How far (in meters) above the earth's surface will the acceleration of gravity be 85.0 % of what it is on the surface?

X = 6910319.7 m

Explanation:

let X be the distance where the acceleration of gravity is 85% of what it is on the surface and g1 be the acceleration of gravity at the surface and g2 be the acceleration of gravity at some distance X above the surface.

on the surface of the earth, the gravitational acceleration is given by:

g1 = GM / (r^2) = [ (6.67408*10^-11) (5.972*10^24) ]/[ (6371*10^3) ^2] = 9.82 m/s^2

at X meters above the earth's surface, g2 = 85/100 (9.82) = 8.35m/s^2

then:

g2 = GM / (X^2)

X^2 = GM/g2

X = / sqrt{GM/g2}

= / sqrt{ (6.67408*10^-11) (5.972*10^24) / 8.35

= 6910319.7 m

Therefore, the acceleration of gravity becomes 85% of what it is on the surface at 6910319.7 m.