Assuming:

Launch Speed: 17,180 mph (Speed of Space Station) [1]

Launch Acceleration: 25g (Based on my understanding of what humans can survive) EDIT [2]

Just using that as a rough estimate and Wolfram Alpha we get 242,250 meters, or about 150 miles.

http://www.wolframalpha.com/input/?i=17,239.2+mph+/25+gs

http://www.wolframalpha.com/input/?i=31.434+seconds+at+17,23...

Please feel free to check my math and yell at me.

[1] http://en.wikipedia.org/wiki/International_Space_Station

[2] cperciva [http://news.ycombinator.com/item?id=1993399] used 3g. Based on that you would have to multiply the launch distance by (25/3) and get 2,018,750 meters, about 1,250 miles. Again please correct me if I got something wrong.

Anyway, aren't you forgetting that most of the time is spent at a lower speed, so that you want to integrate the acceleration rather than simply multiplying by the escape velocity?

d = 1/2 at^2, where t = escape velocity / 3 G?

http://www.wolframalpha.com/input/?i=escape+velocity+%2F+3gs t = 6.3 min

http://www.wolframalpha.com/input/?i=distance+travelled+in+6... d = 2100 km (1300 miles)

While we're at it, I'm not sure what we really should be using for escape velocity. We should probably account for the deceleration after we leave the muzzle while still in the atmosphere. And if we presume a surface mounted rail gun, we'd probably want the escape velocity for a launch aimed just above the horizon, which involves more of this.[1]

Or if we're somehow presuming vertical, we'd want the escape from that height. But if vertical, we'd have to account for the extra gravitational force on the passengers, unless this is already accounted for in the human limits. In any case, I'm not sure that the speed of the Space Station (or the surface velocity I used above) really makes sense here.

[1] I vaguely recall that ignoring air resistance, the velocity is the same regardless of direction, but I don't have confidence in this, and my quick searching hasn't turned up anything definitive. Is this possibly right?

Think of it this way: if you were approaching Earth from space and wanted to enter orbit with the ISS, you need to reach a speed, e.g. 17,600. The speed is the same no matter where you are coming from, such as from the Earth's surface.

On the equator, if you take off east, you already have about 1,000 mph working with you. If you take off west, you need to get an extra 1,000 mph. Towards the poles, the (dis)advantage lessens.

Details aside, the overall conclusion remains that rail guns are not likely to be a useful means of propelling squishy cargo like humans into space.

For cargo a railgun a few miles long would probably be enough, but designing a missile which can survive surface air densities while travelling at orbital velocities would be very difficult.

That said for human payloads you'd need a really long track, tens or hundreds of kilometers. I wouldn't want to live near the launch end. (edit: to get a useful amount of speed; still assuming rockets to get us the rest of the way)

Rockets turn sideways because when you're in orbit, you're moving sideways.

And yes, since the first half of the tunnel is sloping downwards you are using a large fraction of g to accelerate the slug/projectile, thus reducing the power requirements.

Its an engineering problem, not a physics problem.

Unfortunately a project of this scale is beyond the largest organizations that exist today, nation-states. Maybe later?

When the day comes for the need of a super-structure to be built in orbit, this technology will likely be used. This or a space elevator. Getting far enough away from our own gravity well will free us from this rock.

I think our government should build such a launch accelerator in the desert somewhere. (Tilted at 45 degrees or so, to get a cargo drone out of the atmosphere and give it a big boost towards orbital velocity.) The ability to place bulk cargo into orbit would give the US a tremendous economic advantage. (Selling solar power satellites, colonization of Mars...)

A ram accelerator is probably more practical in any case.

Just how long does the shielding have to withstand high temps? We already solve a much harder problem with reentry of orbital vehicles. The problem here is much simpler, as we don't actually want to shed most of the velocity. In fact, we want to lose as little of our kinetic energy as possible. If ablative shielding can handle reentry, it can handle launch. AFAIK, the mass ratio for reentry shielding is quite good.

The rail gun cargo is going in the opposite direction - you start in the thick air at a speed greater than orbital velocity. It's been years since I could grind through the numbers, but it makes a big difference

This speaks to my point. In launch accelerator mode, you want to dissipate as little energy as possible. In reentry mode, you want to dump the majority of a big honking chunk of kinetic energy.

I'm sure there are all sorts of optimizations to consider. How about a mostly vertical trajectory, with most of the horizontal velocity provided by lasers heating a heat-exchanger?

By keeping the trajectory mostly vertical, one can avoid most of the lower atmosphere by stationing the opening to the atmosphere miles above the surface. This also limits the downrange danger-zone for falling malfunctioning cargo pods. Additionally, it makes the cargo pods easier to target with the lasers, since the powered flight would be entirely in the upper atmosphere. It would also make the rocket nozzles easier to design.