Proof that Objects Don't Always Return Under Gravity

So, I told this crackpot over at phys.org that I would be proving what is written in the title.  It is now late, and the proof is turning out to be really long, really because he's a bit of a weasel, and thinks he can get away with saying vague, hand-wavy, things like, "Heisenberg kicks in when velocity gets low enough."  Here is the proof, so far, so that Mr. Crackpot doesn't think I am sitting on my hands.  NOTE: THIS PROOF IS NOT FINISHED, YET: [UPDATE:  Here's some background for this blog posting.  Over as phys.org, I have a long history of arguing with a crackpot I call Clever Hans.  That isn't his screen name, but a nickname that I gave him after realizing that his approach to understanding math and physics was about as deep and methodical as the approach taken by his namesake.  This is not the first time I have attempted to prove this particularly well known property of gravity to him.  My first attempt, however, involved solving differential equations and using limits, something he summarily dismissed as "only a mathematical model," adding that at some point, "Heisenberg would kick in," thereby demonstrating not just a poor understanding of quantum mechanics, but an even poorer understanding of potentials.  Anyway, to make a long story short, I decided to attempt this task once again, this time with just high school algebra, conservation of energy, and conservation of momentum.  I don't know why I bother, though.  I am sure that his response will be something profound, like, "you forgot a comma, therefore, your proof is invalid."  ]

All right, Mr. Crackpot.  Thank you for your patience. Here is my proof.

Before I start, let me establish something about gravitational potential.

Now, the gravitational potential is given by -GmM/r, according to http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html, where r is the distance between the gravitating objects, and m and M are the masses.

But, you have insisted, often, that the laws of gravity as presented by textbooks, are merely approximations.  Therefore, I will allow for the more general potential, $\frac{-GmM}{r} + s\left(r\right)$, where s(r) is an error correcting term.  Now, at normal distances, s(r) must be approximately 0 because, otherwise, we would have detected it by now.  However, we are concerned with what will happen at large distances, and so, s(r) may come to dominate the potential.  More on that in a little bit.

Now, you argue that, in a universe with only two gravitating objects, they will eventually return to each other, regardless of their initial velocities.  Normally, I would say this means that at least one of the objects will slow to 0m/s, and then change direction.  However, I anticipate that you would counter that some kind of hypothetical quantum effects might allow the object to "jump" between moving forward and moving backward without completely slowing to a halt, and so, I'll grant you that, for argument's sake.  Let's say that whenever an object slows, and turns around relative to another object, there must be some lowest speed threshold, $S_q$, such that AT SOME POINT BEFORE TURNING AROUND, the relative speed will be less than or equal to $S_q$.

For example, let's say $S_q = 10^{-12}m/s$.  Then the relative speed between the two objects must at least slow down to $10^{-12}m/s$, if not less, before at least one of the objects change direction.  However, as with s(r), let's not assign it a value just yet.

With this in mind, all I need to do to prove you wrong is find a counter example to your claim.  So, here is what I will do. I will do a proof by contradiction, and assume that you are correct, and then show that it implies physically unreasonable conditions for a specific case.

For this, I have chosen a very simple set up.  We choose two identical, spherical, objects, A and B, of mass, 10^13kg, about the mass of the comet, 67P/Churyumov-Gerasimenko.  Object A starts out at $-10^{6} $ m from the origin.  B starts out at $10^{6}$ m from the origin.  Their initial velocities are equal in magnitude and in opposite directions, so that A is travelling, initially at $-v_0$, and B is travelling initially at $v_0$.  I'll get to the value of $v_0$ in a little bit.  For now, let's at least require that $|v_0|$ is well below the speed of light so that relativistic effects are negligible.  With this in mind, let's calculate the gravitational potential of A and B for this initial condition.

Well,

\begin{equation}-\frac{Gm^2}{2\times 10^6 \text{meters}} = -3369200000 \text{J}\end{equation},

 so, our potential is $-3369200000 \text{J} + s(2\times 10^6 \text{meters})$.

