Gravitational effects on orbits

Allan and I were discussing this offlist and realized there had
been an oversight in a formula, so the numbers given previously
were not correct.  Here are the correct numbers and how to get
them:

>From basic physics (ie the class we took 15 to 20 years ago) we
recall the equation for linear acceleration: d=1/2 at^2.  That
"1/2" is the part that usually gets omitted.  In this case, a =
10m/sec^2 (one gravity, close enough) and t=900 seconds (15
minutes in the default scale), and d works out to about 4000km or
4mu.

The Earth's radius we have determined to be approximately 6mu
(6.375 to be picky, but we'll call it 6).  So at 6mu from the
center, the acceleration is 4mu/turn.  Gravity drops off as the
square of the increase of distance, so for a radius R from the
center of the planet, the gravitational effect is 4mu/turn *
(6/R)^2 .  Thus at 12 mu radius, the gravitational effect is 4 *
(6/12)^2 =1mu/turn.  7mu radius is close to 3mu/turn effect, and
8.5mu radius is about 2mu/turn effect.

Now, the other part of the problem is that if you make the
endpoint of movement the determining factor, you'll tend to
crash.	Let's say you swoop in and end your turn at 7 mu radius.
No matter how fast you're going and no matter what the direction,
gravity grabs you and pulls you 3mu towards the center of the
planet.  7mu (your altitude) - 3mu (gravity effect) abruptly
meets 6mu (planetary radius) with a severe detrimental effect on
your hull.  Let me say it again--no matter how fast or what
direction you're going in, you crash.

Traveler took the critical point to be the midpoint of the move
instead of the endpoint, and I suspect that may work better.