Time dilation for the clock in the orbit

Suppose that we want to compute the total time dilation for a clock located in an orbiting satellite relative to the clock in our cell phone on the ground.

Consider two different approaches below.

1. Use special relativity and compute time contraction due to the relative velocity. Use approximation of general relativity in the Newtown limits and compute time expansion due to the less gravity and then find the total time dilation.

2. Don't use special relativity. Stick to the approximation of general relativity based on the symmetry and find Schwarzschild metric and the geodesic for the Earth limits. Find the time dilation assuming a relative velocity in the metric.

The question is:

Which of them are more justified and provide a better approximation? Are they equivalent? What happens when the relative velocity of the satellite is zero?

How good is the approximation in either of the two approaches above.

When we pick the second approach and use the Schwarzschild metric we get this equation:

$$dt' = \sqrt{1-\frac{3GM}{c^2r}}dt = \sqrt{1-\frac{3r_s}{2r}}dt$$

where $r_s$ is the Schwarzschild radius: $r_s = 2GM/c^2$.

Here we not only assume the asymptotic flat metric to measure $r$ but also switch to Newtown gravity when we want to cancel $v$:

$$v = \sqrt{\frac{GM}{r}}$$

So it appears that in the second approach there are many more approximation assumptions.

• You'll find that these calculations have already been done on this site. You could compare both calculations and see how much difference there is (not much!). – John Rennie Sep 23 '16 at 18:14
• For calculation (2) see: What is the correct formula for gravitational time dilation for a satellite in a circular orbit?. Actually that contains enough info for you to do calculation (1) as well. – John Rennie Sep 23 '16 at 18:17
• @JohnRennie can you how the equation is derived from the metric in the circular orbit case? The $\frac{3}{2}$ factor in particular. – user56963 Sep 23 '16 at 18:37
• I derive the equation at the end of this answer – John Rennie Sep 23 '16 at 19:24
• @JohnRennie I mean how to cancel the velocity $v$ without using the Newtown gravity and only in GR. The factor $\frac{3}{2}$ is there because we combine Newton gravity law into Einstein's General Relativity. If we stick to the GR the velocity should remain. Yet again it is not a four velocity. – user56963 Sep 23 '16 at 20:58