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9.6: Position in a Parabolic Orbit

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    When a “long-period” comet comes in from the Oort belt, it typically comes in on a highly eccentric orbit, of which we can observe only a very short arc. Consequently, it is often impossible to determine the period or semi major axis with any degree of reliability or to distinguish the orbit from a parabola. There is therefore frequent occasion to have to understand the dynamics of a parabolic orbit.

    We have no mean or eccentric anomalies. We must try to get \(v\) directly as a function of \(t\) without going through these intermediaries.

    The angular momentum per unit mass is given by Equation 9.5.28a:

    \[h = r^2 \dot{v} = \sqrt{2G \textbf{M} q}, \label{9.7.1} \tag{9.7.1}\]

    where \(v\) is the true anomaly and \(q\) is the perihelion distance.

    But the Equation to the parabola (see Equation 2.4.16) is

    \[r = \frac{2q}{1 + \cos v}, \label{9.7.2} \tag{9.7.2}\]

    or (see section 3.8 of Chapter 3), by making use of the identity

    \[\cos v = \frac{1-u^2}{1+u^2} , \quad \text{where} \quad u = \tan \frac{1}{2} v , \label{9.7.3a,b} \tag{9.7.3a,b}\]

    the Equation to the parabola can be written

    \[ r = q \sec^2 \frac{1}{2} v . \label{9.7.4} \tag{9.7.4}\]

    Thus, by substitution of Equation \ref{9.7.4} into \ref{9.7.1} and integrating, we obtain

    \[q^2 \int_0^v \sec^4 (\frac{1}{2} v) dv = \sqrt{2G\textbf{M} q} \int_T^t dt. \label{9.7.5} \tag{9.7.5}\]

    Upon integration (drop me an email if you get stuck!) this becomes

    \[u + \frac{1}{3} u^3 = \frac{\sqrt{\frac{1}{2}G \textbf{M}}}{q^{3/2}} (t - T) . \label{9.7.6} \tag{9.7.6}\]

    This Equation, when solved for \(u\) (which, remember, is \(\tan \frac{1}{2} v\)), gives us \(v\) as a function of \(t\). As explained at the end of section 9.5, if \(q\) is in astronomical units and \(t − T\) is in sidereal years, and if the mass of the comet is negligible compared with the mass of the Sun, this becomes

    \[u + \frac{1}{3} u^3 = \frac{π \sqrt{2}(t - T)}{q^{3/2}} \label{9.7.7} \tag{9.7.7}\]

    or \[3u + u^3 - C = 0, \quad \text{where} \quad C = \frac{π \sqrt{18} (t-T)}{q^{3/2}}. \label{9.7.8a,b} \tag{9.7.8a,b}\]

    There is a choice of methods available for solving Equation \ref{9.7.8a,b}, so it might be that the only difficulty is to decide which of the several methods you want to use! The constant \(\frac{1}{3}C\) is sometimes called the “parabolic mean anomaly”.

    Method 1: Just solve it by Newton-Raphson iteration. Thus \(f = 3u + u^3 − C = 0\) and \(f^\prime = 3(1 + u^2 )\), so that the Newton-Raphson \(u = u − f / f^\prime\) becomes

    \[u = \frac{2u^3 + C}{3(1 + u^2)}, \label{9.7.9} \tag{9.7.9}\]

    which should converge quickly. For economy, calculate \(u^2\) only once per iteration.

    Method 2:

    Let \[u = x-1/x \quad \text{and} \quad C = c-1/c. \label{9.7.10a,b} \tag{9.7.10a,b}\]
    Then Equation 9.7.8a becomes

    \[x = c^{1/3} . \label{9.7.11} \tag{9.7.11}\]

    Thus, as soon as \(c\) is found, \(x\), \(u\) and \(v\) can be calculated from Equations 9.7.11, 10a, and 3a or b, and the problem is finished – as soon as \(c\) is found!

    So, how do we find c? We have to solve Equation 9.7.10b.

    Method 2a:

    Equation 9.7.10b can be written as a quadratic Equation:

    \[c^2 - Cc - 1 = 0. \label{9.7.12} \tag{9.7.12}\]

    Just be careful that you choose the correct root; you should end with \(v\) having the same sign as \(t − T\).

    Method 2b:

    Let \[ C = 2 \cot 2 \phi \label{9.7.13} \tag{9.7.13}\]

    and calculate \(\phi\). But by a trigonometric identity,

    \[2 \cot 2 \phi = \cot \phi - 1/\cot \phi \label{9.7.14} \tag{9.7.14}\]

    so that, by comparison with Equation 9.7.10b, we see that

    \[c = \cot \phi . \label{9.7.15} \tag{9.7.15}\]

    Again, just make sure that you choose the right quadrant in calculating \(\phi\) from Equation \ref{9.7.13}, so as to be sure that you end with \(v\) having the same sign as \(t − T\).

    Method 3.

    I am told that Equation 9.7.8 has the exact analytic solution

    \[u = \frac{1}{2} w^{\frac{1}{3}} - 2w^{-\frac{1}{3}}, \label{9.7.16} \tag{9.7.16}\]

    where \[w = 4C + \sqrt{64 + 16C^2}. \label{9.7.17} \tag{9.7.17}\]

    I haven’t verified this for myself, so you might like to have a go.

    Example: Solve the Equation \(3u + u^3 = 1.6\) by all four methods. (Methods 1, 2a, 2b and 3.)

    Example: A comet is moving in a parabolic orbit with perihelion distance \(0.9 \ \text{AU}\). Calculate the true anomaly and heliocentric distance 20 days after perihelion passage. (A sidereal year is 365.25636 days.)

    Exercise: Write a computer program that will return the true anomaly as a function of time, given the perihelion distance of a parabolic orbit. Test it with your answer for the previous example.

    9.6: Position in a Parabolic Orbit is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.