Ask Question
7 July, 09:57

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

+3
Answers (1)
  1. 7 July, 11:33
    0
    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.
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “How far (in meters) above the earth's surface will the acceleration of gravity be 85.0 % of what it is on the surface? ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers