I am using GMAT to propagate satellite motion. At first, I excluded all the propagators besides harmonics(of second degree and 0th order, as shown in the picture):

enter image description here

Here are the results(x,y,z coordinates) of propagating with the time interval of 10 days from the 668 km height:

1207.4141854713, 193.09221015360, 6947.8246951043

Then I added the solar radiation pressure(Spherical model):

enter image description here

Here are satellite coordinates after that:

1207.4150240009, 193.09806743665, 6947.8324711391

After that I excluded the radiation and added relativistic effects, as demonstrated here: enter image description here

New coordinates are provided below:

1207.4097579535, 193.11627451516, 6947.8248222515

The difference between method was estimated by calculating the distance:

$$r_{crad}=\sqrt{(x_{c}-x_{rad})^2+(y_{c}-y_{rad})^2+(z_{c}-z_{rad})^2},$$ $$r_{crel}=\sqrt{(x_{c}-x_{rel})^2+(y_{c}-y_{rel})^2+(z_{c}-z_{rel})^2},$$

where $x_{c},y_{c},z_{c}$ are "clear" coordinates of satellite(i.e. without relativity or radiation), $x_{rad},y_{rad},z_{rad}$ and $x_{rel},y_{rel},z_{rel}$ are "radiation" and "relativistic" ones, respectively. The results of difference estimation: $r_{crad}=$9.8 m,$r_{rel}=$24 m. That means that relativistic factors affect more than solar radiation. I would like to know, does it make sense? And what can be a reason of it, if yes? Can this reason be the height of the satellite(the value of 668 km means that the satellite is in termosphere)?

  • If you post the initial state and the time, I can propagate it with my integrator to see the effect of the relativistic acceleration. I would suggest you to use a more updated gravity model up to degree and order 10 or 15, or try at least degree > 2. – Cristiano Aug 15 at 23:11
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    Solar radiation pressure depends on the mass, surface area, and reflectivity of the satellite. What values did you use? – cms Aug 16 at 1:40
  • @Cristiano, @cms the initial values are -976.3107644649057, -4835.627052558522,-5031.728586125443,-0.7944031487816871,5.474532271429767,-5.094496750907486. Dry mass is equal to 850 kg, the coefficient of relativity - 1.8, SRP area - 1 $m^2$ while drag area is 15 $m^2$(I understand that this value makes little sense but it was given by GMAT, and I will experiment on its changing). – Alex Johnson Aug 16 at 2:34
  • @Cristiano, I have done as you offered(with order and degree 10), but no significant changes in the answer(approx. 10 m and 24 m) – Alex Johnson Aug 16 at 2:46
  • You probably solved the problem, but I post my results, just for comparison. I used: ECI J2000, J2, atmosphere, Sun, Moon and planets. You didn't say the initial time; for MJD= 58331 with relativistic acceleration I get {1226.40563426209, 103.573958084106, 6946.3459010849}; {1.42246382070419, -7.36991301670154, -0.143032319925665}, w/o: {1226.40886395264, 103.557250595093, 6946.34557093811}; {1.42246066885988, -7.36991329522474, -0.143050152062709}; the distance is 17 m. – Cristiano Aug 16 at 14:43
up vote 3 down vote accepted

The reason was SRP area if satellite. If you change it to 10 m, $r_{crad}$ will be 106 meters, and being set to 15 m, it provides 158 m difference.

  • 1
    Very nice! It's always okay to post answers to your own question, and to accept them as well. I usually wait a few days to see if any further answers are posted, then click the best one, which certainly could be your own answer. – uhoh Aug 16 at 13:18

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