At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.785 m/s2 if the acceleration due to gravity at the surface has magnitude 9.80m/s2?
The acceleration due to gravity falls off as the inverse of the square of the distance from the center of the Earth. Thus the distance of interest is 1/(d/R)^2 = 0.785/9.8 d = R·√(9.8/0.785) ≈ R·3.53328 or R·2.53328 above the surface of the earth.
Wikipedia says the radius of the earth is 6371 km, so your distance is (6371 km)·2.53328 ≈ 16,140 km
