How to solve this long distance projectile problem? A projectile is fired horizontally from the top of a mountain of height 'h' at a very high velocity. The projectile almost completes one revolution around the earth and crash lands at the bottom of the mountain (coming from the opposite side after circling the earth). I know that if it was fired at the orbital velocity, it would have orbited the earth in uniform circular motion. Hence, I assume the initial velocity in the question would be lower than the orbital velocity. The path taken by the projectile in this scenario seems to be a spiral. And there would be tangential acceleration as well, apart from centripetal acceleration, I believe.
At what velocity should we fire the the projectile for taking such a path? What would be the final velocity? (Assume the earth to be a perfect sphere and neglect atmospheric drag)
My approach was to find the work done by centripetal force at an instant (F.dh), then integrate it and equate it to mgh. (For uniform circular motion, work done by the centripetal force would be zero). But, it's a bit complicated method since both 'v' and and 'r' in mv^2/r are variables, and has to be expressed in polar co-ordinates.

 A: The path described is impossible. Without air drag, the projectile should move along an ellipse. So if you trace it around it should end up where it started due to conservation of energy.
Here is a picture of the limiting case for going around I think.

You might be tempted to make the following case, but this is also physically impossible.

Although the projectile is following an ellipse, another rule of orbital motion is that the center of mass (center of earth in this case) has to be on one of the ellipse foci. The foci is offset from the center of the ellipse, and in the diagram above the ellipse is concentric to the earth.
So the rule is, if it misses the earth on the other side exactly, it will continue and return back to the launch point.
I suggest the op lookup some references to orbital mechanics, and see if they can work out the problem on their own, as it presents some great learning experiences.
A: Lossless (iso-energetic) orbits:
The sum of energy kinectic + potential remain constant - no airdrag or such: these orbits are elliptical. If shot horizontally, there is one distinct speed, for which the orbit is circular. Slower speeds will mean that the starting point is the apogee, and the projectile will first drop altitude & gain velocity until perigee, and then slow down again and gain potential energy, until it's at the starting point (unless impact on earth / other stuff earlier). Higher speeds mean that the starting point is the perigee. That is until escape velocity.
These orbits are also called Kepler orbits, and there are Kepler's laws governing it. For your case - you could calculate total energy states (potential+kinetic).
For Kepler orbits you wouldn't need integration (see also [conservative vector field][1])
lossy orbits (with energy dissipation):
The spiral you depicted would not be an iso-energetic orbit - your projectile would need to lose energy, so e.g. airdrag helps. How much? Here is where the fun starts: airdrag is not a linear function of velocity, so I'd point you to numerical integration.
