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# 9.7: Position in a Parabolic Orbit

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

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 $w = 12C + \sqrt{64 + 144C^2}. \label{9.7.17} \tag{9.7.17}$

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 an elliptic 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.