Twin paradox on opposite sides of the Earth The earth is rotating at about 460m/sec at the equator, so relative to me, someone on the other side of the earth is travelling at about 920m/sec. This means that their clock is running slower than mine due to special relativity. Of course, to them I am moving at 920m/sec so they see my clock running slow.
If Elon has drilled a tunnel through the middle of the earth, we can both meet in the middle and synchronise our clocks, then go back to our surfaces, wait 1 month and then meet again and compare our clocks. What would we see when we meet? I assume the clocks would be the same, and then we would both be wondering why the special relativity calculations gave us the wrong result that the others clock should be behind.
 A: You would be the same age, and if you thought special relativity gave the wrong answer that would be because you didn't understand special relativity.
All the effects of SR are symmetric between any two inertial frames, so if you and I are moving toward each other, your time is dilated in my frame and my time is dilated to the same degree in yours.
The example you gave is more complicated because you and Elon are continually switching between interval frames, but since you are doing so symmetrically the result will be symmetric too.
You might find yourself wondering, as other have, how time dilation can be symmetric. The key to understanding it is to realise that it arises because the time in two moving frames of reference are out of synch everywhere along their mutual direction of motion- that is what causes the time dilation effect, as I will show in an example now.
Suppose you and I are exactly the same age, and we are some distance apart in the frame of the Earth where it is time $\mathrm{T_{earth}}$ everywhere. Suppose at some agreed time $\mathrm{T_{earth}}=0$ we each accelerate instantaneously then coast towards each other at a speed where each of us thinks the other is time dilated by $20 \%$. When we meet we compare watches and find that each shows $4$ minutes has passed. Given that we have each been time dilated, according to the other, by $20\%$, that means each of must have spent $5$ minutes travelling in the other's frame, but how could that be when both our watches show that only $4$ minutes have passed?
The answer is that at the moment we started our respective journey we each moved from the Earth's reference frame, where it was $\mathrm{T_{earth}}=0$ everywhere, into our own frames at $\mathrm{T_{you}} =0$ where you started your journey and $\mathrm{T_{me}}=0$ where I started mine. However, because we were in frames moving relative toward each other, our times where out of synch everywhere. Specifically, where you started your journey at $\mathrm{T_{you}}=0$ the local time in my frame was $\mathrm{T_{me} =-1\ m}$. Conversely, where I started my journey at $\mathrm{T_{me}}=0$ the local time in your frame was $\mathrm{T_{you} =-1m}$.
Now you can see how the effect of time dilation arose. I went from a place where it was $\mathrm{T_{you}=-1m}$ to meet you at a place where it was $\mathrm{T_{you} =4\ m}$ , thus moving through a total of $5 \mathrm{m}$ in your frame even though there was only a $\mathrm{4\ m}$ difference in mine. Likewise you went from a place where it was $\mathrm{T_{me}=-1\ m}$ in my frame and met me where it was $\mathrm{T_{me}=4m}$, so moving through $5 \mathrm{m}$ of time in my frame.
It is the lack of synchronisation, which is know as the relativity of simultaneity, which caused each of us to think the other was time dilated, even though both our clocks ticked at exactly the same rate throughout.
A: By symmetry, you would have to be the same age!
The special relativity argument you present fails here because both of you are in an non-inertial frame of reference.
A: The factor for time dilation can be understood from the Minkowskian metric with one spatial dimension for inertial frames:
$$c^2d\tau^2 = c^2dt^2 - dx^2 \implies \frac{d\tau^2}{dt^2} = 1 - \frac{v^2}{c^2} \implies \gamma = \frac{dt}{d\tau} = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$
In the situations with 2 inertial frames, the metric can be used for both, so the time dilation is symmetrical.
For the case of the question, it is more convenient to use the Minkowskian metric in polar coordinates:
$c^2d\tau^2 = c^2dt^2 - dr^2 - r^2d\theta^2 - r^2sin^2(\theta)d\phi^2$
In the case, there is no radial movement, so $dr = 0$. Also the movement is around the equator, so $\theta = \frac{\pi}{2}$ and $d\theta = 0$
$$c^2d\tau^2 = c^2dt^2 - r^2d\phi^2 \implies \frac{d\tau^2}{dt^2} = 1 - \frac{\omega^2r^2}{c^2} \implies \frac{dt}{d\tau} = \frac{1}{\sqrt{1-\frac{\omega^2r^2}{c^2}}}$$
Now, only the non rotating frame is inertial. As can be seen from the formula, there is a time dilation between this frame and any rotating frame. As it is function of $\omega$ and $r$, the time dilation is the same for both in this case, in relation to the non rotating frame.
Of course, the Schwartzschild metric is more appropriate for the earth, but the conclusions don't change by taken gravity in consideration.
A: TL/DR: The other clock appears to be running slower because of your relative speed, but appears to be running faster with the same factor because you are in an accelerated reference frame. The two effects cancel each other out.
If you are both moving with speed $v$ wrt. the centre of the Earth, then you have a relative speed $w$ wrt. each other according to the relativistic velocity addition formula:
$$w = \frac{2v}{1 + v^2/c^2}$$
That means that the time dilation due to the relative speed is equal to the Lorentz factor:
$$\gamma = \frac{1}{\sqrt{1 - w^2/c^2}} = \frac{1}{\sqrt{1 - \frac{4v^2/c^2}{(1+v^2/c^2)^2}}}$$
Setting $\beta = v/c$, this simplifies to:
$$\gamma = \frac{1+\beta^2}{1-\beta^2}$$
Ignoring factors of $\beta^4$ and higher, that is equal to:
$$\gamma \approx 1 + 2\beta^2$$
So, the other clock would appear to be ticking $1 + 2\beta^2$ slower than yours.
But, you are both circling around. That means that, according to Special Relativity, you are not in an inertial reference frame. Each of you are constantly accelerating towards the centre of the Earth. Imagine that the Earth was not there, then, in order to keep circling, you would need a rocket strapped to your back accelerating you in the direction of the centre with an acceleration equal to $g = 9.8m/s^2$. That is, you are in an accelerated reference frame with an acceleration of $g$. In such a reference frame the other clock would appear to run faster with a factor
$$T_d = e^{gh/c^2}$$
with $h = 2r$ the distance between you, and $r$ the distance to the centre. The centripetal acceleration needed to keep your rotating is equal to:
$g = v^2/r$, and therefore
$$T_d = e^{2v^2/c^2}$$
Again setting $\beta = v/c$ and ignoring factors of $\beta^4$ and higher, that is approximately equal to
$$T_d \approx 1 + 2\beta^2$$
So the speed-up factor due to the accelerated reference frame is approximately equal to the slow-down factor due to the relative speed. They cancel out, and the other clock would appear to tick at the same rate as your own clock.
A: You don't need a twin halfway around the world to address this issue, because you yourself spend a lot of time halfway around the world (if you happen to live on the equator) or at least halfway around your circle of latitude.
At noon every day, you will say (using your current instantaneous frame) that your own clocks have run slow every moment of every day of your life, except at noon each day, when they run exactly right.  At 1PM every day, you will say that your clocks have always run slow at noon (and at all other times other than 1PM).
There is no contradiction there.  A clock that runs slow in one frame is not required to run slow in another frame.
If you and your friend meet in the middle of the tunnel, your clocks will agree.  You will say that in the month you've been apart, your own clock and your friend's clock have both run slow most of the time, and that although there have been times when yours was slower, and times when your friend's was slower, the total amount of slow running was the same for both clocks.  So of course they agree!
