Schwarzschild Metric & Satellite Orbits: A Question

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TrueBlue1990
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Thanks in advance - this problem has been bothering me for a while!

I'm working with an unpowered spaceship orbiting a large mass M. The orbit is circular and it is following the geodesic freely. It has an orbit radies of r = R.

My question is this. The metric of the space-time curvature is the Schwarzschild metric:

ds^2 = - (1 - 2GM/(r*c^2) )*(c^2)dt^2 + [(1 - 2GM/(r*c^2) )^-1]*dr^2 + (r^2)dθ^2 + (r^2)*(sinθ^2)d∅^2 -(1)

But I keep seeing references to ds^2 = - (c^2)*d(tau)^2 -(2)

I understand the angular terms and dr disappear as the d(tau) means we are observing the orbit from the reference frame stationary with the satellite. But I can only how equation (1) = equation (2) in a Schwarzschild curved space if r (distance between mass and satellite) tends to infinity. But, in the case of the satellite's reference frame r=R.

I will be eternally grateful to anyone that can shed light on my error of understanding. Many thanks.
 
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I would like to answer but I have no idea what you mean. The circular orbit has

[tex]\frac{dt}{d\tau}=\frac{\sqrt{r}}{\sqrt{r-3\,m}},\ \ \frac{d\phi}{d\tau}=\frac{\sqrt{m}}{r\,\sqrt{r-3\,m}}[/tex] in case that helps.

Remember that the proper-length s in the metric is also cτ, so ds2= c22

[Edit]As Bill_K points out below this is true up to a sign, depending on the metric signature.
 
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Also r > 6m for orbits to be stable.
 
But I keep seeing references to ds^2 = - (c^2)*d(tau)^2
This is just a matter of notation. ds2 is called the spacetime interval. It's positive for a spacelike distance and negative for a timelike one. For a timelike interval it's more convenient to use in place of ds2 the proper time. So you pull out a minus sign, and pull out a factor of c2 to give it the right dimensions. Otherwise ds2 and dτ2 are interchangeable.