starthaus
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Yes, this is correct. Here is a more precise (but with same results) calculation:Passionflower said:A scientific formula is right or wrong there is no middle way. I certainly would appreciate it if my formulas or calculations were found wrong, that's the way to learn, you apply, make mistakes, and hopefully someone else takes the trouble of telling you you are right or wrong.
To make things easy let's consider the simplest case a satellite on a circular polar orbit and a clock at one of the poles on Earth. The time ratio is:
\frac{\tau_S}{\tau_E} = 1+\frac{GM}{Rc^2} - \frac{3GM}{2rc^2}
In geometric units it is even simpler:
\frac{\tau_S}{\tau_E} = 1+\frac{M}{R} - \frac{3M}{2r}
G = 6.6726E-11
M = 5.9742E+24 kg (0.004435407 m)
R = 6378000.1 m
So the clock in the satellite is faster if r >\frac{3}{2}R > 3189 km otherwise it is slower.
Start with the Schwarzschild metric for dr=d \theta =0
(cd\tau)^2=\alpha (cdt)^2-(r d\phi)^2
valid for your satellite
and:
the Schwarzschild metric for dr=d \theta =d \phi=0
(cd\tau)^2=\alpha (cdt)^2
valid at the pole
where \alpha=1-2m/r
and m=\frac{GM}{c^2}
\frac{d \tau}{dt}=\sqrt{1-2m/r-(r \omega/c)^2}<1 for any r
The above gives:
\frac{d \tau_s}{d \tau_p}=\sqrt{\frac{1-2m/r-(r \omega/c)^2}{1-2m/R}}
So:
\frac{d \tau_s}{d \tau_p}>1 if
2m/r+(r \omega/c)^2<2m/R
But, from the Kepler's law, it can be shown that :
(r \omega/c)^2=m/r
so, the above becomes:
3m/r<2m/R i.e., your condition r>\frac{3R}{2}
The only problem I see with your proof is that : \frac{3R}{2}=3*6400/2=9600km :-)
Now, GPS satellites are moving at about 20,000km above the Earth, so , if it weren't for the frequency precompensation at launch, their clocks would be faster than the ones left on Earth.
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