How much Fuel would the Space Shuttle Save Launching at 100,000 ft? Assuming it was possible?

That is a really complicated calculation and would all depend on where you started. Obviously the pull of gravity from the earth would be considerably less. As a reference just to let you knwo the escape velocity is about 11 km/s. This is the minimum speed you need to obtain to escape the pull of the Earth if starting off on the ground. Figuring the earth is 12,000 milles in diameter and when they launch they are approximately 6,000 miles from the center. The formula for the force of gravitational pull between two objects is = (G(gravitational constant) x mass of the first object (Earth) x mass of space shuttle)) / the radius of the distance from the center of mass of one to the center of mass to the other. As the radius goes up the force pulling in drops by a factor of r^2 which means it would be only a fraction of what it is on the ground as opposed to if it was 100,000 ft up. The logistics of it seem rather impossible right now though. So in answer to your question it would save quite a large portion. I would estimate 30% or more.

To launch that high it will need to be launched off of another “vehicle”. Assuming that there is a such vehicle that could carry this shuttle, it might save a good amount of fuel. However, the “vehicle” carrying the shuttle would be using a lot of fuel. Ergo, my answer would be that the margin of profit gained by launching this way would be minimal.

radius of earth 20,900,000 ft
altitude of shuttle about 1,000,000 ft
gravitational potential energy -GMm/r
a fair approximation would be the ratio of the change in the potential energy from launch height to orbit
PE from 1.000.000 to orbit -GMm(1/21,900,000 – 1/21,000,000)
PE from ground to orbit -GMm(1/21,900,000 – 1/20,900,000)
You can work out the math yourself.