In a cartesian reference frame (O,X,Y,Z), put two circles centered at the origin O, of radius a. The blue circle is perpendicular to Z. The red circle makes an angle i with the blue (cf. Fig below).

enter image description here

Call $\Omega$ the angle between X and the line (O,R), R being one of the two intersections between the circles. The angles $\Omega$ and $i$ define the red circle uniquely in the reference frame. Without loss of generality, set $\Omega=0$ so that the line (O,R) is aligned with X.

Let’s assume a particle p is moving along the red circle with uniform angular speed $n=2\pi/T$, where $T$ is by definition the period in seconds. Without losing generality, let’s set time origin, t=0, when p is at position R for « the first time ». The cartesian coordinates of p as a function of time are: $$ p_x(t)= a \cos nt $$ $$ p_y(t)= a \sin nt \cos i $$ $$ p_z(t)= a \sin nt \sin i $$ Let’s put another particle p’ at the same position R at time t=0, and set its period T’ to smaller than T (p’ moves faster than p).

Call $\text{err}_x= p’_x – p_x$, and same for y and z axes.

If we plot the linear « errors » on the 3 axis with respect to time, we obtain three curves like this.

enter image description here

Now, recall that in his answer to his own Question Orbit propagation with J2..., @chia-jiun-wei provided plots that are very much similar to our curves. His plots compared a 2body orbit to an SGP4-propagated orbit, for a given set of parameters setting.

This similarity begs the following question:

Could it be that, for nearly circular near-Earth orbits, to a first-order approximation and within a certain time limit, the « errors » between the 2body and the SGP4 models can be described simply as the fact that one model makes the object move faster/slower than the other one?

Related to this problem, I had posted the following question: Is the formula for nodal period of a near Earth satellite given by Wikipedia correct?.

  • 1
    $\begingroup$ IMO, for point 2, there is no reconciliation required since 2 different angular velocity for the same radius circle means that the law of motion was not derived from the $\frac{1}{r^2}$ gravitational field, so doesn't fit the Newton's law to begin with. $\endgroup$
    – AJN
    Commented Nov 1, 2021 at 12:34
  • $\begingroup$ @AJN, I concede to this ( a perturbed gravitational field is by definition no longer the $1/r^2$ gravitational field) $\endgroup$
    – Ng Ph
    Commented Nov 1, 2021 at 13:58
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    $\begingroup$ As a starter: your two statements of "and set its period T’ to smaller than T" and "(p’ moves faster than p)." are in conflict with each other. the one, OR the other can be true, not both. For the period of the second orbit to be shorter, it must be a smaller ellipse than the circular orbit at the same altitude. Thus, it has to be moving slower at apogee, where it intersects the first circle. $\endgroup$ Commented Nov 1, 2021 at 14:16
  • $\begingroup$ @Pcman, if p' is increasingly ahead of p, then it must come back to R before p does. Hence T'<T and vice-versa. It is a mathematical model to see what we would observe in the plots of the differential coordinates, when 2 particles trace exactly the same path, but one is moving faster. Your objection would be correct it the two particles were moving under a common force potential. I didn't make such assumption in the description of the movements. Never mind my 2nd question which I think is resolved (by AJN's comment) $\endgroup$
    – Ng Ph
    Commented Nov 1, 2021 at 18:35
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    $\begingroup$ I would say that these sort of errors can be described as two "orbits" having components of different frequencies, but it is not limited to the case you describe (orbits in the same plane with different periods). If one orbits stays in the same plane while the other precesses (but has the same nodal period), you get similar errors for x- and y-components. $\endgroup$
    – Litho
    Commented Nov 1, 2021 at 19:21


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