Show that when r changes by ?
r; the change in the acceleration due to gravity,
g; is given by ?g ˜ -2g ( ?r / r)

Question

The acceleration due to gravity, g; is given by
g = GM / r² ;

where M is the mass of the earth, r is the distance from the center of the earth,
and G is the universal gravitational constant.

Show that when r changes by Δ
r; the change in the acceleration due to gravity,
g;
is given by

 

Δg ≈ -2g ( Δr / r)

[Hint: Let g be a function of r; that is: g (r) = GM / r² ]:

Answer 1

Let g=g(r)=GM/r^2;
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)