Well, "outer space" and the moon are not the same. In one case you have near zero gravity and a vacuum, and on the other hand you are subjected to nearly normal gravitational forces (1/6 earth G) as opposed to nearly 0 G in space. 1/6 sounds like a small number, but compared to zero, which you cannot divide by, it's a nearly infinite increase in G as compared to "outer space".
So there are three questions:
1: How does a
projectile act in 0 G?
2: How does a
projectile act in reduced G?
3: How does a
firearm and it's
cartridge act in a vacuum?
So, let's examine each:
In zero G: In space, a projectile would travel indefinately, slowed only by a few small physical effects: Gravity (there is *always* gravitational pulls in space, towards *something*), and the drag effects of subatomic particles impacting on the object. But basically it would keep on moving in a straight line until it eventually was either captured into orbit by some sort of body, or drawn into reentry onto the surface of some sort of body. For an astronaut firing a frearm in 0 G: Newton tells us that if you fire a firearm, half of the energy is ALWAYS in an equal and opposite vector to the projectile fired. If YOU are on a frictionless surface (IE: are an astronaut on an EVA), the force applied will drive you rearwards at a rate governed by the foot pounds of force applied, and your mass. That's a simple equation. You will stop when either drag or gravitational forces come into equalibrium with your new vector. Drag will be small, gravitational forces also small, and your distance travelled will be large. Velocity would not be very large since force = 1/2 mass times the square of the velocity of the projectile, and that mass would be small.
In reduced G: The moon has a gravitational pull of 5.2 ft/sec2, as opposed to the earths 33 ft/sec2, meaning that a projectile fired parallel to the moons surface will fall towards the surface and impact the surface about after about 6 times the length of time as the same projectile fired on the surface of the earth. The ballistic coefficient of the projectile will be a major player in decelleration on earth, but this will not be a large factor on the moon. The bottom line is that between taking 6x the length of time to fall to the surface, and the fact that the projectile will not be decellerating very much as it goes downrange, a projectile on the moon would be a LOT further downrange when it struck the surface, and it would have a LOT more retained energy when it did so. The astronaut firing the firearm would sense normal recoil forces as the kinetic energy imparted as recoil would be the same.
In a vacuum: A firearm fired in a vacuum will behave normally. This has nothing to do with zero G or reduced G (space = 0 G and moon = 1/6 G). A semiauto would cycle normally. Propellant would burn normally. Velocities would be normal, but would not drop off due to frictional effects (drag).
OK, you want to shoot at the passing bad-guys space vehicle?
In orbit: Firing a projectile from one space vehicle toward another in a dissimilar orbit is a problem of orbital mechanics. It's not nearly as simple as aiming and shooting. Google orbital rendezvous equations for some answers. Start here:
http://www.braeunig.us/space/orbmech.htm
What else? Yes it
is rocket science.
(And greetings from Edwards AFB where we teach this stuff at the test pilots school).
Willie
.
