I am studying low-thrust maneuvers, in particular a change of inclination only. According to (Ruggiero et al., 2011)[1], if one writes the perturbation force in the Gauss planetary equations in terms of yaw and pitch angles $\alpha$ and $\beta$, the expression for the change in inclination is as follows:
$$ \frac{\operatorname{d}\!i}{\operatorname{d}\!t}=\left|\vec{f}\right| \frac{r}{h} \cos{(\omega + \nu)} \sin{\beta} $$
(hence depending only on the out-of-plane angle)
If one further derives this equation with respect to $\beta$, arrives to this expression for the optimal out-of-plane angle for the maximum instantaneous change of the inclination:
$$ \beta = \frac{\pi}{2} \operatorname{sgn}\left(\cos{(\omega + \nu)}\right) $$
Therefore, we should change the direction of the thrust vector every half orbit.
My problem with this result is that, intuitively, I would expect a change after every crossing of the line of nodes, and therefore depending on $\sin{(\omega + \nu)}$. In this way I would think a net torque around the node would be exerted and hence the inclination should change. Instead, the change is shifted 90 degrees and I don't understand why.
Can somebody provide some physical explanation, simulation, or alternative derivation that helps me clarify why changing the direction in this way produces a net change of inclination?
[1]: Ruggiero, A., P. Pergola, S. Marcuccio, and M. Andrenucci. "Low-thrust maneuvers for the efficient correction of orbital elements." In 32nd International Electric Propulsion Conference, pp. 11-15. 2011.
examples
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