Scott T. Jensen wrote

Please bear with me on this. I'm trying to understand the mechanics
of what
you're saying.

Don't try to understand orbital mechanics using the normal concept of
speed and distance, like you know them from everyday travelling
situations. Orbital mechanics are often counter-intuitive in this
regard.

If you want something in circular orbit around earth to enter the
atmosphere, you must change its velocity so that it enters a elliptical
orbit where the lowest point is inside the atmosphere, say at 100km just
to keep the numbers simple.

The most effecient way of doing this (i.e. requiring lowest change of
speed) is to "break" by altering the velocity directly opposite the
current velocity wrt to earth. If you are in a normal prograde
equatorial orbit the orbital velocity vector is pointing east and you
therefore need to change the velocity towards west, which means you want
to shoot your rifle towards west if you want the bullet to go low.

However, beside direction, your bullet also need a certain minimum speed
in order to be able to reach the atmosphere. The higher the orbit the
more speed is required. For instance, at 300km altitude the bullet only
needs around 60 m/s, but at 36000 km altitude (GEO) the required speed
is around 1500 m/s. So depending on the gun, it should be able to
re-enter a bullet at least up to some fairly high altitude, but probably
not so high as geostationary altitude.

The math is as follows: If you are in a cirular orbit at altitude h, you
are r = h + R from the center of earth (with R = 6378 km) and you are
moving with respect to the center of earth at a speed of

v = sqrt(k/r),

where k = 398601 km^3/s^2 is the gravitational constant for earth (also
called GM). Now you want to change your orbit into an elliptical one
with the highest point at radius r, and the lowest at s = 100km + R. In
such an orbit you would at the highest point need an orbital speed of

w = sqrt(k*(2/r-1/s))

which means that you must change your speed with

dv = v - w

in order to change from the circular orbit to the elliptical orbit. If
you use the numbers as I have written them above, then the speed unit
will be km/s.


Regards,
--
Filip Larsen