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Two orbits are given with the same orbital elements, except the argument of perigee (AoP). How to calculate the true anomaly (TA) of the intersection of the orbits?

Currently, I calculate it numerically, going through the [0,360] interval.

Definitions

AoP is the angle from the ascending node to the periapsis.

TA is the angle between the direction of periapsis and the current position of the body, as seen from the main focus of the ellipse.

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The problem is to find the points at which two orbits with the same semi-major axis length, eccentricity, right ascension of ascending node, inclination but different arguments of periapsis intersect. Measuring angle from the ascending node, this means one wants to find the angles $\theta$ for which $r_1(\theta)=r_2(\theta)$ where $$r_1(\theta)=\frac{a(1-e^2)}{1+e\cos(\theta-\omega_1)}\\r_2(\theta)=\frac{a(1-e^2)}{1+e\cos(\theta-\omega_2)}$$ where $a$ is the semi major axis length, $e$ is the eccentricity, and $\omega_1$ and $\omega_2$ are the different arguments of periapsis for the two orbits. Assuming eccentricity is not zero, this reduces to solving $\cos(\theta-\omega_1)=\cos(\theta-\omega_2)$ for $\theta$. Naively taking the inverse cosine of both sides yields $\theta-\omega_1 = \theta-\omega_2$. This has no solutions if the two arguments of periapsis $\omega_1$ and $\omega_2$ differ.

Another route toward a solution is to recognize that cosine is an even function, which suggests $(\theta-\omega_1) + (\theta-\omega_2) \equiv 0 \pmod {2\pi}$. This has two solutions for $\theta$ in the range $[0,2\pi)$. One is the angle that bisects the two arguments of periapsis, $\frac12(\omega_1+\omega_2)$. The other is diametrically opposed to this bisector. The solutions $\theta$ are measured from the ascending node. The true anomalies are found by subtracting the arguments of periapsis from these solutions.

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  • $\begingroup$ If I have 310 and 270 degrees AOP. the θ values will be 110 and 290? And then I should convert θ to true anomaly? $\endgroup$ – Leeloo Nov 19 '18 at 19:02
  • $\begingroup$ @Leeloo - Correct. $\endgroup$ – David Hammen Nov 19 '18 at 19:04
  • $\begingroup$ Is it correct, that (taking θ=110) TA values will be 20 and 60? ω1+θ and ω2+θ $\endgroup$ – Leeloo Nov 19 '18 at 19:12
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    $\begingroup$ Incorrect. You need to subtract the arguments of periapsis, not add them. $\endgroup$ – David Hammen Nov 19 '18 at 19:28

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