5
$\begingroup$

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)?

Improve this question
$\endgroup$
7
  • $\begingroup$ 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. $\endgroup$ – Cristiano Aug 15 '18 at 23:11
  • 1
    $\begingroup$ Solar radiation pressure depends on the mass, surface area, and reflectivity of the satellite. What values did you use? $\endgroup$ – cms Aug 16 '18 at 1:40
  • $\begingroup$ @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). $\endgroup$ – Alex Johnson Aug 16 '18 at 2:34
  • $\begingroup$ @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) $\endgroup$ – Alex Johnson Aug 16 '18 at 2:46
  • $\begingroup$ 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. $\endgroup$ – Cristiano Aug 16 '18 at 14:43
3
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$ – uhoh Aug 16 '18 at 13:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.