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 .

c. rb

explanation:

the atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons.