Raymond L. answered • 05/20/15

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Math, SAT Math, Physics

By Taylor Expansion, g(r+Δr)=g(r)+g

^{'}(r)Δr+(g'(r)Δr^2)/2+higher order terms

Therefore, g(r+Δr)-g(r)=Δg=g'(r)Δr+(g'(r)(Δr)^2)/2+higher order terms

Use Linear approximation( from Calculus 1),which ignores contributions from 2nd order term and all higher order terms,

g(r+Δr)-g(r)=Δg≈g'(r)Δr

g(r)=GM/r^2 and derivative of g(r) respect to r is g'(r)=-2GM/r^3

So, Δg≈g'(r)Δr=(-2GM/r^3)Δr, but g=GM/r^2.

Therefore,Δg≈-2g(Δr/r)