl.washington.edu/eotwash/experiments/shortRange/sr.html

You are claiming that the inverse square law is invalid, I think. I don't actually know what you're claiming because you haven't said. The escape velocity equation that I wrote down comes from assuming the validity of the inverse square law and conservation of energy. (Your equation was pulled from your ass.) You claim not to violate conservation of energy, so then you must think that the inverse square law is violated. If this is true, then you must explain why the precision experiments at University of Washington have confirmed the inverse square law to tenths of milimeters.

Note that I have said nothing about GR in this post. Your equation is not consistent with Newton's laws---it doesn't reduce to a familiar (well-tested) form. If I assume that your equation should be derived in the same manner, it says, then, that the gravitational potentail energy between two bodies M and m, separated by R, is given by:
\( U = \frac{Mm}{2M + R}\).
This implies that the force between two bodies is (take a derivative and check my math):
\(F = -\frac{Mm}{(2M + R)^2}\).
Now let's estimate the acceleration due to gravity on the surface of the earth:
\(R\sim10^6, M\sim10^{24}, G\sim 10^{-11}\).
Now write F = mg (which I hope you are familiar with), lose the negative sign, and calcuate g:
\(g = \frac{GM}{(2M + R)^2} \sim \frac{10^{-11}10^{24}}{(10^6+10^{24})^2} \sim 10^{-35}\),
which is about 36 orders of magnitude away from the tested value of 9.8m s^-2, which many high school students are aware of.

Appendix:
In case you don't believe that my calculations are correct, check MY formula for g...
\(g = \frac{GM}{R^2}\sim \frac{10^{-11}10^{24}}{(10^6)^2}\sim10^{-11}10^{24}10^{-12}\sim 10\),
in fine agreement with the experimental result.

 

Again, I have presented all of the steps in my derivation so that you can check my work, and point to the place where I have made a mistake.

 

Hopefully by now you see that YOU'RE the one trying to dictate nature's behavior. In post 28 I showed how your escape velocity gives a value of g that is 36 orders of magnitude wrong.

 

You sure like to quibble over "small" errors don't you.

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Those tests are in relatively extremely weak gravity only (“short range”, like tenths of millimeters, doesn’t mean “small r”). They do not otherwise confirm the predictive equations in your derivation.

GR also predicts a violation of the inverse-square law. GR’s prediction of gravitational length contraction (also referenced as excess radius) predicts that violation. The violation becomes more apparent with stronger gravity, and may be deemed negligible in weak gravity. From here: “However, in the limit of low velocities and weak gravitational fields, Einstein's theory still predicts that the gravitational force between two point objects obeys an inverse-square law.”

Two related mistakes: You double-counted G, which is already factored into M in both of the equations (GR’s and mine) in the OP. Geometric units are used in the OP, as noted there. And the mass of the Earth in geometric units is 0.00444 meters. I already showed in the OP that the new equation is fully experimentally confirmed to all significant digits. Then I know you will not be able to show a damning difference between GR’s equation and mine for the surface of the Earth. Any analysis that attempts to show such a difference will have a problem.

 

Yes zanket you are right. I did screw up in inserting factors of c and such. Here is the correct derivation.

\(F = \frac{G Mm}{(R+\frac{2GM}{c^2})^2}\),
which implies
\( g = \frac{GM}{(R+\frac{2GM}{c^2})^2} \)

Then, the inverse square law gives:
\(\frac{GMm}{(R+\frac{2GM}{c^2})^2}=\frac{GMm}{R^2} \frac{1}{(1+\frac{2GM}{Rc^2})^2} \sim \frac{GMm}{R^2}(1-\frac{4GM}{Rc^2}+\cdots)\)

So you DO have deviations from the inverse square law, which (presumably) even survive in the weak field approximation. So one should be able to derive a new version of Kepler's law... (Well, I can...)

\(T^2=\frac{4\pi^2R^3}{GM}\left(1+\frac{4GM}{Rc^2} + \cdots\right)\)

Someone with a calculator and google should check to see that zanket's new law of gravity doesn't violate the length of the year, both on earth and on the other outer planets.

 

I should point out that with accurate enough observations, even the weak field limit of your theory should be measurable. In particular, I think these things have been tested very accurately.

 

I confirmed that this equation is fully experimentally confirmed. It is a little strange though, since it seems to “top out” around 0.8c. I’d call my equation simpler since it returns a full range from a limit of 0 to a limit of c.

Well, this is just an escape velocity equation. A full theory of gravity is coming from me later.

The experimental confirmation section in the OP covers any test at r>=100000M, which includes every test to date.

 

Alright zanket. I've been crunching some numbers and I think the orbital period of venus rules out any variations of this inverse square law in the way that you want to modify it. The agreement is pretty good, up to the last decimal place or so, but (accounting for uncertainties) you should find that the precision measurements of the Venusian year rule out your model.

I won't show you the calculation, though, because I don't feel I need to.

Also, you should be aware of this paper, which finds no deviations of the inverse square law on large scales:
http://www.physics.upenn.edu/~raulj/paper_75.pdf
I want to make it clear that your formula, as presented, is general. This means that any deviations on large scales, as well as small scales, are fair game. You never derived your equaiton, so we can only guess when it is valid.

 

I didn’t say there is an error in the derivation of Einstein’s escape velocity equation. I said “I showed that a theory that incorporates SR and Einstein’s equation for escape velocity will be self-inconsistent”. There does not need to be an error in the derivation of that equation for GR to be self-inconsistent when that equation is incorporated into it.

If you’re still so sure that I could not have shown a self-inconsistency of GR, why not show what is wrong with my proof of that? I even boiled it down to a short paragraph with a diagram. If you are right then it should be a simple matter to refute it.

 

Nice try. Then I don’t feel the need to refute you.

The paper discusses deviations in weak gravity. Weak gravity = large r. GR’s predicted deviation of the inverse square law may be deemed negligible in weak gravity. I already showed you a link that says that GR obeys the inverse square law only in the limit of low velocities and weak gravity. When these scientists look for deviations in weak gravity, they are looking for deviations that exceed the deviations that GR predicts. They are looking for something that might reject the theory.

An equation’s validity is not determined by its derivation. It’s determined by whether the equation is self-consistent and agrees with observations, and nothing more. You’re making an invalid assumption about the value of a derivation. I ask you (and it’s not rhetorical), can you prove that an equation’s validity is determined by its derivation? Can you give an example?

 

Now you agree with GR? I'm confused.

It doesn't matter if the variations are small---they're still calculable. This is why I said "The deviations survive into the weak field approximation". I showed you how to prove your model wrong---look at the precision data in Venus's orbit.

Maybe I should ask you---where does your formula apply? IS it not universal? If it is universal, then one can calculate the deviations from Newtonian gravity and compare to experiment. If it isn't universal, then it only applies in some regime. But, again, you haven't derived this equation, so there's no way for anyone (including you) to know this.