accelerationcelestial-mechanicsnewtonian-gravitynewtonian-mechanicsorbital-motion
Why do orbital speeds decrease further away from the focus? A simple question, but I want to make sure I am understanding this correctly: Is it ONLY a function of the gravity well? As in, the gravitational field is weaker as you move away from the massive body, so the speed decreases? What if the gravitational field was constant through space? Would the orbit's speed then be constant?
This should be a home-run for someone.
Yes the free body moves outward, but there are two critical things you have to know to interpret this statement correctly.
First, this is the effective potential, taking into account gravity and centrifugal force. It has this form because we went into the non-inertial frame co-rotating with the two masses. Mathematically, the potential is $$ \Phi_\mathrm{eff}(\vec{r}) = -G \left(\underbrace{\frac{M_1}{\lvert \vec{r}-\vec{r}_1 \rvert}}_\text{potential from mass 1} + \underbrace{\frac{M_2}{\lvert \vec{r}-\vec{r}_2 \rvert}}_\text{potential from mass 2} + \underbrace{\frac{M_1+M_2}{2\lvert \vec{r}_1-\vec{r}_2 \rvert^3} \lvert \vec{r} \rvert^2}_\text{centrifugal component}\right), $$ and it only decreases far away because of that last term.
Physically, this is because placing an object "at rest" in this frame corresponds to having it move with the same angular frequency as $M_1$ and $M_2$ about the center of mass. If you initialize an object $5\ \mathrm{AU}$ on a tangential path having the same angular velocity as the Earth, it will be moving too fast for a circular orbit at that distance, and so it will move away from the Sun.
This does not mean the object will go away forever, and that brings us to the second point, explained in Chay's response: Not all effective forces have been accounted for; in particular, the Coriolis force does not arise from $\Phi_\mathrm{eff}$. The Coriolis force depends on velocity, so it has no scalar potential depending solely on position, and so it is not included in the analysis so far. Once your test object starts moving in your rotating frame, it will experience a perpendicular deflection that will eventually force it to turn around.
The Moon's orbit would be nearly Keplerian were it not for the perturbing effects of the Sun. The time from perigee to perigee and from apogee to apogee wouldn't change, and the time from perigee to apogee would be exactly half the orbital period. What you are seeing are perturbing effects of the Sun on the Moon's orbit.
If you use push the site you found, http://www.timeanddate.com/astronomy/moon/distance.html, to just beyond 2015, you'll see that only 24.69 days transpire between the last lunar perigee of 2015 and the first lunar perigee of 2016. That's almost three days less than the average value of 27.55455 days. There's something curious going on here!
Below is a plot of the times between successive perigees (red curve) and successive apogees (blue curve). Note that the perigean anomalistic month is subject to significantly greater variations than is the apogean anomalistic month. Also note that the two curves have the same frequency (about 7 months) but are almost 180 degrees out of phase. These variations are the direct cause of the phenomenon you observed. You'll see a marked difference between perigee-to-apogee versus apogee-to-perigee when the time from perigee to successive perigee is changing fastest.
It's the Sun that makes these variations occur. The Sun makes the lunar apse line (the line from Earth to the Moon at perigee) precess. This, in combination with the way tidal forces themselves vary, makes the Moon's orbit more eccentric that normal when the lunar apse line (the line perigee to apogee) aligns with the syzygies (new moon or full moon), less eccentric than normal when the apse line is more or less orthogonal to the line from the Earth to the Sun. The perigean months are longest when perigee more or less coincides with a new or full moon, and shortest when perigee more or less coincides with the moon in its quarter phases.
Best Answer
"Focus" is an inconvenient word if you're thinking of changing the potential, because if you do then the orbits are no longer conics and the word kind of loses its meaning. That aside, let me see if I understood your question correctly:
In that case, the answer is that orbital speeds decrease because $V$ itself increases at longer distances. This is simply conservation of energy: $$\frac12m\mathbf{v}^2+mV(|\mathbf{r}-\mathbf{r}_0|)=E,$$ and if $V$ becomes less negative then $v^2$ must be smaller. Thus, potentials where this doesn't happen must have regions where the potential is repulsive from the origin. One such example is $$V(r)=-\frac1r -r,$$ though of course there's no physical system with that behaviour.