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# Relativity and Cosmology

This is a Blog on relativity and cosmology for engineers and the like. You are welcome to comment upon or question anything said on my website (http://www.relativity-4-engineers.com), in the eBook or in the snippets I post here.

Comments/questions of a general nature should preferably be posted to the FAQ section of this Blog (http://cr4.globalspec.com/blogentry/316/Relativity-Cosmology-FAQ).

A complete index to the Relativity and Cosmology Blog can be viewed here: http://cr4.globalspec.com/blog/browse/22/Relativity-and-Cosmology"

Regards, Jorrie

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### Orbiting Clocks Puzzle

Posted October 29, 2007 12:00 AM by Jorrie

We know from experience that accurate atomic clocks on-board the space station (ISS) tick slower than the same clocks on Earth. Uncorrected ISS clocks lose about 25 μs per day to the ground clocks. This is due to the time dilation caused by the orbital velocity and the gravitational time dilation at the height of the ISS, averaging at about 330 km above sea level.

However, the clocks on-board the GPS satellites tick at exactly the rate as clocks on Earth, because they have been adjusted (rate corrected) to do so. The puzzling thing is that if they were not rate corrected, they would have gained 38 μs per day on the ground clocks. The GPS satellites are at about 20 thousand km above sea level.

How is it that uncorrected ISS clocks lose time and uncorrected GPS clocks would gain time relative to ground clocks? If this is true, there must be an intermediate height where an uncorrected orbiting clock runs at the same rate as a ground clock. Any idea how high that is?[1][2]

Jorrie

[1] If you want to make a precise calculation, see equations 1.16 and 1.17 in the technical article (pdf) 'Introducing engineers to relativity', down-loadable from my web page: What is Relativity? If you have the eBook, it's on page 25.

[2] Solution posted in reply #12 below.

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#1

### Re: Orbiting Clocks Puzzle

10/29/2007 4:21 PM

Yes, it has to do with the ratio of orbital velocity versus proximity to the Earth's center of mass.

The ephemeris data should be available for the ISS and any GPS SV on the web (we use the USAF database for the GPS SVs), but the long and short of it is:

GPS SV Altitude above geoid = ~20,000 km

GPS SV Orbital Velocity = ~14,000 km/hr

ISS Altitude above geoid = 330 km

ISS Orbital velocity = 27,744 km/hr

While the velocity is nearly 2X for the ISS, the altitude above the geoid is 60 times lower than the GPS SV.

I don't have time to plug in the numbers, but the two different orbital parameters for the two SVs demonstrates the effects of special and general relativity and their dominance for those parameters.

So, Special Relativity is concerned with the relative velocities of two observers (i.e., Earth and the SV). General relativity is concerned with the proximity to Earth and the effect of gravity.

The faster we go, the slower the clock gets compared to a "stationary" observer's clock. The closer the clock is to the Earth, the slower the clock runs.

So, the GPS orbital altitude dominates over the Special Relativity effects of the SV's velocity. The converse is true for the ISS, where proximity to Earth combines with Special Relativity's effects to slow the clock down.

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#2

### Re: Orbiting Clocks Puzzle

10/30/2007 5:31 AM

You concluded with: "So, the GPS orbital altitude dominates over the Special Relativity effects of the SV's velocity. The converse is true for the ISS, where proximity to Earth combines with Special Relativity's effects to slow the clock down."

The final sentence may be a bit confusing to others, because it may be interpreted that both velocity and gravity slows the ISS clocks down relative to the ground.

Any guess on the altitude for earth-orbit clock rate equality?

Jorrie

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#3

### Re: Orbiting Clocks Puzzle

10/31/2007 12:36 PM

Essentially, setting the equation for general relativity equal to the equation for special relativity and solve for v should give you the orbiting velocity where they cancel out.

sqrt(1-(2GM/Rc^2)) = sqrt(1-v^2/c^2)

However, doesn't that yield the escape velocity of the Earth (11.2 km/s)? So I don't think there is an orbit where this happens, right?

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#4

### Re: Orbiting Clocks Puzzle

10/31/2007 1:32 PM

Hi Hero.

No, your equation is not quite applicable. One must find what happens at geoid level (~terrestrial clocks) and what happens at orbit level clocks, each with height and velocity effects and then equate the equations for the clocks.

Jorrie

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#5

### Re: Orbiting Clocks Puzzle

10/31/2007 3:09 PM

Opps. I forgot to normalize for the reference point!

I also did not consider the Earth's rotation, which adds Special Relativity effects of about 138 ns per day (clock runs slower by 138 ns).

