@Krumple,
Krumple wrote:Yes, but why does everyone keep ignoring the fact that time adds upon itself? Microseconds would go on to become seconds, then minutes, then hours then days, so on and so forth.
For ~ a year at that altitude I was estimating. Sure, if you stay at the top of a tower for long enough you'll see a noticeable lag. A year isn't going to do it.
Krumple wrote:
In the first instance you have a total of 20 micro seconds.
In the second instance you have a total of 10 micro seconds.
This is a constant for altitude right? So as you continue with time the micro seconds would add upon themselves accordingly. This is what people neglect with the math.
Uh... no, it doesn't work like that. It's not 20 microseconds at bottom of tower = 10 microseconds at top. For a simple, perfectly spherical system, it's more like:
t = t0 SQRT(1 - 2GM/rc^2)
t0 is time measured in zero gravity, t is measured at the distance r from the centre of the Earth which has mass M. G and c are constants. So if t1 is time measured at the top and t2 at the bottom of the tower:
t1^2 = t0^2 (1 - 2GM/r1c^2) => t0^2 = (t1^2)/[1 - 2GM/r1c^2]
t2^2 = t0^2 (1 - 2GM/r2c^2) => t0^2 = (t2^2)/[1 - 2GM/r2c^2]
=> (t1^2)/[1 - 2GM/r1c^2] = (t2^2)/[1 - 2GM/r2c^2]
=> t1^2 = t2^2 c^2 r2/r1 ([r1c^2 - 2GM]/[r2c^2 - 2GM])
t2 = 2t1 when t2^2 = 4t1^2, giving:
r1 = r2 [8GM/(3r2c^2 - 2GM)]
G = 6.674 x 10^-11
M = 5.974 x 10^24
c= 3 x 10^8
giving:
r1 = 2.5 x 10^20 r2.
So in the Earth's gravitational field, 1 microsecond at the top will correspond to 2 microseconds at the bottom when the tower is approximately 250 000 000 000 000 000 000 metres high.