GPS time measurement is not affected by Earth’s Orbital motion around sun. Reply

This is an original research work of Manmohan Dash, published here, 30 November 2011, 1.20 pm.

Why orbital motion of earth around sun, does not matter to GPS time measurements?

GPS-IIRM

Photo Credit; wikimedia

Because, it is not only earth, the satellite is also marching in a circle around the Sun and the Sun is simply too far.

Sun and earth distance is 1 AU ~ 149,597,871 kms — 15 karod kms, mere pyare bhaion aur behenon.

Earth-Satellite distance is about 26,000 kms = 0.000174 AU ~2 milli AU.

Photo Credit; Bibalex.org

Photo Credit; Bibalex.org

The angular speed of earth and satellite, due to Sun, will be almost same; w = r^(-1.5) = sqrt (1/ r.r.r), what difference it makes, if we change r, by 0.0002, when r is 1, the result is 1 changes to 0.9997.

So, if the angular speed of earth, is 1, the angular speed of satellite, assuming it is 26000 kms closer to sun-center, would be 0.9997. Thats a difference in the 4th digit. Now angular speed of earth itself is of 10^-5 order. So, this is 10^-9 order change. Why do we worry when earth spin itself is 1% of static effect, which is 1000 times at-least, smaller, than 1 nanosecond.

In other words, Sun’s effect, will be slightly more or less, between earth and satellite, by 10^-14, a digit in the 14th figure. If we adjust our digits, it will come to order 13 or go to order 15. So, we can safely say, the effect of Sun because of earth and satellite orbiting about Sun is, < 10^-12 nano-second = 10^-21 seconds. Anything at 10^-8 seconds is important.

We knew this distance very closely, even in 1672, Cassini computed Sun-earth to be 14 Karod kms = 140 million kms, the truer, as we know today, is, 15 karod kms = 150 million kms.

Tyco Brahe measured: 8 million kms, Kepler measured 24 million kms, but Cassini’s was approximately correct 140 million kms, read the NASA account here.

Photo Credit; Tyco Brahe.

Photo Credit; Discovery.com, Tyco Brahe.

Photo Credit; Bigthink.com

Photo Credit; Bigthink.com

cnrs.fr photo credit, Cassini.

cnrs.fr photo credit, Cassini.

Note that: w = (G.M_sun)^0.5 / r^1.5 = sqrt (G.M_sun / r.r.r). so v^2 = v.v = G.M_sun / r — where M_sun is large, but r is 1 AU. v.v is what gives the order of time, which is 1 / r, so r = 1 and r = 0.99983, have, a difference of 0.00017. That is, the first change in time, appears in the 4th decimal place — between earth and satellite.

We have to put the angular speed due to orbit around Sun, in the right order, before we can see, the order of time — so I am reviewing the above, but seems, if not 10^-21, it is at-least 10^-16 and only 10^-8 seconds matter.

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