# A mass M is split into two parts m and (M-m) which are then separated by a certain distance. What ratio of m/M maximise the gravitational force between the parts?

$(a)\;\frac{2}{3} \quad (b)\;\frac{3}{4} \quad (c)\;\frac{1}{3} \quad (d)\;\frac{1}{2}$

Let r be the direction between the parts
$F= \large\frac{Gm(M-m)}{r^2}$
$\qquad= \large\frac{G}{r^2} $$(mM-m^2) The force will be maximum \large\frac{dF}{dm}$$=0$
ie $\large\frac{d}{dm} \bigg[\large\frac{G}{r^2} $$(mM-m^2)\bigg]=0 \large\frac{G}{r^2}$$(M-2m)=0$
$M-2m=0$
$M=2m$
$\large\frac{M}{m}$$=2$
or $\large\frac{m}{M}=\large\frac{1}{2}$
Hence d is the correct answer.

edited Feb 17, 2014 by meena.p