Satellite Clocks: General & Special Relativity Time Variations

[Zack K]

Something that crossed my mind recently; I know that satellites have to adjust their clock due to their relativistic time variations in relation to us. I was wondering do they adjust their times in accordance to general relativity or special relativity or both? Or is their speed so insignificant since they are traveling way too slow for special relativity to be factored in. If the former is the case, how would one go about and add the time dilation due to special and general relativity? Is it just a simple addition of the time dilation due to both gravity and speed?

 

[phinds]

Something that crossed my mind recently; I know that satellites have to adjust their clock due to their relativistic time variations in relation to us. I was wondering do they adjust their times in accordance to general relativity or special relativity or both? Or is their speed so insignificant since they are traveling way too slow for special relativity to be factored in. If the former is the case, how would one go about and add the time dilation due to special and general relativity? Is it just a simple addition of the time dilation due to both gravity and speed?

GPS satellites need to account for -7 microseconds/day due to SR (motion) and +45 microseconds/day due to GR (gravity):
https://en.wikipedia.org/wiki/Error...tioning_System#Special_and_General_Relativity

Remember, Google is your friend.

 

[Ibix]

I know that satellites have to adjust their clock due to their relativistic time variations in relation to us. I was wondering do they adjust their times in accordance to general relativity or special relativity or both?

I think what you probably intend to ask is if there is a correction needed for gravitational time dilation and a separate correction for velocity-related time dilation. The answer is that it depends how you want to look at it. It is perfectly possible to just use GR tools to calculate the appropriate correction to clock rates for a clock in a particular orbit without considering it as two separate elements. So you could say "no". Alternatively, you could split the correction into a correction for gravitational time dilation and a correction for velocity-related time dilation (as I see @phinds has done) and handle them separately, in which case the answer is "yes" - although I would say that it's a mistake to call the former a GR correction and the latter an SR correction because that would imply that GR doesn't handle velocity-related time dilation. It does.

Note that satellites in general don't bother. It's only where a really high precision clock measurement is needed - i.e., the GPS system - that anyone makes such a correction.

 

[phinds]

... I would say that it's a mistake to call the former a GR correction and the latter an SR correction because that would imply that GR doesn't handle velocity-related time dilation. It does.

I agree. I was being simplistic for a first-cut answer.

 

[lomidrevo]

the correction is to be implemented in the user's device when the calculation of the position is being performed

 

[Janus]

Something that crossed my mind recently; I know that satellites have to adjust their clock due to their relativistic time variations in relation to us. I was wondering do they adjust their times in accordance to general relativity or special relativity or both? Or is their speed so insignificant since they are traveling way too slow for special relativity to be factored in. If the former is the case, how would one go about and add the time dilation due to special and general relativity? Is it just a simple addition of the time dilation due to both gravity and speed?

The equation for time dilation for a clock in circular orbit(as measured by a distant observer) is:
$$ T = \frac{t}{\sqrt{1-\frac{3GM}{rc^2}}}$$
where r is the radius of the orbit.
This could also be expressed as
$$ T = \frac{t}{\sqrt{1-\frac{2GM}{rc^2}- \frac{GM}{rc^2}}}$$
Orbital velocity for a circular orbit is:
$$ v = \sqrt{\frac{GM}{r}}$$
Thus
$$ \frac{GM}{r} = v^2$$

Substitute into the second equation and you get:
$$ T = \frac{t}{\sqrt{1-\frac{2GM}{rc^2}- \frac{v^2}{c^2}}}$$

For a non-circular orbit, you would have to add another term under the radical which contains the radial motion component of the orbital velocity at the point of the orbit you are interested in.