Could you explain? Centripetal acceleration is v^2/r, so naively I'd think something moving at LEO orbital velocity in a radius of 1 mile requires ~4,000g because LEO has a radius ~4k miles under 1g.
[1] https://en.wikipedia.org/wiki/Low_Earth_orbit#Orbital_charac...
(Which makes sense — if you managed to accelerate something to orbital velocity at sea level, it would (1) shed much of that speed before it actually reached a near-zero-atmosphere altitude and (2) burn up.)
Further the rocket equation is a bitch so your saving even more fuel. The problem is you also end up with a lot of atmospheric drag and heat so the final savings are not as huge.
Agreed about the rocket equation though - that first 6.25% of payload energy is much harder than the rest
Rockets get energy from their fuel directly from burning it, but also from the kinetic energy of their fuel. So in space a rocket that can add 100km per hour of delta V from fuel before running out can do that at 0MPH , 1000 MPH, or -10,000MPH all the way up until relativity becomes an issue.
So, first find out how much speed/energy you need as a baseline it's velocity squared (100% velocity)^2 = (1v)^2 = 1e = 100% energy. Now instead of that we need to go from V1 to V2 you need (V2 - V1)^2 energy. That's (100%v - 25%v)^2 = (1v-0.25v)^2 = (0.75v)^2 = 0.5625e so we still need 56.25% as much energy, and we gained 100%e - 56.25%e = 43.75% energy.
PS: Unfortunately, these things don't start in space so we need to consider air resistance.