Ignoring air resistance, gravitational pull of the moon, etc.
It doesn’t really land at the take off point, it’s just that the elliptical path takes it around to exactly where it started (position, speed and direction of travel). If you launch from a rail gun pointed up, the object will hit the ground before it finishes one orbit (imagine tracking backwards from the launch, the path will go below the surface).
If you launch from a horizontal rail gun, the object will travel halfway around the Earth, climbing until it reaches its highest point (the apogee, where its velocity is the slowest), then it comes back down, picking up speed until it gets back to where it was launched from. If you launch again a little faster, the object reaches a higher apogee but still comes back down to where it started.
Sorry, Russell. It is an orbit, but we are not explaining it perfectly. No matter how fast you launch it, the object has two choices - to be in a closed (elliptical) orbit, or to be in a hyperbolic orbit. If you are really careful you can get it into a parabolic orbit, but that’s close enough to hyperbolic as makes no odds.
As has been said, in a hyperbolic orbit the object will leave Earth entirely, eventually travelling in a straight line away from Earth.
In an elliptical orbit, the object will follow an ellipse once it leaves the rail gun. Unless it comes back to the starting point, it would not be an ellipse, it would be a spiral. Of course, the air resistance would make it a spiral, but once you get something out of the atmosphere it will be orbiting in an ellipse.
If your spiral idea worked, it’d be really easy to get to, say, the moon. Just launch the space shuttle and wait for it to spiral up to the moon.
Of course, any elliptical orbit that is small enough will lie nearly entirely within the surface of Earth. This makes it tough to get back to the starting point, but it is not because the orbit is not elliptical, it is simply because of impact with some large object that happens to be in the way.
Or we can phrase it this way: The final portion of the track is a tangent of the orbital ellipse. Despite the earth’s rotation (or because of it) that point on the ellipse will someday have the track intersecting it, at which point the satellite will crash.
I seem to recall reading some research on the feasibility of launching objects into space with railgun technology. Maybe about 10-12 years ago. One thing that stands out in my mind was that for “short” rail guns (less than 5 miles or something like that) the magnetic forces required were so strong that they would deform the rail. Hence you could only use it to launch one object, then you’d have to build a new rail. This realization was apparently one of the more significant in concluding that the idea wasn’t exactly cost-effective. Wish I could remember the source better, but it was so long ago…
RussellM said "Assuming linear deceleration, the average velocity over the final 5cm will be 0.7ms-1, so the time taken will be
t = d / (average)v = 0.071 seconds"
Looks like you’ve dropped a decimal point somewhere. Linear decelaration from 14 m/s to zero gives an average velocity of 7 m/s, giving 200 G’s. Then if we stop in 5 mm rather than 5 cm, which I don’t think is unreasonable, we get roughly 2000 G’s. Getting into the right ballpark anyway!
Are you telling me you can make an object that can be given escape velocity (plus the impulse needed to overcome air resistance) and make it through the atmosphere without disintegrating? Where can I find this object?
Maybe, and I don’t know. Anyone know the fastest anyone has got a solid object to travel through the atmosphere? How fast does an APFSDS tank shell go?
Now I’ve done some sums. Assuming the figure of 30 km/s launch velocity is correct, and orbital velocity is 8 km/s, and all this velocity loss is due to drag (potential energy to get to orbit height is only a few percent of the kinetic energy to reach orbit velocity so I’m ignoring it):
you have to dissipate about half a billion joules of KE per kilogram of payload, in about a second. Same energy you get from burning 3.3 US gallons of gasoline (2.75 british gallons). Maybe an ablative heat shield could handle this, I’ve really no idea! What we need now is a rocket scientist.
Um, yeah. The only problem is, you aren’t supplying the energy from a rail gun, you are supplying energy from a rocket. Once the bullet leaves the gun, it will return to the same spot (neglecting friction and collisions with inconveniently non-zero sized planets.)
In order to go from one elliptical orbit to another elliptical orbit or to a hyperbolic orbit, you must use a rocket (or other propulsion not rail-gun related.)
I think maybe there is some confusion here about what the projectile is. If it is just a bullet, and has no propulsion other than that imparted by the gun, it cannot leave the orbit it attained upon leaving the rail/barrel. Only if you put a rocket or other propulsion unit on it can it change the orbit to no longer intersect with the point in space where the end of the barrel was.
And to cover the final point in the OP: The optimum angle. If you’re launching from an airless world, such as the Moon, then the angle doesn’t matter. It’s escape speed, not escape velocity: If you give an object enough kinetic energy, it’ll escape. For Earth launches, however, you want to minimize the air resistance, and you do this by going through less air. This means straight up, which is why rockets don’t launch horizontally. (There’s usually some angle to the eastward, too, to take advantages of the Earth’s rotation, but it’s mostly up.) Unfortunately, launching up with a railgun provides other problems: It’s very difficult to construct a vertical track several miles tall. You might be able to build in some slope, and run it up the side of a steep mountain, but you’re going to have to be fairly close to horizontal, if you hope to get the thing built. This means that you’ll be slicing through a lot of atmosphere, and losing a lot of energy to friction.
The long and the short of it is that an electromagnetic catapult might be practical for sending stuff from Luna to Terra (or elsewhere in the solar system), but not from Earth.
To get from the surface of Earth to Low Earth Orbit (LEO) requires a delta-V (velocity change) of 9 km/s. To reach escape velocity from the surface takes a delta-V of 12 km/s.