Orbiter-Forum Integrating the satellite position
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 07-17-2018, 09:41 PM #1 cristiapi Orbinaut Integrating the satellite position If I integrate the orbital speed (magnitude of the velocity vector) of a satellite as a function of time, I get the distance traveled. If I integrate the satellite position as a function of time, what do I get? Is there any physical meaning of that number?
 07-17-2018, 10:41 PM #2 boogabooga Bug Crusher If the are no maneuvers, you get 0, with the physical significance that the trajectory is symmetric about the major body.
 07-17-2018, 10:47 PM #3 cristiapi Orbinaut If I integrate the magnitude of the position of SFERA 2 (no engines) as a function of time (in seconds) for exactly 1 orbit I get 36974363.45. What does that number represent?
 07-17-2018, 10:55 PM #4 Nicholas Kang Spaceflight News Reporter Is this what you are looking for?
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 07-17-2018, 11:06 PM #5 cristiapi Orbinaut Good! Thank you. I also found this link: http://wearcam.org/absement/examples.htm but is not clear to me how the absement (very strange word for me!) can be applied to the satellite motion. In MKS units that number is 36974363455 m*s, what does it mean? EDIT: according to this link: http://wearcam.org/absement/Derivati...splacement.htm "-1st derivative (integral) of position is absement", SFERA 2 is absent from the start of the orbit for a duration of 36974363455 / 3600 = 10270656,52 hours (?) Last edited by cristiapi; 07-17-2018 at 11:13 PM.
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 07-18-2018, 11:46 AM #7 cristiapi Orbinaut I was wrong in saying that SFERA 2 is absent from the start of the orbit for that duration because I integrate the radius vector and not the displacement (the in-track distance). If I google for "integral of the radius vector", I get just 1 result where I see "saying that the integral of the radius vector in the ellipse, with respect to the eccentric anomaly as independent variable, gives the area swept out.", but when I click the link, that text is not showed. Probably the integral of the radius vector as a function of time is just a meaningless number, but it seems that if I integrate as a function of eccentric anomaly, I get the area swept out.
 07-20-2018, 01:48 PM #8 Nicholas Kang Spaceflight News Reporter I am not very familiar with both integration of vectors and Kepler's equation. (Don't be surprised. I am still in my Pre-U studies and will only enter university September/October next year :p The good news is I will be studying for a degree in Physics. ) I think the text is right. (Do you mind sharing the link with me? Thanks) After all, when you derive the formula for 2-body problem, it is the polar equation of conic sections that contains the magnitude of the radius vector, r and the true anomaly, . So, if you treat as the radius vector, I think you should still be able to integrate a vector wrt to a scalar. (I think you are good at that.) That said, if you integrate the magnitude of the radius vector, you should get a constant straight line graph for mag(radius vector) against time graph. Hence, integrating mag(radius vector) wrt time should give you a linear graph with no y-intercept. I don't see any significant physical interpretation in the absement for satellite motions though. If it were significant, I am pretty sure we would have come across it in Pre-U/undergraduate courses.
 07-20-2018, 09:37 PM #9 cristiapi Orbinaut Now I see two results: https://www.google.com/search?q=%22i...ient=firefox-b this is the first: https://www.sciencedirect.com/scienc...83665675901397 Quote: Originally Posted by Nicholas Kang  I don't see any significant physical interpretation in the absement for satellite motions though. If it were significant, I am pretty sure we would have come across it in Pre-U/undergraduate courses. I surely agree. My best wishes for your studies!
 07-21-2018, 12:31 AM #10 Nicholas Kang Spaceflight News Reporter Uh, I only see one result now. Never mind. I don't think we can access the article in Science Direct unless we pay for it. Quote: Originally Posted by cristiapi  I surely agree. My best wishes for your studies! Thanks!
 08-14-2018, 09:42 AM #11 perseus Addon Developer I think that it has no meaning in terms of value of defined physical magnitude, It would have the meaning of multiplying the space traveled by a mobile object by the time used in said displacement. # ~ =. e * t or the same magmitud that speed by time squared

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