The
Shape of a Newtonian Orbit |

We'll be making some (very) slight use of Lagrangian mechanics, and we'll need to use a slightly clever substitution in order to integrate a differential equation, but aside from that it's mostly going to be a straightforward "equation grind".

(I'm going to refer to the thing being orbited as "the star" throughout this page.)

(1)

Potential energy is, of course,

(2)

and the Lagrangian is

(3)

Note, again, that we've taken K to be positive here. We now find the partials,

(4a)

(4b)

Since (4a) is

(4c)

(4d)

Putting together (4d) and (4c), and turning (4b) around, we get the equations of motion we'll need to solve:

(5a)
(5b) |

We're going to concentrate on (5a). The terms in 1/r are a little nasty, so the first thing we'll do is try to get rid of them. To do that we'll substitute

(6)

With

(7)

This doesn't look much better, but let's move on and see where we get. We've got time derivatives of

(8)

Substituting (8) into (7) leads to this:

(9)

Now we finally want to get rid of the explicit θ terms, which we do by using:

(10)

Substituting (10) into (9) leads to this:

(11)

Collecting terms we finally arrive at:

(12)

That's certainly solvable! Just to make it even easier, we substitute:

(13)

which leads to:

(14)

which has the obvious solution

(15)

We observe that θ

Substituting

(16) |

(17a)

In case (17a) the radius oscillates around the value m

Figure
2
-- Parabola, focus at origin: |

In case (17b) the orbit goes

(17c)

Finally, in case (17c) there is an entire forbidden sector; the radius is only positive and finite as long as

(18)

Note that the argument to arccos() in (18) is always

Figure
3 -- Hyperbola, focus at origin: |

(19)

We can use (19) to recast conditions (17) in terms of

An elliptical orbit:

(20a)

A parabolic orbit:

(20b)

A hyperbolic orbit:

(20c)

We can also see that the farthest point on an elliptical orbit -- the aphelion -- must be:

(21)

First we observe that

(22)

To reduce clutter we introduce

(23)

Substituting (22) and (23) into (16) and cross-multiplying, we obtain:

(24)

Shifting the last term to the left and the 1 to the right and then squaring, we obtain:

(25)

Collecting terms,

(26)

If

(27)

If we shift the origin by substituting

Figure 4
-- Parabola, touching origin: |

(28)

as shown in figure 4, we get the standard form

(29) |

Comparing this with equation (4) on our parabola focus page, we see that it's a parabola with the focus at

(30)

Looking back at (28), we see that, before we shifted the axes, the focus of the parabola was indeed at the origin.

(31)

Again, to put this in standard form, we shift the origin by the substitution

(32)

to obtain:

(33) |

Comparing with equation (8) on the ellipse focus page we see that, if N

(34) |

Figure
6 -- Hyperbola, origin between the foci: |

Confirming that the focus really is in the middle of the star is a little messier for the ellipse and hyperbola, and we'll do that in the next section.

To start with, we have too many "

(35)

Comparing (33) and (35) we equate the second terms:

(36)

And we equate the right hand sides:

(37)

Substituting (37) into (36) we obtain:

(38)

Multiplying out, we obtain:

(39)

Sum the fractions:

(40)

Take square roots and we're done:

(41)

That takes care of the elliptical case. We still need to do the hyperbolic case.

(42)

Comparing with (34), we equate the second terms:

(43)

Dividing the top and bottom of the right side by

(44)

We also equate the right hand sides of (42) and (34):

(45)

Combining (45) with (44) we obtain:

(46)

which was to be shown.