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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 `
2 years ago

```							We have g = GM/R^2Acceleration due to gravity, g at height h from the surface is: g’ = GM / [R+h]^2Dividing g with g’ gives: g/g’ = [R+h]^2 / R^2 = (1+ h/R)^2Substituting the value g’ = g/4g / [g/4] = (1+ h/R)^24 =  (1+ h/R)^2Squaring both sides: 2 =  (1+ h/R)h/R = 2-1 ----> h/R = 1h = R This shows that when h = R, the value of g becomes g/4...So the value of X is one... Ans: X = 1
```
2 years ago
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