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

User Canova
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1 Answer

5 votes

Answer:

X = 6910319.7 m

Step-by-step 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 .

User Yehia Awad
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