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The value of acceleration due to gravity at a point P inside the earth and at another point Q outside the earth is g/2 (g being acceleration due to gravity at the surface of earth). Maximum possible distance in terms of radius of earth R between P and Q is?

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4 years ago

```							For (for above the surface),g/2 = GM/(R+h)^2For below the surface, g/2 = GM/(R-d)^2Equating both we get,R^2 = h^2 + 2RhandR^2 = d^2 – 2Rdsolve quadratic for both h and d terms as:h = (root2 – 1)Randd = (root2 + 1)Rhence answer is:h + d = (root2 + 1)R + (root2 - 1)R => 2root(2)R meter.
```
4 years ago
```							 (for above the surface),Forg/2 = GM/(R+h)^2For below the surface,g/2 = GM/(R-d)^2Equating both we get,R^2 = h^2 + 2RhandR^2 = d^2 – 2Rdsolve quadratic for both h and d terms as:h = (root2 – 1)Randd = 3/2Rhence answer is:h+d= R/2(2*Root2+1)
```
2 years ago
```							Dear Student,Please find below the solution to your problem.For(for above the surface),g/2 = GM/(R+h)^2For below the surface,g/2 = GM/(R-d)^2Equating both we get,R^2 = h^2 + 2RhandR^2 = d^2 – 2Rdsolve quadratic for both h and d terms as:h = (root2 – 1)Randd = (root2 + 1)Rhence answer is:h + d = (root2 + 1)R + (root2 - 1)R=> 2root(2)R meter.Thanks and Regards
```
4 months ago
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