Robert Clark wrote:
....
Acceleration a= v^2/r, v the speed , r the radius for a body traveling
in the circle. So r = v^2/a. For v = 10,000 m/s and a = 40 m/s^2, r =
2,500,000 m, or 2500 km.
This is worse than going in a straight line at constant a = 40 m/s^2 :
s= v^2/2a = 1,250,000 m = 1250 km.


- Bob

However, we might be able to reduce the distance by using a
combination of a circle and a straight-line. You would use a circle to
build up to a certain velocity then travel in a straight-line for the
rest of the distance:

You gradually build up the speed going in the circle. As you are
buiding up speed in the circle the acceleration will be both along the
circle as well as the usual radial acceleration for a constant speed
around the circle. Since the total accleration has to be less than 40
m/s^2, you would keep the tangential acceleration small but the radial
acceleration close to but less than 40 m/s^2. Then when the speed
finally reaches the highest speed you want in the circle, you make the
acceleration be purely radial at 40 m/s^2. Then r = v^2/40.
As for the part of the trajectory in a straight-line, you would direct
this part radially inward into the circle you were just using. Then no
additional distance need be used as far as the distance from the launch
point is concerned, as long as this straight-line stays inside the
circle.
Then what we have is that the speed is v when going into the
straight-line. We want to build up the speed again at 40 m/s^2 to
10,000 m/s. Let's calculate the distance required to build up to a
speed given some initial speed. The formulas for constant acceleration
travel in a straight-line a

v = v0 + at, and
x = x0 +v0t +(1/2)at^2

For our scenario we can set x0 = 0. However, the v0 has to be the
speed we get from the circle portion of the trajectory, so we keep that
in the equations. Then t = (v-v0)/a and plugging this into the equation
for x we get: x = (v^2 - v0^2)/2a .
So if v is the initial speed you get coming from the circle, a = 40
m/s^2, and 10,000 m/s is the speed you want to reach, then x =
(10,000^2 - v^2)/80 is the distance it takes to reach the speed of
10,000 m/s. Now you want this distance to remain inside the circle, so
you want (10,000^2 - v^2)/80 = 2r = 2(v^2/40) , 10,000^2 -v^2 = 4v^2
, 10,000^2 = 5v^2, and
v^2 = (10,000^2)/5 . Now r = v^2/40 , and you want r as small as
possible so you take the smallest allowable v, which means v^2 =
10,000^2/5, and r = 500,000m = 500 km.
Then this would provide a shorter distance then traveling in a
straight-line alone.
...


This won't work, at least not as written. I'm trying to direct the
craft into the circle while using the velocity it attained in
travelling around the circle. However, this velocity vector around the
circle would be directed *tangentially* to the circle. So I couldn't
use this to give an additional boost radially into the circle.
What might do it is if this high speed took place while still in
significant atmosphere. Then we might be able to use lifting surfaces
on the craft to provide a velocity component perpindicular to the
direction of motion. Indeed the possibility of using lifting surfaces
would significantly increase the range of accelerations permitted
beyond that allowed by the beamed propulsion itself.


Bob Clark