What's a decent formula for the average distance of an orbiting body, as a function of periapsis, apoapsis?

The answer is a(1 + e^2/2), where a is the semi-major axis and e the eccentricity.

That is calculated in here: http://scitation.aip.org/getabs/servlet/GetabsServ... which you can find in a library.

Online the formula is used in this article: http://iopscience.iop.org/0004-637X/592/1/555/full... see formula 8.

So for an extreme example as Halley's comet, e= 0.93, the time averaged distance is 50% larger than the semi-major axis.

Assuming an inverse square central field (force) acting on the bodies orbiting...

f(r) = -k/r² k = Gm1m2

given the differential equation of the orbit of a mass moving under a central force is:

d²u/dθ² + u = - [/mh²u²][f(u^-1] where u = 1/r for a central force and h is the angular momentum per unit mass. h = L/m

then the equation above becomes:

d²u/dθ² + u = k/mh²

A solution is:

u = A cos(θ - θo) + k/mh²

Constants of integration A and θo set up the initial conditions, so let θo = 0

and we get subsituting u = 1/r that

r = 1/(Acosθ + k/mh²)

This is the polar equation of an orbit and can be written as:

r = ro [1 + e]/(1 + e cosθ) where e = eccentricity and is equal to Amh²/k

and ro = mh²/k(1 +e) when θo = 0

Given the constant e is the eccentricity then for

e < 1 --> the orbit is an ellipse

e = 0 --> the orbit is a perfect circle

e = 1 --> the "orbit" is parabolic (never comes back)

e > 1 --> the "orbit" is hyperbolic (never comes back)

So long as e is zero or <0 then you should be able to use the perimeter of a circle or an ellipse to calculate the distance traveled. After that all bets are off! :^)

I remember using this stuff to calculate "TLI" for the Saturn V space program. TLI = trans lunar injection, the burn they did to increase their circular orbit to an ellipitical orbit with the moon at apogee.

We found that if the burn provided 1% too much energy, the near stationary circular orbit about the earth would go from circular to hyperbolic, meaning they'd never come home! Of course they could correct any such factor, but it showed how delicate this stuff is.

Unfortunately, the arc-length of an ellipse is an incredibly difficult problem to solve. In fact, there is no closed-form solution.

Here is a short formula that gives a numeric approximation if you know the two "radii" of the ellipse (semimajor and semiminor axes a and b).

http://www.efunda.com/math/areas/EllipseGen.cfm

For any eccentricity of 93% or less, the most error this formula could result is a 5% error.

Most orbits of the solar system, both planets and moons, are much less eccentric than this.

orbiting bodies dont actually orbit their primaries, thats a common mistake.... both bodies orbit the barycenter located "halfway" between the sum of their masses.

Time is only EVER a factor of density