RE: FT: Carriers

To be more accurate, when you execute a deceleration burn in a circular
orbit, it puts you into an elliptical orbit with the point of your burn
at
the apogee (highest point of the orbit).  So you lose some altitude, but
pick up velocity as you drop, and at the perigee (lowest point of the
orbit) you have too much velocity for a circular orbit at that altitude.
If you don't change anything, you'll continue on the elliptical orbit
and
rise back up again to the point where you originally decelerated.
However, at the perigee you can decelerate again and maintain a lower
circular orbit.

To re-enter, you decelerate enough at the apogee so that elliptical
orbit
has a higher eccentricity (flatter ellipse) and the perigee is actually
within the atmosphere at which point you can use aero braking to lose
the
rest of your excess velocity.

To drop straight down, you need to cut all the forward velocity of your
orbit.

In any case, your most efficient use of thrust is pushing you in the
opposite direction of your orbital velocity, rather than thrusting down.

This is actually one of the flaws in orbitally dropped munitions (like
Thor from Renegade Legion).  It takes just as much work to drop
something
from orbit as it takes to get it up there (aside from the fact that on
the
way down you can use atmospheric drag to do some of the work).

Randy Wolfmeyer
Dept. of Physics
Washington University

On Mon, 11 Mar 2002, Brian Bilderback wrote:

> B Lin rote:
>
> >The carrier would be pointing backwards and downwards, effectively
> >decelerating the fighters relative to the carrier.  This would have
two
> >effects - it puts them in a lower orbit and slows them down.
>
> Makes sense
>
> >If the object is slower than orbital speed, it falls out of orbit. 
So by
> >decreasing speed you drop, by dropping you increase the required
orbital
> >speed and if you don't thrust forward you drop.
>
> Which is exactly what you want to do if you'ree trying to enter the
> atmosphere (referring to the fighters).
>
> 2B^2
>
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