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The acceleration due to gravity is reduced to a value one-fourth of that on the surface of earth at a height h, where h is equal to Xradius of earth. Find the value of X

The acceleration due to gravity is reduced to a value one-fourth of that on the surface of earth at a height h, where h is equal to Xradius of earth. Find the value of X 

Grade:10

1 Answers

Sanju
106 Points
3 years ago
We have g = GM/R^2
Acceleration due to gravity, g at height h from the surface is: g’ = GM / [R+h]^2
Dividing g with g’ gives: g/g’ = [R+h]^2 / R^2 = (1+ h/R)^2
Substituting the value g’ = g/4
g / [g/4] = (1+ h/R)^2
4 =  (1+ h/R)^2
Squaring both sides: 
2 =  (1+ h/R)
h/R = 2-1 ----> h/R = 1
h = R
 
This shows that when h = R, the value of g becomes g/4...So the value of X is one...
 
Ans: X = 1
 
 
 

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