AT WHAT DISTANCE ABOVE THE SURFACE OF THE EARTH IS THE ACCELERATION DUE TO EARTH'S GRAVITY 0.980M/S² IF THE ACCELERATION DUE TO GRAVITY AT THE SURFACE HAS MAGNITUDE 9.80M/S²

Answer 1

radius of earth = 6378.1 km

Assuming g is not affected by other heavenly bodies, then Newton's law of gravitation applies:
g=GMm/r^2=k/r^2 (k=a constant)
=>
g'r'^2=gr^2
r'^2=r^2(g/g')
r'=r sqrt(g/g')
=6378.1 sqrt(9.80/0.980)
= 2.02*10^4 km approx.