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#6

### Re: Orbiting Clocks Puzzle

10/31/2007 5:23 PM

Hero, you wrote: "Opps. I forgot to normalize for the reference point!"

The simplest reference point to use is, interestingly enough, a distant observer at rest with respect to Earth. Then the equations become simple Schwarzschild and one can easily equate them and solve for radial distance to obtain height.

Jorrie

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#7

### Re: Orbiting Clocks Puzzle

10/31/2007 8:08 PM

I tried that and I think I did something wrong with the equation because the numbers I got were 4 to 5 times larger than expected. Not sure what I did wrong.

If I get a little more time I will try to play with it tomorrow.

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#8

### Re: Orbiting Clocks Puzzle

11/01/2007 12:00 PM

Hi Hero.

One don't need the full relativistic equations for the low gravity and low speed orbits around Earth. A nice simplifying approximation is this:

1/dt ≈ 1 - GM/r1 - ½v12/c2 = dτ2/dt = 1 - GM/r2 - ½v22/c2,

where dτ is the time interval near Earth and dt is the time interval for a distant static observer.

Multiply both sides by 2c2, cancel out and we have:

2GMc2/r1 + v12 = 2GMc2/r2 + v22

For Earth, GMc2 = 3.48 x 10^31. Take r2 as Earth's radius and v2 as the equatorial rotational speed. Now the trick required is to solve the left hand side for r1. (I'll leave a little challenge).

Jorrie

PS: it seems that few engineers have the time and energy to struggle with this 'esoteric' stuff. We are a down-to-earth lot, it seems. Look at the response to the floor tiles challenge...

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#12

### Re: Orbiting Clocks Puzzle: Solution

11/06/2007 1:34 AM

I wrote: "One don't need the full relativistic equations for the low gravity and low speed orbits around Earth. A nice simplifying approximation is this:

1/dt ≈ 1 - GM/r1 - ½v12/c2 = dτ2/dt = 1 - GM/r2 - ½v22/c2,

where dτ is the time interval near Earth and dt is the time interval for a distant static observer.

Multiply both sides by 2c2, cancel out and we have:

2GMc2 /r1 + v12 = 2GMc2/r2 + v22"

Those c2's were actually erroneous 'red herrings' and should be ignored (a dimensionality analysis shows up the error!). The equations should simply be:

1/dt ≈ 1 - GM/r1 - ½v12 = dτ2/dt = 1 - GM/r2 - ½v22

and

2GM/r1 + v12 = 2GM/r2 + v22,

because GM/r already has the dimension m2/s2.

The final 'trick' is to realize that v12 is the orbital velocity squared: vo2 = GM/r1, giving:

3GM/r1 = 2GM/r2 + v22

from which is easy to see that the required radius from Earth's centre of mass (c.o.m.) is:

r1 = 3GM/(2GM/r2 + v22)

where v2 is the surface velocity due to Earth's rotation at the location of the clock and r2 is the distance of that clock from Earth's c.o.m.

Jorrie

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#9

### Re: Orbiting Clocks Puzzle

11/01/2007 12:09 PM

I think my numbers are right!

Altitude = 3189 km

Velocity = 6456 m/s

That would be a circular orbit.

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#10

### Re: Orbiting Clocks Puzzle

11/01/2007 12:31 PM

Yep, close enough! (It seems we happened to post more or less at the same time.)

Did you consider the Earth clock's movement due to the rotation of Earth?

I got some 3170 km for an equatorial clock, but it depends a bit on what geoid radius one assumes and so on.

Let's see if someone else bites on this cherry...

Jorrie

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#11

### Re: Orbiting Clocks Puzzle

11/01/2007 1:18 PM

"PS: it seems that few engineers have the time and energy to struggle with this 'esoteric' stuff. We are a down-to-earth lot, it seems. Look at the response to the floor tiles challenge...

That's too funny!

I did consider both GR and SR for all positions. After my first effort to try to solve for both equations equal to zero in a previous post, I then did this in a hack way and I just regrouped and slogged through it bit by bit to make sure I wasn't missing something. In the end what I did was created a spreadsheet in Excel and the resulting columns were altitude and the sum of the GR and SR effects. I could see where the convergence was pretty easily once I found the error of my ways with the GR calculation. I am a visual type of guy.

I really need to improve my math skills. Unfortunately, I lost all my first year college text books for physics and calculus. I saved all my other books for the remaining years, but switched my math/physics major to premed, then at the very end of that (after a 4.0 average) I just punted and joined the workforce in the engineering field! ;-)

I keep telling myself that when things slow down a little I will pick up some of the more advanced math studies to round out my knowledge a little more. Then again, there are so many other subjects of interest that compete that it is hard to get anything done!