GPS SV velocity mismatch

Question

When using classic Newton's second law we have

$$ \vec{F} = m \vec{a} $$

$$ \vec{F} = G \frac{M m}{r^2}{e_{r}} $$

$$ \vec{a} = \left( \frac{d^2 r}{dr^2} - r \left( \frac{d \phi}{dt} \right)^2 \right) \hat{e_{r}} $$

Because eccentricity of a GPS orbit is around e = 0.005 I believe we can treat GPS orbits as circular and write

$$ G \frac{M m}{r^2} = m r \left( \frac{d \phi}{dt} \right)^2 $$

Replacing angular velocity

$$ \frac{d \phi}{dt} = \frac{v}{r} $$

finally we get

$$ v = \sqrt{\frac{G M}{r}} $$

Now, GM ($\mu$) taken from GPS specification is $3.986005 * 10^{14} [m^3/s^2]$, $r$ taken from decoded ephemeris is about $26414660 [m]$. All this gives us

$$ v = 3885 [m/s]. $$

But the same ephemeris data, when decoded according to GPS Interface Specification, Table 30 (SV Velocity) gives values around

$$ v = 2774 [m/s] $$

for SV velocities.

At the beginning I thought that my procedure to calculate SV velocity is wrong. But later on found on the web (https://www.gps.gov/technical/icwg/meetings/2019/09/GPS-SV-velocity-and-acceleration.pdf) some other calculations where SV velocity is in the range 2600 to 3100 [m/s] with avarage 2880 [m/s]. And the question is. Can someone explain why I get such big difference when using classic Newton's law (3885 m/s) and when using ephemeris data (2774 m/s)? Most likely my understanding is wrong, but still cannot figure out what is wrong.

Satellite position, velocity and acceleration as decoded from ephemeris can be found here https://github.com/lukasz-wiecaszek/gr-gnss/blob/master/examples/kepler.dat

Answer 1

Let's go ahead and make a full answer from the comments.

Both the satellite and the reference frame are rotating from west to east. So you have to take your orbital velocity from Newtonian mechanics and subtract the velocity corresponding to the rotation of the reference frame. This reference frame velocity is not that of the surface of the Earth, but what would be obtained if the rotation were projected radially outward to the satellite's orbit. Since the radius of the orbit is four times that of Earth, the projected reference frame speed will be four times as fast as the speed on the ground, and that is enough to create a significant impact.

Let us compute the speed at two points in the orbital, first at the peak latitude which matches the orbital inclination and then at the Equator.

For the first computation, a simple scalar computation is sufficient because both the satellite and the reference frame are heading due east. Wikipedia gives an inclination angle of "approximately $55°$", and Earth's sidereal rotation period is $86164\text{ s}$, so (remembering that the reference-frame rotation velocity is proportional to the cosine of the latitude) we have

$|v|=3885-(2\pi×26414660/86164×\cos 55°)=2780\text{ m/s}.$

For the equatorial case, we need to take vector components because the satellite us heading in a northeasterly or southeasterly direction while the reference frame us still moving due east. We take vector components to the east and then to the north or south:

$v_e=3885\cos55°-2\pi×26414660/86164=+302\text{ m/s}$

(the satellite is still heading eastwards in the rotating frame)

$v_{n/s}=3885\sin55°=3182\text{ m/s}.$

So

$|v|=\sqrt{302^2+3182^2}=3197\text{ m/s}.$

This corresponds roughly to the range of speeds quoted in the pdf file. Examining the file we find it uses a slightly higher altitude ($26.56\text{ Mm}$ versus $26.41\text{ Mm}$ in the problem statement) and a slightly lower peak latitude, both of which would lower the relative velocities somewhat. Also the pdf file accounts for the orbital eccentricity. As with the calculation above, the relative speed is faster at the Equator than at the peak latitude due to the non-alignment of satellite and Earth-rotational directions.