Special Relativity in Free Fall I've been thinking about the following situation in the theory of relativity which leads to a paradox:
Two spaceships orbit a planet, on the same circular orbit, moving in opposite directions. For an external observer the two spaceships behave the same way and their clocks stay synchronized. The spaceships are in free fall so to each of them one can attach an inertial frame and Lorentz transformations should apply.
An observer in say spaceship #1 will see spaceship #2 travelling on a circular orbit, passing by periodically. According to special relativity the clock in spaceship #2 will appear ticking slower.  However after completing a trip in orbit at the next meeting the clocks of the 2 spaceships should still be synchronized. How can this be explained?
 A: Your insights are basically correct and indeed this is a nice illustration of an aspect of physics.
First I want to be clear about what is observed.

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*Say rocket 1 takes 60 minutes of its own proper time to complete an orbit. Then rocket 2 will also take 60 minutes of its own proper time to complete an orbit.


*Say the rockets pass one another at relative speed $v = 0.8 c$. Then the Lorentz factor is $\gamma = 5/3$ so each considers that life on the other rocket is proceeding at a slower rate (by a factor $5/3$) as the rockets pass one another, just as predicted by special relativity. However one can only assert this when the rockets are sufficiently close to one another that a single inertial frame is adequate to describe physics on board both of them. This will not be true when they are on opposite sides of the planet from one another.


*When the rockets are far apart, special relativity is no longer adequate to describe both rockets together or to relate timing on one of them to timing on the other. Special relativity proposes to map spacetime using a flat space. Compare this to mapping the surface on a sphere using a flat steel plate. You can do it ok for parts of the sphere close to one another. But for parts far apart you will need two such flat plates and then you need some sort of mathematical framework to relate one such plate to the other. Special relativity simply does not know how to relate one of these spaces to the other. In the spacetime context we are talking about tangent spaces and special relativity entirely lacks the mathematical tools to relate one tangent space to another. The transformation it can handle---the Lorentz transformation---actually maps a single tangent space back to itself, a bit like a rotation within the plane of the steel plate in my illustration.
So although it is true that at each stage of its journey there is a convenient tangent space in which to discuss each rocket, and in that tangent space the inertial motion of the rocket corresponds to a straight worldline, nevertheless the sequence of tangent spaces can only be handled by some further mathematical tools: the ones which general relativity provides. When one does this can have a case where a clock which was in one sense going "slow" relative to the one you are holding can make a journey relative to you and when it comes back it can have registered the same amount of proper time as your own clock. This does seem a bit paradoxical but perhaps not so much when you consider that the journey made by the other clock took a different route through spacetime.
The reason a traveler aboard one of the rockets may say that the other is ticking more slowly is that they observed this to be so as the rockets passed one another, and then they might argue by symmetry that there is no reason to suppose this to no longer hold for the rest of the orbit. This is the reasoning which leads to the paradox, and indeed it is a good paradox (I mean a genuine puzzle, not just someone reasoning in an obviously unsound way). The puzzle is to show why this reasoning should not lead to the conclusion that at the next meeting the two clocks will have advanced by different amounts. I would say the best way to avoid this incorrect conclusion is to point out the danger of comparing rates at different places in spacetime with any sort of simple approach. One must be much more careful to specify clearly what is being compared with what when one rate is at A and the other at B, with A and B separated by intervals of the order of the local radius of curvature of spacetime.
A: The Lorentz transformations, from where the calculations of time dilations of SR comes, are based in the Minkowiskian metric. If they are orbiting a planet, a Schwarzschild metric should be used instead.
As they are following geodesics, the equation for $\frac{\partial^2 t}{\partial \tau^2}$ is:
$$\frac{\partial^2 t}{\partial \tau^2} = \frac{k}{r^2} \frac{1}{1 + \frac{k}{r}}\frac{\partial t}{\partial \tau}\frac{\partial r}{\partial \tau}$$
For a circular orbit, $\frac{\partial r}{\partial \tau} = 0 \implies \frac{\partial t}{\partial \tau}$ is constant.
That means: both clocks ticks at the same rate for a distant observer whose clock follows "t" coordinate. So, if they are syncronized in one meeting event, they will be also syncronized in the other ones.
A: There is a simple answer to your question which I will first give but then there is a deeper interesting question that you are actually trying to ask but came up with the wrong thought experiment to ask it ;)
OK, so you are considering three (two) perspectives: that of a distant observer and that of either of the satellite.