Now, let's see what conservation of momentum tells us.  Well, the initial total momentum is $mv_0 - mv_0 = 0 \textrm{kg m/s}$.

That means that if vA and vB are the velocities of A and B respectively in the future, m*vA + m*vB = 0, so that vA = -vB.  So, again, vA, and vB will be equal in magnitude an opposite in direction.  Therefore, we can dispense with this notation, and instead assume that at some point in the future, A is travelling at $-v_1$, and B is traveling at $v_1$.  Currently, we don't know what $v_1$ is.  It could be negative or positive or 0.

Suppose that A is at position xA and B is at position xB.  Since A and B are always moving at the same speed, we can also assume that xA = -xB.  As above, I can dispense with the "A" and "B" notation, and instead say that A is at position $-x_1$, and B is at position $x_1$.  Again, we DON'T YET KNOW what $x_1$ is.

Without loss of generality, assume that B turns around at some point in the future.  Now, since the initial conditions are that A and B start out moving away from each other at the same speed, we can assume that A and B will have moved even further away from each other by the time that their relative speed, $2x_1$, slows down to Sq or lower.  So, we can DEFINITELY assume that this happens when B's position, $x1$ is $> 10^6 \text{meters}$.

Now, let's use conservation of energy to get an idea of how large $x_1$ needs to be so that its magnitude is between 0meters/s and $S_q/2$.

The initial total energy is

$mv_0^2 - 3369200000 \text{J} + s(2\times10^6 \text{meters}) = (10^{13 }\text{kg})\times v_0^2 - 3369200000 \text{J} + s(2\times 10^6 \text{meters})$

Well, $\frac{-Gm^2}{2x1} = -\frac{3.3692\times10^{15} \textrm{kg meters}^3/\text{s}^2}{x_1}$

So, the total energy is $(10^{13}kg)\times v_1^2 -(3.3692\times 10^{15} \textrm{kg meters}^3/ \text{s}^2)/x1 + s(2x1)$

and conservation of energy tells us that

$S_q^2 = v_1^2 = v_0^2 - 0.00033692 \text{J\kg} + \frac{s(2\times 10^6 \text{meters}) }{10^{13}\text{kg}}+ \frac{ 3.3692\times10^{15} \textrm{kg meters}^3/\text{s}^2}{x_1\times10^{13}\text{kg}} - \frac{s(2x_1)}{10^{13} kg} = v_0^2 - 0.00033692 \text{J/kg} + \frac{s(2 \times 10^6 \text{meters})}{10^{13}\text{kg}} + \frac{336.92 \text{meters}^3/\text{s}^2}{x1} - \frac{s(2x_1)}{10^{13}\text{kg}}$

Or more succinctly,
$S_q^2 = v_0^2 - 3.3692\times 10^{-4} \text{J/kg} + \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}} + \frac{ 336.92 \text{meters}^3/s^2}{x_1} - \frac{s(2x_1)}{10^{13}\text{kg}}$

Now, there are two cases to consider: that s(r) is bounded as r increases, or that s(r) is unbounded as r increases.  I assert that it is physically unrealistic if s(r) is unbounded as r increases.  I'll assume that Mr. Crackpot agrees with this since his argument is that as distances grow, quantum effects "kick in", meaning that s(r) becomes quantum energy fluctuations.  In other words, the only two reasonable contributors, in this non-relativistic case, to the gravitational potential energy are the classical gravitational potential energy and quantum energy fluctuations.  Now, I am aware that Mr. Crackpot, being unabashedly bad at physical intuition, might, indeed, protest that I haven't considered the case where s(r) is unbounded.  If he does, I have a proof for that, too, but in the interest of not making this post any more of the novella it is, I will assume that it is acceptable to require that s(r) is maximally bounded.

Thus, at most, s(r) is bounded in magnitude by a particular energy scale.  Let's call this scale $s_{ub}$, so that $-s_{ub} \leq s(r) \leq s_{ub}$.  Note that $s_{ub}$ is nonnegative.

