9.6: Position in a Parabolic Orbit

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.