Ever wonder how astronomers can figure out a comet's orbit in three dimensions so quickly, even though you can't necessarily see how far the comet is at first glance? It turns out that the mathematics behind it are elementary, but complex. By that I mean that nothing beyond the first weeks of trigonometry is required, but the derivation is not all that easy to follow.

We will begin by assuming that our observations can be reckoned as if they were taken from the sun. As long as the comet is still far away, this will be a harmless enough assumption for us to obtain an approximate orbit.

We will also assume that the comet is approaching the sun on a
parabolic orbit. Technically, Kepler's first law of planetary motion
states that all planets and gravitationally bound comets travel in
ellipses, but the long-period comets that venture into the inner
solar system are so eccentric (*e* very close to 1) that they
are for practical purposes parabolic (*e* equal to 1). In fact,
some of them are perturbed sufficiently that they end up with a
hyperbolic orbit (where *e* is greater than 1).

Kepler's second law of planetary motion states that a planet or comet sweeps our equal areas in equal times. Another way of saying the same thing is to say that the comet's angular velocity is inversely proportional to the square of its distance from the sun. For example, if it is twice as far, it must cover one-fourth the angle per unit of time, in order to sweep out an equal area. In algebraic terms, we can write this as

where *r* is the comet's distance from the sun,
*w* is the comet's angular velocity, and *K* is some constant
which we have to determine. Now, consider our three observations.
They will reveal that the comet is not moving at a constant angular
velocity, and this is the key to the enterprise. If it were travelling
at a constant angular velocity, then it would have to be orbiting the
sun in a circle, and that is counter to our original assumption. Instead,
it will gradually be increasing its angular velocity, and the rate at
which it does this will indicate how fast the comet is approaching, and
determine the parameters of its orbit.

Suppose that the first observation *O*_{1} is taken
at time *t*_{1} = 0 and at position angle
*ß*_{1} = 0, and the second observation
*O*_{2} is taken at time *t*_{2} and at
position angle *ß*_{2}. We can then obtain a
*derived observation* *D*_{1.5} at time
*t*_{1.5} at position angle *ß*_{1.5},
where

In other words, we pretend that we saw the comet at a place exactly
halfway between *O*_{1} and *O*_{2}, at
a time exactly halfway between as well. Similarly, if we say that the
third observation *O*_{3} is taken at time
*t*_{3} at position angle *ß*_{3},
we can obtain a second derived observation *D*_{2.5} at
time *s*_{2.5} at position angle *b*_{2.5},
where

We note that the average angular velocity between the first two observations is

and the the average angular velocity between the second and third observations is

and finally we also observe that the angular separation between the two derived observations is just

If we denote by *r*_{1.5} and *r*_{2.5}
the distances of the comet from the sun at the moments of these two
derived observations, then taking into account our first equation, we
can write

w_{2.5} / w_{1.5} | = | (K / r^{2}_{2.5}) /
(K / r^{2}_{1.5}) |

= | (r_{1.5} / r_{2.5})^{2} |

Consider now the triangle made by the sun *S* at one corner,
and our two derived observations *D*_{1.5} and
*D*_{2.5} at the others. If we denote by *d* the
linear distance between the two derived observations, the law of
cosines give us

Dividing both sides of this equation by
*r*^{2}_{2.5}, we get

(d / r_{2.5})^{2} | = | 1 + (r_{1.5} / r_{2.5})^{2} -
2 (r_{1.5} / r_{2.5})
cos ß_{D} |

= | 1 + u - 2 sqrt (u) cos ß_{D} |

where *u* = *w*_{2.5} / *w*_{1.5}.
Now, the component of the motion along
*D*_{1.5} *D*_{2.5} that is
"sideways" is given by *ß _{D}r*

v | = | sin^{–1}
(ß_{D}r_{2.5} / d) |

= | sin^{–1} (ß / sqrt (1 +
_{D}u - 2 sqrt (u) cos ß) )_{D} |

From simple algebra, we know that the angle which *SO*_{2}
makes with *D*_{1.5}*D*_{2.5} is

Finally, the segment *D*_{1.5}*D*_{2.5}
is approximately parallel to the parabola at *O*_{2}, and
using the geometry of the parabola, we can determine that the perihelion
point of the comet's orbit is located at position angle

ß_{q} | = | ß_{2} + pi -
2v_{adj} |

= | pi - 2v |

Now, we must determine the actual perihelion distance *q*.
We know that the perihelion is separated from the observation
*O*_{1} by an angle of *pi* - 2*v*.
We can find the actual distance *r*_{1} at
*O*_{1} by solving the simultaneous equations

This can be solved using the quadratic equation and some simple trigonometric identities to yield

The *distance proportion p* is the ratio between the distance
at *O*_{1} and the perihelion distance *q*, and
is given by

p | = | sqrt (x^{2} + y^{2}) |

= | y sec 2v |

at the proper intersection solution point defined by the line and the parabola. Again, using Kepler's second law, we know that

w_{q} | = | K / q^{2} |

= | K / (r_{1} / p)^{2} | |

= | p^{2} (K /
r^{2}_{1}) | |

= | p^{2} w_{1} |

where *w*_{1} is the derived angular velocity at
*O*_{1}

By equating observed centripetal acceleration with the predicted acceleration due to gravity, we can write

from which we can easily derive

where *G* is the universal gravitation constant, *M*
is the mass of the sun, and cbrt is the cube root function. The orbit
is now completely determined, with the perihelion at distance *q*,
coplanar with the three observations, located at the angle
*ß _{q}*.

Here is an approximate but simple way of accounting for the earth's
offset in computing the orbit. Using the parabolic orbit determined
above, find the predicted location of the comet in three-dimensional
space at the three observation points, *O*_{1},
*O*_{2}, and *O*_{3}.

Next, determine the earth's position relative to the sun at the time
of the observations, and imagine an *anti-earth E'* such that the
sun is precisely between the earth and *E'*. Compute the new
angles *ß'*_{1} = 0, *ß'*_{2} =
*O*_{1}*E'O*_{2}, and
*ß'*_{3} =
*O*_{2}*E'O*_{3}. Redo the orbit
determination with the new angle values, and remember to reckon the
orbital axis to the new *ß'*_{1}. The new orbit
should be approximately corrected for the earth's offset.

Copyright (c) 2000 Brian Tung