With this in mind, we can say that

$\frac{s(2x_1)}{10^{13}\text{kg}} = v_0^2 - 3.3692\times 10^{-4} \text{J/kg} + \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}} + \frac{ 336.92 \text{meters}^3/s^2}{x_1} - S_q^2$

implies the inequalities,

$-\frac{s_{ub}}{10^{13}\text{kg}} \leq v_0^2 - 3.3692\times 10^{-4} \text{J/kg} + \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}} + \frac{ 336.92 \text{meters}^3/s^2}{x_1} - S_q^2 \leq \frac{s_{ub}}{10^{13}\text{kg}}$

We only need the right inequality,

$v_0^2 - 3.3692\times 10^{-4} \text{J/kg} + \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}} + \frac{ 336.92 \text{meters}^3/s^2}{x_1} - S_q^2 \leq \frac{s_{ub}}{10^{13}\text{kg}}$

This is equivalent to writing

$\frac{s_{ub}}{10^{13}\text{kg}} +  S_q^2 \geq v_0^2 - 3.3692\times 10^{-4} \text{J/kg} + \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}} + \frac{ 336.92 \text{meters}^3/s^2}{x_1}$

However, since $x_1$ is positive, it is also true, if we just simply drop the term containing it.

$\frac{s_{ub}}{10^{13}\text{kg}} +  S_q^2 \geq v_0^2 - 3.3692\times 10^{-4} \text{J/kg} + \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}}$

Now, I will reveal the value of $v_0$.  I have assigned it the hugely nonrelativistic value of 340.29 meters / s.

So, we have

$\frac{s_{ub}}{10^{13}\text{kg}} +  S_q^2 > 115797.2841 \text{meters}^2/\text{s}^2 - 3.3692\times 10^{-4} \text{J/kg}+ \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}}$

which is equivalent to

$\frac{s_{ub}}{10^{13}\text{kg}} +  S_q^2 > 115797.28376308\text{meters}^2/\text{s}^2+ \frac{s(2\times 10^6 \text{meters})}{10^{13}\text{kg}}$

Now, $2\times10^6\text{meters}$ is not very large.  It's in the neighborhood of 1000km. We know, then that $s(2\times 10^6 \text{meters})$ is too small to detect, here at home, so that we can simply say that its value, compared with $115797.28376308 \text{meters}^2/\text{s}^2$, is approximately 0.  It is, therefore, acceptable to write,

$\frac{s_{ub}}{10^{13}\text{kg}} +  S_q^2 > 115797.28376308 \text{meters}^2/\text{s}^2$

Now, the above inequality implies that

\begin{equation}\frac{s_{ub}}{10^{13}\text{kg}} \geq 57898.64188154\text{meters}^2/\text{s}^2\end{equation}

OR

\begin{equation}S_q^2 \geq 57898.64188154 \text{meters}^2/\text{s}^2\end{equation}

(they can't both be less than half of $115797.28376308 \text{meters}^2/\text{s}^2$)

If inequality (2) , then

\begin{equation}s_{ub} \geq 57898.64188154\times 10^{13} \text{J}\end{equation}

This is clearly absurdly large.  To give you an idea of how large this is, this is about 6 times as energetic as the 2004 Indian Ocean Earthquake.  Obviously, $s_{ub}$, and hence s(r) is not just a quantum energy fluctuation.

On the other hand, if (3) is true

$S_q \geq 240.621 \text{meters \ s}$.

In human terms, this is a very large velocity, and is equal to 538.252 miles per hour.

Again, this is unacceptable.  Normal objects like cars and humans go much slower than this before turning around.  So, again, we arrive at an absurd physical requirement.

The only possibility, assuming s(r) is due to quantum fluctuations is that your initial premise, that objects always return under gravity is horribly flawed.

The ball is in your court.  Now, I spent a long time writing this up.  Let's see if you are honorable enough to respond accordingly rather than replying with a terse, irrelevant comment, like "it's just math!"