Re: GPS calculations

From: jahn (suzysewnshow_at_yahoo.com.au)
Date: 09/29/04 

Date: Wed, 29 Sep 2004 06:32:58 -0400

"Henri Wilson" <H@..> wrote in message news:3aojl0d0l2usb2s61je64lplrgbkhvtlrh@4ax.com...
> On Tue, 28 Sep 2004 14:49:01 -0500, EjP <nospam@hackers.are.bad> wrote:
>
> >xxein wrote:
> >> For those confused by 46 us/day and 38 us/day, It depends on wether
> >> you think the satellites orbit once a day or twice a day.
> >>
> >> It seems there is a lot of confusion both on this ng and the general
> >> science archives.
> >>
> >> End.
> >
> >There is no confusion amongst those who know what they're talking
> >about. The correction is 38 usec a day, based on a perturbative
> >treatment which allows you to separate the "SR" and "GR" compoenents.
> >If you do this, you get a 45 usec advance from the gravitational
> >potential difference and a 7 usec slowing form the SR component,
> >giving a net correction of 38 usec. This was calculated *long*
> >before the satellites were launched and match observation exactly.
> >
> >Since the orbital period of GPS satellites is about 12 hours,
> >the correction per period is about 19 usec. I'm not sure
> >where your 46 usec comes from, except that it's close to the
> >45 that one gets from a partial calculation.
> >
> >The calculation is actually pretty simple. I've appended it
> >below. This version is a bit long because I was arguing with
> >some kook that claimed all sorts of "extra" factors had been
> >added, and I was showing that you can calculate the whole thing
> >if you know the altitude and period of the satellite and
> >the radius of the Earth.
> >
> > -E
> >
> >
> >===============Details of GPS Calculation========================
> >The altitude of the GPS sats are about 20,000 km, so their
> >radius is 27,000 km. The observed period is 12 hours. From this you
> >can work out that the velocity is
> >
> > v = 2*PI*R/T = 3.93 x 10^3 m/s
> >
> >This means that \beta = v/c = 1.31 x 10^(-5). The velocity at the
> >surface
> >of the earth is negligible compared to this, so the SR clock correction
> >would come from
> >
> > \gamma = 1/sqrt{1-\beta^2} ~ 1 + .5*\beta^2 ~ = 1 + (8.5 x 10^(-11))
> >
> >so the orbiting clock would tick "slowly" by 1/gamma, or a fractional
> >change of
> >
> > 8.5 x 10^(-11)
> >
> >and the clock would lose (86400)*(8.5 x 10^(-11)) = 7 x 10^(-6) s/day
> >
> >
> >Now, move on to GR. Use
> >
> > v = sqrt(GM/R) to get GM = 4.2 x 10^14
> >
> >The GR gravity-only time dilation "gamma" is (look it up)
> >
> > \gamma = 1/sqrt{1 - 2GM/(Rc^2)} ~ 1 + GM/(Rc^2)
> >
> >So the difference between a clock ticking on Earth (R=Re=6.7E6) and one
> >ticking in orbit
> >(R=Ro=27E6) would be
> >
> > 1+GM/(Re*c^2)
> > ------------- ~ 1 + (GM/c^2)(1/Re - 1/Ro)
> > 1+GM/(Ro*c^2)
> >
> >since the earth clock is deeper in the gravitational well, it would tick
> >slowly and relative to it, the GPS clock would tick fast by a factor
> >
> > (GM/c^2)(1/Re - 1/Ro) = 5.23 x 10^(-10)
> >
> >so the clock would gain
> >
> > 86400*(5.23 x 10^(-10)) = 45 x 10^(-6) s/day
> >
> >so the net gain would be 45 - 7 = 38 microseconds, which is (strangely
> >enough) EXACTLY the value they use, see
> >http://www.phy.syr.edu/courses/PHY312.98Spring/projects/GPS/GPS.html
> >
>
> So what is it about movement or gravity wells that actually affects the way a
> clock ticks?
>
> You CDEFs all talk crap.
>
> Just because light increases speed and incurs a doppler shift as it falls down
> a gravity well, you think 'clock ticks' do the same.
>
> Maybe PA's famous tick fairies are at it again....!
Yikes!
Do moving clocks tick slower?
Hint: Motion is relative.
<< Numerically integrating the satellite
motion equations relative to the center of the earth is done for
numerical reasons,where the equations that are integrated are
the formal difference (not a Lorentz transformation) of the
equations of motion of the GPS satellite and earth in the
barycentric frame.>>
http://www.draper.com/publications/digest03/paper203.pdf

Do photons Fall?
Hints:
You can't see an event before it happens.
The values of mu and epsilon for all known materials is greater than the value in a vacuum, so the speed of light has a maximum
value in a vacuum.
Are atomic clock affected by gravity?
Hint: The accuracy of Earth-bound cesium atomic clocks is limited by the length of time each cesium atom can be observed -- on
Earth, gravity quickly removes the atoms from the observation region. In the microgravity environment of space, each atom can be
observed for many seconds.

Kind regards,
Sue...
>
>
> HW.
>
> www.users.bigpond.com/hewn/index.htm

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