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March 1, 2015

March 1, 2015

Posted by **fatima** on Wednesday, June 18, 2014 at 2:04am.

- physics -
**Steve**, Wednesday, June 18, 2014 at 5:38amg = GM/r^2

at r+h, g1 = GM/(r+h)^2

at r-x, g2 = GM/(r-x)62

∆g = GM(1/(r-x)^2 - 1/(r+h)^2)

= GM/((r-x)^2(r+h)^2) ((r+h)^2-(r-x)^2)

= GM/((r-x)^2(r+h)^2) (2rh+h^2 - 2rx+x^2)

Since r-x and r+h are both approximately r, we have

∆g = GM/r^4 (2r(h-x) + h^2+r^2)

= 2GM/r^3 (h-x)

The h^2+r^2/r^4 is negligible.

h-x = r^3∆g/GM

Not sure just what kind of answer you want.

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