Question

The value of acceleration due to gravity $\left(g\right)$ at height $h$ above the surface of earth is given by $g^{\prime }=\frac{g{R}^2}{(R+h{)}^2}$. If $h<<R$, then prove that $g^{\prime }=g(1-\frac{2h}{R})$.

Answer 1

We are given
$g^{\prime }=g{\left(\frac{R}{R+h}\right)}^2=g{\left(\frac{1}{1+\frac{h}{R}}\right)}^2=\frac{g}{{(1+\frac{h}{R})}^2}=g{(1+\frac{h}{R})}^{-2}$
As we are given $h<<R$, hence, $h/R<<1$
We can write $g^{\prime }=g(1-\frac{2h}{R})$.