gonegahgah said:
If you had a tall tower on the equator and the tower had a clock at its base and a clock at the top would those clocks travel through time differently too each other?
Let's say for example that one clock is at 1r from the centre of the Earth and the other is at 2r - just to get a significant distance between them..
The clock at 1r would be at 1g and the clock at 2r would be at 1/4g (approximately).
The clock at 1r would also be traveling at 1s while the clock at 2r would be traveling at 2s.
Would both these SR+GR effect the relative time for both clocks?
What effect would they have?
Although as others have pointed out, we shouldn't call these SR+GR effects, we can split the time dilation into velocity and gravitational potential components. For a spherical non rotating massive body you can use:
t = t_0 \sqrt{1-\frac{v^2}{c^2}} \sqrt{1-\frac{2GM}{rc^2}}
where v is the instantaneous tangential velocity as measured by a non rotating observer at altitude r. The Earth is not exactly spherical and neither is it non-rotating, but even so, the above formula is good enough to work out the time dilation of GPS satellite clocks (for example) to a reasonable degree of accuracy.
For a very tall tower, it is easy to see that if it was tall enough the velocity at the top of the tower would be approaching light speed and this places a limit on how tall the tower can be, for anybody with non-zero rotation. This indicates that the velocity component that increases the time dilation with increasing altitude, overwhelms the gravitational time dilation which decreases with increasing altitude.
For a small orbiting object the the orbital velocity is given by v=\sqrt{(GM/r)} and the above time dilation formula can be written in terms of altitude only as:
t = t_0 \sqrt{1-\frac{GM}{rc^2}} \sqrt{1-\frac{2GM}{rc^2}} = \sqrt{1-\frac{3GM}{rc^2}+\frac{2G^2M^2}{r^2c^4}If I recall correctly, it works out that whether or not a given satellite clock ticks faster than a clock on the surface of the Earth, depends on the altitude of the satellite.
P.S. Short answer to your question is... yes
If you use the first formula, plug in the values for G, c, M (the Mass of the Earth), r (radius of the Earth at the equator) and v (the velocity at the equator of the Earth) to obtain the time dilation for a clock at the base of the tower. For the clock at the top of a tower of height h, replace v with v*(r+h)/r.