3
$\begingroup$

Consider two sun-synchronous orbits with similar orbital elements, except for their times. For example 8h and 9h orbits. Their semi-major axes (a), inclinations (i), and eccentricities (e) are the similar and the orbits have no requirements of RAAN. (Though the rate of change of RAAN is same as they're both SSOs)

I am trying to find maneuvers that could bring a satellite from one of these orbits to the other. The way I look at it is that the satellite needs to catch up, with the other orbit, and shift its argument of perigee by ±15°. So that it reaches above a location, in ±1hr of the other.

The general approach I can think of is to change a, i or both so that the satellite goes into an intermediate orbit, and uses apsidal precession to achieve the ±1 hr orbit. I can see here how J2 affects the nodal and apsidal drifts. So I'm not sure how to define the intermediate orbit so that constraints on both ends are met.

Is there a prescribed way of solving such a problem, or maneuvers that are commonly used for this?

$\endgroup$
12
  • 1
    $\begingroup$ Are you trying to make the orbits intersect, or are you looking for a rendezvous? $\endgroup$
    – GdD
    Commented Apr 5, 2022 at 16:05
  • 1
    $\begingroup$ iirc, MRO (and other Mars assets) have done just this (shift ascending node LMST), there are some papers on the JPL TRS on their maneuvering $\endgroup$ Commented Apr 5, 2022 at 16:14
  • 1
    $\begingroup$ Plane changes (which is what you are asking about) are ridiculously expensive in terms of delta V. $\endgroup$ Commented Apr 5, 2022 at 17:06
  • 1
    $\begingroup$ @DavidHammen yes, plane changes are often very expensive. However, I was thinking that I could change the semimajor axis of the first orbit, changing its nodal drift rate, and when the required time is near, change the semimajor axis again to make it SSO. Does this make any sense or am I wrong in assuming that drift rate could be used to change SSO time? $\endgroup$ Commented Apr 6, 2022 at 10:03
  • 1
    $\begingroup$ changing the inclination will also change the drift rate $\endgroup$ Commented Apr 6, 2022 at 11:21

1 Answer 1

1
$\begingroup$

Using the following equation (from David Hammen's answer) for RAAN rate of change:

$$\dot\Omega = - \frac{3}{2} J_2 \left(\frac R p\right)^2 n \cos i$$

Rearranged a bit:

$$Const. = -\frac{3}{2} J_2 R^2, n \propto \frac{1}{T}, T \propto a^{3/2}, p = a(1-e^2)$$

$$\dot\Omega \propto \frac{1}{a^{3/2}} \cdot \frac{1}{a^2(1-e^2)^2} \cdot \cos i$$ $$\dot\Omega \propto \frac{\cos i}{a^{7/2}(1-e^2)^2}$$

Finding the partial derivatives w.r.t inclination and semi-major axis will reveal that for a ~polar orbit (and probably most orbits) it is best to change inclination to affect nodal drift rate:

$$\frac{\partial \dot\Omega}{\partial i} \propto -\frac{1}{a^{7/2}} \cdot \sin i$$

$$\frac{\partial \dot\Omega}{\partial a} \propto \frac{1}{a^{9/2}} \cdot \cos i$$

$\endgroup$
4
  • 1
    $\begingroup$ Calculus is scary, hopefully I've done this right $\endgroup$ Commented Apr 6, 2022 at 13:10
  • 1
    $\begingroup$ Wow thanks, this looks pretty solid! I'm going through the papers you previously pointed me to and they did infact change inclination, it makes sense. I will try this, and maybe some astro software too, to compare other effects. J2 affects drift of apsidal line as well and I think that will also change the time at which the spacecraft ascends across equator. I suspect using both drifts in the 'right' amounts will be cheapest in terms of dV. $\endgroup$ Commented Apr 7, 2022 at 7:57
  • 1
    $\begingroup$ Also could you please put a negative sign on the partial derivative wrt i? I can't edit since its just 1 character. $\endgroup$ Commented Apr 7, 2022 at 8:00
  • 1
    $\begingroup$ @pathfinder_EOS SSO's tend to be circular (e ~= 0) so argument of periapsis drift might not be a concern $\endgroup$ Commented Apr 7, 2022 at 11:12

Your Answer

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

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