Eh, I might as well just run through it. It's a good exercise. (P.S. If you can't follow this, don't worry about it. Just skip to the part after the quote.)
The coordinate system of choice will be a rotating frame with angular velocity Ω. Naturally, we'll chose Ω such that the satellite isn't moving, but we can work it out later. I'll also completely ignore gravity. It's a minor effect. One can use all the same logic used here to add correction to Schwarzschild metric that would allow you to do gravity from a rotating frame, so long as center of rotation matches center singularity of Schwarzschild metric.
Anyways, we need the distance element. Let's first work out what the distance is in spatial coordinates. We really need to only consider r and \small \varphi, so I'm going to drop the 3rd spatial coordinate all together. Because the frame is rotating, an object that did not move in \small \varphi is actually Ωdt from where it started out. So the square of spatial distance between two points separated by dt is dr² + r²(Ωdt - d\small \varphi)². Separation in time is still just dt. So the line element works out to the following.
ds^2 = c^2dt^2 - dr^2 - r^2(d\varphi^2 + \Omega^2 dt^2 - 2\Omega d\varphi dt)
And so we have the metric tensor.
g_{\mu\nu}=\left( \begin{array}{ccc}<br />
1-\frac{\Omega^2 r^2}{c^2} & 0 & \frac{\Omega r^2}{c}\\<br />
0 & -1 & 0\\<br />
\frac{\Omega r^2}{c} & 0 & -r^2 \end{array} \right)
With 3-vector \small x^{\mu} given by (ct, r, \small \varphi). The inverse will come in handy.
g_{\mu\nu}=\left( \begin{array}{ccc}<br />
1 & 0 & \frac{\Omega}{c} \\<br />
0 & -1 & 0 \\<br />
\frac{\Omega}{c} & 0 & \frac{\Omega^2 r^2}{c^2}-\frac{1}{r^2} \end{array} \right)
In this coordinate system, a stationary object will have velocity \small u^{\mu} = (u^t, 0, 0).
c^2 = u_{\mu}u^{\mu} = g_{\mu\nu}u^{\mu}u^{\nu} = (1-\frac{\Omega^2r^2}{c^2})(u^t)^2
u^t=c\left(1-\frac{\Omega^2r^2}{c^2}\right)^{-\frac{1}{2}}
This is actually sufficient for time dilation. If you note that in inertial frame of reference tied to Earth, v=Ωr for any circular orbit, and that c dt/d\small \tau = \small u^t, you have the time dilation formula in hand. It results in no time dilation for Earth and time dilation consistent with SR for the satellite.
Of course, having metric and 3-velocity, it might be interesting to find the acceleration. Because metric depends on radius only, it is not too difficult.
a^\mu = \nabla_u u^{\mu} = \frac{\partial u^{\mu}}{\partial x^{\nu}}u^{\nu} + \Gamma^{\mu}_{\rho \sigma}u^{\rho}u^{\sigma}
That leaves just three relevant Christoffel Symbols.
\Gamma^t_{tt} = 0
\Gamma^r_{tt} = -\frac{1}{2}g^{rr}\frac{\partial g_{tt}}{\partial r} = -\frac{\Omega^2 r}{c^2}
\Gamma^{\varphi}_{tt} = 0
And that means we only have one surviving term in acceleration, \small a^{\mu} = (0, \small a^r, 0).
a^r = \Gamma^r_{tt}u^tu^t = -\Omega^2r\left(1 - \frac{\Omega^2 r^2}{c^2}\right)^{-1}
Note that so long as Ωr << c, the term in parentheses is ~1. In which case, acceleration is simple centripetal acceleration from classical mechanics. As Ωr goes to c, acceleration becomes infinite.
Thank! But is it possible to resolve this 'dilemma' without using any equations? Just qualitative explanation is enough. Thanks again!
There is no dilemma. Special Relativity formulas are correct in Earth's frame of reference. The only reason you have "disagreement" is because you tried to apply SR formulas to satellite. They won't hold. The simple way way to figure out what happens to satellite is to simply consider it from Earth's perspective. You can use SR equations to describe accelerated objects, so long as your frame of reference isn't accelerating.
If you compute satellite's time dilation and acceleration from Earth's perspective, you'll get exactly the same result as all that mess above gives you. That mess resolves the dilemma, but you already were able to compute the results using Special Relativity.