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*The perspective of a deep-space observer who is in a natural inertial frame by being far away from the influence of gravity and by managing to not be pushed around by other forces. Assuming nice symmetrical things about the gravity of the earth, you are correct that this asymptotic observer will see both the clocks synchronized. In other words, when the two satellites meet, their clocks will read the same.  We can be rigorously sure that this is the correct prediction because the calculations are straightforward in this frame. The metric reads
$$d\tau^2=dt^2\Big(1-\frac{2M}{r}\Big)-\frac{dr^2}{1-\frac{2M}{r}}-r^2d\phi^2$$
Now, the proper time along a trajectory is simply $\int d\tau$ along that trajectory. We know that the trajectories of the two satellites in orbit at radius $R$ can be taken to be $\phi = \phi_0 \pm \omega t$ where $\omega = \frac{1}{R}\sqrt{\frac{M}{R}}$. Along this trajectory, $dr=0$. You plug in $d\phi=\pm\omega dt$ and you get that $$\int d\tau = \int\sqrt{dt^2\Big(1-\frac{2M}{R}\Big)-\frac{R^2}{R^2}\frac{M}{R}dt^2}=\Big(1-\frac{3M}{R}\Big)\int dt$$
As you can see, the $\pm$ sign does not matter, and thus, we can be sure that the two clocks would read the same time when they meet.

*In the perspective of one of the satellites, the other satellite would not be in an inertial frame -- because the orbiting satellite's own rest frame is an inertial frame only locally, i.e., it will see gravity on the other satellite. Thus, there would be two (three) time-dilations that the satellite would see in the clock of the other satellite: the special-relativistic due to their relative motion, the one due to the gravity of earth, and the third due to the pseudo-force. Notice that the pseudo force that one satellite would see on the other satellite would not cancel out the gravity that it sees on the other satellite. Moreover, one also has to notice that the satellite would not be using the same inertial frame itself over the time, its local inertial frame at one instance of time is not the same inertial frame as its local inertial frame at another instance of time. OK, so if you do a calculation and carefully take into account all of these things (if you write down the Schwarzschild metric, go to the local inertial coordinates of an orbiting object, rewrite the metric in new coordinate, and use that new metric to make predictions about the satellite's own clock and the clock of the other satellite, you would be automatically taking care of all of this) then you would arrive at the same conclusion that the asymptotic observer arrives at, i.e., it would lawfully predict that their clocks will read the same when they meet.


However, now onto the real question. What if we can somehow make the twins meet each other in flat spacetime, i.e., without gravity -- so that we can be sure that each twin will lawfully predict the other twin's clock to be reading something different when they meet. Well, we don't need curved spacetime to do that, we only need a compact spacetime! Are there flat compact spacetimes? Absolutely, cylinder $\times$ time, donut $\times$ time, etc. Since cylinders, donuts, etc. are absolutely flat, this is purely in the realm of special relativity. Imagine two twins going around a cylinder in opposite directions, each moving at a constant velocity. We have the same paradox that you propose but we can use special relativity all along and nicely arrive at the contradiction!
Well, of course, not really. While the spacetime is flat, unlike in traditional special relativity on $\mathbb{R}^3\times t$, on a compact spacetime, there is a preferred frame of reference in "special relativity" (more appropriately, "the physics of flat spacetime"). In particular, this preferred frame is picked apart by the specific frame of reference in which the compactification of space is defined. For example, a cylinder is a line for which the identification $x \equiv x +a$ is made. Well, the natural question arises, at what time? Because, $x$ and $x+a$ are distant points and simultaneity of an event happening at $x$ and an event happening at $x+a$ is dependent upon the frame! So, when we speak of a spacetime where the space has a cylindrical geometry, we have picked out the specific frame in which the events that are identified are of the form $(x,t)$ and $(x+a, t)$. In another frame, the events being identified via the compactification would not be simultaneous and thus, there is a clear asymmetry  among inertial frames.
There is a nice paper by Phillips Peters called ``Periodic Boundary Conditions in Special Relativity'' where they explain this point in much more detail (with nicely illustrated diagrams).
