# Why is not the $g$-field of the sun considered in the GPS-settings?

I have read this SE-Physics question and its answers, however, I could not find anything regarding the effect of Sun's $$g$$-field on the GPS clocks. When a GPS clock orbits earth, depending on whether it is in superior or inferior conjunction, the $$g$$-potential of the sun changes. In the traditional approach to the GPS-settings, which possibly proves GR, it is claimed that two kinds of time rates are involved:

1- Due to SR by factor $$\sqrt{1-v_{Sat}^2/c^2}$$, and

2- Due to GR by factor $$\sqrt{1-2GM_E/r_{Sat}c^2}=\sqrt{1-\Delta\phi/c^2}\approx 1-\Delta\phi/2c^2$$.

The key term in the above equations is the difference in the $$g$$-potentials having the value:

$$\Delta\phi=\frac{2GM_E}{r_{Sat}}=\frac{2×6.67×10^{-11}×6×10^{24}}{2×10^7}=4×10^7$$

However, if we consider the gravitational field of the sun, and if we consider the difference in the $$g$$-potentials, for when the satellite is between the sun and earth and when the earth is between the satellite and sun, we reach:

$$\Delta\phi=GM_S \left (\frac{1}{R_{SE}-r_{Sat}}-\frac{1}{R_{SE}+r_{Sat}} \right)\approx GM_S\frac{2r_{Sat}}{R_{SE}^2}$$

$$=6.67×10^{-11}×2×10^{30}\frac{2×2×10^7}{(1.5×10^{11})^2}$$

$$=3×10^5,$$

where $$R_{SE}$$ is the distance between the sun and earth. Although the recent value is nearly one hundredth the previous one, we can make them of the same order by assuming that the GPS clocks orbit the earth at slightly greater distances, say, $$r_{Sat}=2×10^8\space m$$ (ten times greater than the orbital height of today's GPS devices). Does this mean that if we want to send a GPS device at larger distances from the earth, we need to consider the gravitational potential of the sun as well? Is the obtained value for $$g$$-potential of the sun small enough that can be neglected as the GPS error?

• What is the impact of the Sun’s gravity on clocks on Earth? – Jon Custer Jul 26 '20 at 14:02
• @JonCuster It is about the difference in the $g$-potentials. This difference between the earth and the satellite, in a superior or inferior conjunction, is $$\Delta\phi=GM_S \left (\frac{1}{R_{SE}\pm r_{Sat}}-\frac{1}{R_{SE}} \right)\approx \mp GM_S\frac{r_{Sat}}{R_{SE}^2}$$, which has the same order as mentioned above. – Mohammad Javanshiry Jul 28 '20 at 7:03

OP is forgetting that the Earth is also moving (as opposed to being held still) in the gravitational field of the sun. So the potential difference from the Sun gravitational field must be compensated by the potential of Earth frame that is accelerating toward the Sun. As a result the influence of the Sun on the clocks of satellites are given by terms with second derivatives of the potential: $$\partial_i \partial_j \phi$$ and so the measurable potential difference from the Sun would be of order $$\Delta' \phi \sim \frac{G M_\odot r_\text{sat}^2}{R_\text{SE}^3 },$$ and the frequency shift from this effect would be of order $$10^{-16}$$ and so could be neglected (at the current level of precision).