Integrating the satellite position

[cristiapi]

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?

 

[boogabooga]

If the are no maneuvers, you get 0, with the physical significance that the trajectory is symmetric about the major body.

 

[cristiapi]

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?

 

[Nicholas Kang]

Is this what you are looking for?

 

[cristiapi]

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/Derivatives_of_displacement.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 (?)

 

[Nicholas Kang]

I didn't know there is such a thing called absement either. When I ask my Physics teacher what is the integral of displacement, she replied that there is no such thing as it does not seem to have any common applications, that is, until you ask the same question and I decided to Google it.

(I ask here about the derivative of acceleration and she doesn't seem to know jerk either.)

So, apparently my Physics teacher is wrong.

That said, I couldn't think of any physical representation of absement.

And @cristiapi, the unit for absement should be m*s. Assuming that your satellite is in a circular orbit, and assuming a 2-body problem, the linear speed of your satellite should be constant, so

[MATH]s(t) = vt[/MATH]
Integrating both sides wrt to t, let [MATH]a(t)[/MATH] denote the absement function of the satellite.

[MATH]a(t) = \frac{1}{2}vt^2[/MATH]
I think your absement function is a quadratic one if your orbital speed is linear. The duration in which the object is absent should be the same as your original domain for the [MATH]s(t)[/MATH] function, i.e. the duration for which you calculated the linear distance covered by the satellite. So, the domain for [MATH]a(t)[/MATH] shouldn't be simply dividing the absement by [MATH]3600\ s/h[/MATH].

 

[cristiapi]

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.

 

[Nicholas Kang]

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 The good news is I will be studying for a degree in Physics. :lol

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, [MATH]\theta[/MATH].

[MATH]r = \frac{a(1-e^2)}{1-e\cos\theta}[/MATH]
So, if you treat [MATH]r[/math] 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.

 

[cristiapi]

Now I see two results: https://www.google.com/search?q="integral+of+the+radius+vector"&ie=utf-8&oe=utf-8&client=firefox-b

this is the first: https://www.sciencedirect.com/science/article/pii/0083665675901397

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!

 

[Nicholas Kang]

Uh, I only see one result now.

:hmm:

Never mind. I don't think we can access the article in Science Direct unless we pay for it.

I surely agree.
My best wishes for your studies!

Thanks!

 

[perseus]

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