## 6 Propagation Time for a Signal, Emitted and Intercepted by Satellites on a Space-Oriented Orbit
### Numerical Calculation of the First \(O(\frac{1}{c})\) Correction in the Propagation Time
The relativistic observables, defined for the \(GRAIL\) mission in [6], are also accurate to 1 \(\upmu\)m. No doubt that the numerical calculation of the integral over the Shapiro delay term in (128) will give a more realistic result. Propagation Time of the Signal Compared to the Celestial Time--Consistency of the Numerical Calculation  
Let us compare the calculated propagation time of the signal with the celestial time \(t_{cal}\), which can be calculated from \(M=n(t-t_{P})\) (\(M\) is the mean anomaly, given in the numerical data), where \(t_{P}\) is the initial time of perigee passage. The idea is to calculate the celestial time of motion for the given value \(M\) of the mean anomaly and, thus, to compare the distance travelled by the satellite with the distance travelled by the light (laser or radio)signal. Since this calculation, also, will be approximate, we have taken only the value \(T_{1}^{(E_{6})}\) for the propagation time from the integral (123) and (125):  
\[T_{1}^{(E_{6})}=\frac{a}{c}\int\limits_{0}^{E_{(6)}}\sqrt{1-e^{2}\cos^{2}E}dE=-0. 0281342485273829\;\text{[s]},\]  
which does not account for the second term, responsible for the Shapiro delay. Let us take the ratio of the propagation time to the celestial time:  
\[\text{propagation time for electromagnetic signals}\;=\;\frac{0.0281342485273829\; \text{[s]}}{37.508256148\;\text{[s]}} \tag{130}\]  
\[=0.0007500814864964887. \tag{131}\]  
In other words, the propagation time for the signal is about \(10^{4}\) times smaller than the celestial time, which will mean that for the celestial time \(37.5082561\;\text{[s]}\), the satellite moves 145,125 [km] (the velocity of the satellite is taken to be \(v_{S}=3.874\;\text{[km/s]}\)) and the light signal will move at a distance of  
\[26,558.1510169176263505735093268096\;\text{[km]}. \tag{132}\]  
This value is obtained after multiplying the light velocity \(c=299,792.458\;\;\text{[}\frac{\text{km}}{\text{s}}\text{]}\) with the calculated propagation time, taking into account the sixth iteration \(T_{1}^{(E_{6})}\) according to (125). It is interesting to compare this value for the distance travelled by the signal with the value, obtained after taking only the first term in (128), for the propagation time, where the term \(\frac{a}{c}\sqrt{1-e^{2}\cos^{2}E}\) has not been integrated. This term does not account for the second term for the Shapiro delay, which contained 10 zeroes after the decimal dot and was of the order 0.299 ns (nanoseconds). The result will be  
\[26,558.151025899950855259224944504\;\text{[km]}. \tag{133}\]  
This number is greater (after the fourth digit after the decimal dot) than the preceding number (132), so the integration of the expression (125)  
\[\frac{a}{c}\int\limits_{0}^{E_{(6)}}\sqrt{1-e^{2}\cos^{2}E}dE\]  
in fact diminishes the propagation time (only the first \(O(\frac{1}{c})\) correction) in comparison with the first term \(\frac{a}{c}\sqrt{1-e^{2}\cos^{2}E}\) in the expression (128). The difference between the numbers (133) and (132) is  
\[0.0000089823245046857156176944\;\text{[km]}\approx 0.898\;\text{[cm]}\; \approx 8.98\;\text{[mm]}. \tag{134}\]  
In both cases of the numerical calculation of the numbers (132) and (133), the numerical result confirms that the distance travelled by the light signal (numbers (132) and (133)) is much greater than the distance of 145.125 [km] travelled by the satellite. Respectively, for the calculated propagation time \(0.0281342485273829\;\text{[s]}\) (only the first \(O(\frac{1}{c})\) correction \(T_{1}^{(6)}\) (125)), the satellite will move at a distance of  
\[\frac{145.125\;\text{[km]}}{37.5082561\;\text{[s]}}\times 0.0281342485273829\;\text{[s]} \tag{135}\]\[\frac{a}{c}T_{1}^{(E_{6})}=-0.0281342485273829[\mathrm{s}],\]0  
This distance is considerably smaller than the distance travelled by the signal.