Re: [GZG] Orbital mechanics (was List-iquette (was Ping))

[Roger Burton West <roger@xxxxxxxxxxxxx>]

On Mon, May 29, 2006 at 11:37:49AM -0500, Doug Evans wrote:
>Crap, prolly over my head, but using the example of Earth to Mars, where we
>accept that there's a close-enough-for-government-work ratio of one to two
>for respective years, how would it be best to visualize the possible paths?
>Assuming constant, handwavium-amounts of thrust, of course.

If you can do that, you can pretty much ignore solar gravity completely.
Let's assume an Earth-to-Mars trip on the simple profile:

(1) go from Earth's orbital speed to standstill.
(2) go directly from Earth to Mars.
(3) accelerate to Mars' orbital speed.

This is obviously not the most efficient method in many cases, but it's
a useful approximation.

Stage 1 needs a delta V of 29.8km/s; stage 3 needs a delta V of
24.1km/s. (Orbital speed = sqrt (G * mass of primary / radius).) For a
1g ship, that's a total of about 1.5 hours' thrust. That's small enough
to be more or less ignorable.

A constant-thrust burn over distance S, with acceleration A and a speed
of zero at each end, takes T=sqrt(4*S/A). For the Earth-to-Mars trip,
that's between 49.7 hours and 109 hours depending on the relative
position of the planets. Even in the worst case, therefore, you're
looking at less than five days of travel time.

If you drop the acceleration to 1/4g, you double the stage 2 time and
quadruple the stage 1 time. You're still looking at a maximum of less
than ten days. Neither Earth nor Mars moves so far in that time that
it's worth taking their orbits into account.

The complicated trajectories are the really low-thrust ion-drive types
of orbit, where you've got 1/100g or less for an extended period.

>Also,  I was considering allowing prograde or 'standing still' movement in
>orbit, with the image of the ship 'standing on it's head or tail' with
>relation to the center of the orbit. How much handwavium is necessary to do
>so compared to changing orbits?

Not quite sure what you mean here. Prograde means "in the direction of
the orbit". Do you mean the sort of thing I've been talking about above,
"hovering" with zero orbital speed? That's really no problem at all -
once you've burned down to that speed, all you need to do is exert
enough thrust to counteract solar gravity, and even a lightsail is
capable of that - all you need at 1 AU is 0.0006 gravity.

Does this answer the questions you were asking? :-)

R
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