Gravitational redshift with time dilation?

[EdisT]

Hello, thanks for reading this. I need help trying to find a way to calculate the time dilation due to gravity, from a satellite 50,000km above the surface of the Earth, traveling 10,000km/h relative to a stationary observer on Earth.

A signal is being sent from the satellite to the observer, the frequency of the signal is 1.7 gigahertz (1.7E9 Hz).. Now i calculated the change in frequency due to gravitational blueshift through the conservation of energy formula, but i don't know how to continue from there and turn the change in frequency to time dilation.

Any help is much appreciated! I would love to see the step by step calculations if you do decide to help, thank you!

 

[Dale]

Hi EdisT, welcome to PF

Assuming no radial motion then you can use this formula:
$$\sqrt{1-\frac{2U}{c^2}-\frac{v^2}{c^2}}$$

This formula gives the time dilation relative to a stationary clock at infinity (U=0), so you will have to do it twice, once for the ground observer and once for the satellite.

The derivation can be seen here:
http://en.wikipedia.org/wiki/Time_dilation#Time_dilation_due_to_gravitation_and_motion_together

 

[EdisT]

So for the ground observer it would be:
√(1-((2GMhf/c*r)/c^2) - 10000^2/2997924582^2

where m = hf/c

 

[Dale]

U=GM/r. m never enters into it.

 

[EdisT]

Ah my bad, but if the rest is correct then I would like to thank you very much!

 

[EdisT]

U=GM/r. m never enters into it.

one of the observers is relatively stationary, does that mean that for one of the calculations, v=0?

 

[Dale]

Yes.

 

[PeterDonis]

one of the observers is relatively stationary, does that mean that for one of the calculations, v=0?

Yes.

Actually, an observer who is "stationary" on Earth is not motionless for purposes of this calculation, because the Earth is rotating, and the formula DaleSpam gave is relative to a frame which is not rotating with the Earth. So the correct $v$ for the ground observer is his velocity due to the Earth's rotation (about 450 meters per second if he's on the equator, it gets smaller towards the poles and is zero at the poles).

 

[PeterDonis]

for the ground observer it would be:
√(1-((2GMhf/c*r)/c^2) - 10000^2/2997924582^2

As DaleSpam said, $U = GM / r$, the other factors shouldn't be there in the first term. In the second term, first of all, this looks like the $v$ for the satellite, not the ground observer ($v$ for the ground observer should be what I said in my previous post), and it's in the wrong units: in your OP, you said 10,000 km/h, but $v$ should be in meters per second since you gave $c$ in meters per second; you have to use consistent units throughout. (Also there seems to be an extra $2$ at the end of your value for $c$, $c$ should be 299792458 meters per second.)

 

[Dale]

an observer who is "stationary" on Earth is not motionless for purposes of this calculation, because the Earth is rotating

Yes, good point.