Asked by fatima
If the change in the value of 'g ' at a height 'h ' above the surface of earth is the same as at a depth 'x ' below it. When both x and h are much smaller than the radius of the earth. The relation between x and h is?
Answers
Answered by
Steve
g = 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.
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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