## Elliptical orbits: given time, find position

For a change, I decided to split two problems into two posts.

We have found time from position; now let’s reverse that and find position from time. The following problem is Example 2.1 on p. 31 of Prussing & Conway, “Orbital Mechanics” (see my bibliography page).

“An Earth satellite orbit has a semimajor axis a = 4R, and a perigee radius 1.5R, where R is Earth radius. Find the true anomaly at t=4 hours after perigee passage.”

OK, we have a and rp… and t:

ps = {a->4 R, rp->1.5 R, t->4 Hour}

In general, we might want to confirm that 4 hours is less than the period of the orbit. (Kepler’s equation is for a fraction of the orbital period – you’ll know that after we derive the equation in the next post. It requires that the time be taken modulo the period… we can’t wrap around the orbit multiple times.)

Let’s compute it. I never remember this, but I know where to find it: the period is

$T = 2\pi \sqrt{a^3/\mu}\$

and $\mu = G\ M$, so here’s G, M, and $\mu\$:

Oh, we’re going to need the semi-major axis a in physical units, and that means we need the radius of the earth…

(Either remember that Mathematica® supplies the answer is in meters, or check it, as I do. And I am accumulating values I compute in “sol”.)

Since we will need n, and

$n = \frac{2\ \pi}{T} = \sqrt{\mu/a^3}\$

and then

$T = 2\ \pi / n\$, so

Good: 4 hours is less than the period; we haven’t wrapped around.

Now let’s get the eccentricity, e = c/a… so I need c. For this, there’s no reason not to work in canonical units, where R = 1:

(I’m going to keep those solutions separate because they’re in different units. Mostly… time t is in physical units.)

The arithmetic mean of ra and rp is a, so write that… plug in rp and a… get ra…

Then I remember that b is the geometric mean of the apsides ra and rp:

It’s also carved into my brain that for elliptical orbits we have

$a^2 = b^2 + c^2\$,

(along with the similar but different Pythagorean Theorem) so I get c from a and b:

so the general equation for the orbit (which I don’t remember yet) is… and for our data it specializes to…

We have (M =) n t = E – e Sin E, and we know n, t, and E, so we solve for E:

All that’s left is to get from E to f.

Before we do that, let me remark that for small values of E, or for small values of e, and small values of t (i.e. near periapse) Kepler’s equation can be unreliable. In

n t = E – e Sin E,

for small values of t, the RHS becomes the difference of two very small quantities, and solving for E is difficult. Still, it seems adequate for simulations, and I’ll stay with it until I have to switch.

Here’s one of the alternative equations relating E and f — another thing I have to look up. It relates the tangents of the half-angles. It is particularly convenient because f/2 and E/2 are always in the same quadrant (first or second). I expect to show you all these in the next orbital post.

Convert from radians to degrees by dividing by “Degree”:

… and, yes, that’s the answer in the book.

And that’s all for now, folks.