If an object is placed in orbit 10,000 miles above earth how long would it take before it crashes to earth?

in orbit, never. Stationary 10 k miles (1.61e7 m) it will fall to earth. Note that use of 9.8 m/s² is not correct, as that value applies only near the surface of earth.

assume mass of the object is 1 kg (or much less than that of the earth, doesn't matter)

t = (arccos√(x/r) + √((x/r)(1–(x/r)) (R^3/2)) / √(2µ)

µ = G(m₁+m₂)m₁, m₂ masses of the two bodies

earth radius    R = 6,371 km = 6.37e6 meters

r = 1.61e7 m

x = 6.37e6

the rest is up to you.

You would need access to air drag tables or a high powered algorithm to calculate the decay rate at that altitude, and there would also be tidal effects from the Sun and Moon that would slowly degrade the orbit, and effects from the solar wind.

Without looking up any data I would make a guestimate that the object would remain in orbit for a timespan ranging from a few thousand year up to perhaps several tens of thousand years. 

"orbit" implies that it is in motion.  It's speed, then, will keep it in orbit and it will not crash.  If you are asking how long will it take for an object to fall 10,000 miles under the force of gravity alone - near the Earth, an object in a vacuum will accelerate at approximately 9.8 m/s2, independent of its mass.  (I will ignore the effect of air resistance and complete the calculation.)

10,000 miles is 17,000 km or 17,000,000 m.  Divide by 9.8 asnd take the square root and I get a bit over 1,300 seconds.  About 22 minutes