3
$\begingroup$

According to (Burt, 1967)[1], the transverse component of the perturbing force (that is: contained in the plane of the orbit and perpendicular to the radius vector) is the only one that affects secular changes of both the semimajor axis and the eccentricity (eqs. (17) and (15)):

$$ \widetilde{\frac{\operatorname{d}\!a}{\operatorname{d}\!t}}=\frac{2 a^{3/2}}{\mu^{1/2}} \left(1 - e_0^2 \left( \frac{a_0}{a}\right)^{3/2} \right)^{1/2} f_2 $$

$$ \widetilde{\frac{\operatorname{d}\!e}{\operatorname{d}\!t}}=-\frac{3}{2} \left(\frac{p}{\mu}\right)^{1/2} e f_2 $$

On the other hand:

  1. Later in the same paper it's stated that tangential thrust $f_T$, which is effectively a combination of radial $f_1$ and transverse $f_2$ thrust, "leads to a slightly more rapid enlargement of the orbit".

  2. It is well known that the quasi-optimal strategy for eccentricity change involves exerting thrust along a inertially fixed direction in space perpendicular to the semimajor axis (Edelbaum 1961, Pollard 1997, Ruggiero 2011). Moreover, (Pollard, 1997)[2] states that the secular variation of the eccentricity with such a program is

$$ \left|\widetilde{\frac{\operatorname{d}\!e}{\operatorname{d}\!t}}\right|=\frac{3}{2} f \sqrt{\frac{a}{\mu}} \sqrt{1 - e^2} $$

which is very close to the Burt result except for an $e$ factor.

How do these results not contradict the fact that only transverse $f_2$ thrust produce secular changes on semimajor axis $a$ and eccentricity $e$? If $f_2$ is the only component producing secular changes, shouldn't the radial component present in $f_T$ and the inertially fixed thrust program be immaterial to the evolution of $a$ and $e$?

[1]: Burt, E. G. C. "On space manoeuvres with continuous thrust." Planetary and Space Science 15.1 (1967): 103-122.

[2]: Pollard, James E. "Simplified approach for assessment of low-thrust elliptical orbit transfers." 25th International Electric Propulsion Conference, Cleveland, OH. 1997.

$\endgroup$
7
  • $\begingroup$ Should $f$ in your third equation have a T subscript? $\endgroup$
    – Chris
    Jan 30, 2017 at 0:55
  • 1
    $\begingroup$ @Chris no, I copied it literally from Pollard - I didn't intend to be consistent with my previous notation. $\endgroup$ Jan 30, 2017 at 9:33
  • $\begingroup$ @uhoh I think I already know what they are (I double and triple checked) but obviously there's something I still don't understand, hence my question. Would you expand your knowledge into an answer? $\endgroup$ Jan 30, 2017 at 9:34
  • $\begingroup$ I'll just speculate until then. Is there a possibility of a single word error? Can you search for a published errata? Would changing "normal" to "tangential" in the first line of your question fix everything for example? Drag would be an example, see the bottom of page 6 here "In addition, we can see that the perturbing force due to drag will have an effect in both the semi-major axis and the eccentricity." $\endgroup$
    – uhoh
    Jan 30, 2017 at 9:58
  • 1
    $\begingroup$ Exactly - drag is just another example of what I want to understand, also producing secular changes . My question is more general that these particular examples. $\endgroup$ Jan 30, 2017 at 10:28

1 Answer 1

3
$\begingroup$

The formulas (15) and (17) are given under the assumption that both $f_1$ and $f_2$ (the radial and the transverse components of the perturbing force) stay "sensibly constant over one orbital period". In particular, it means that the angle between the force and the radius vector stays sensibly constant. The two cases you are talking about do not satisfy this assumption: if the force is tangential and the orbit is not circular, then the angle between the force and the radius vector depends on the body's position on the orbit. If the force is applied in an inertially fixed direction, the angle varies even more.

$\endgroup$
1
  • $\begingroup$ Spot on! That is exactly the assumption I was missing. Thank you very much! $\endgroup$ Jan 31, 2017 at 11:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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