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Physics LibreTexts

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=r2˙v=2GMq,

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=2q1+cosv,

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

cosv=1u21+u2,whereu=tan12v,

the Equation to the parabola can be written

r=qsec212v.

Thus, by substitution of Equation 9.7.4 into 9.7.1 and integrating, we obtain

q2v0sec4(12v)dv=2GMqtTdt.

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

u+13u3=12GMq3/2(tT).

This Equation, when solved for u (which, remember, is tan12v), gives us v as a function of t. As explained at the end of section 9.5, if q is in astronomical units and tT is in sidereal years, and if the mass of the comet is negligible compared with the mass of the Sun, this becomes

u+13u3=π2(tT)q3/2

or 3u+u3C=0,whereC=π18(tT)q3/2.

There is a choice of methods available for solving Equation 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 13C is sometimes called the “parabolic mean anomaly”.

Method 1: Just solve it by Newton-Raphson iteration. Thus f=3u+u3C=0 and f=3(1+u2), so that the Newton-Raphson u=uf/f becomes

u=2u3+C3(1+u2),

which should converge quickly. For economy, calculate u2 only once per iteration.

Method 2:

Let u=x1/xandC=c1/c.
Then Equation 9.7.8a becomes

x=c1/3.

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:

c2Cc1=0.

Just be careful that you choose the correct root; you should end with v having the same sign as tT.

Method 2b:

Let C=2cot2ϕ

and calculate ϕ. But by a trigonometric identity,

2cot2ϕ=cotϕ1/cotϕ

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

c=cotϕ.

Again, just make sure that you choose the right quadrant in calculating ϕ from Equation 9.7.13, so as to be sure that you end with v having the same sign as tT.

Method 3.

I am told that Equation 9.7.8 has the exact analytic solution

u=12w132w13,

where w=4C+64+16C2.

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

Example: Solve the Equation 3u+u3=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 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.


This page titled 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 the style and standards of the LibreTexts platform.

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