Question

A mass $M$ at rest is broken into two pieces having masses $m$ and $(M-m)$. The two masses are then separated by a distance $r$. The gravitational force between them will be the maximum when he ratio of the masses $[m:(M-m\left)\right]$ of the two parts is

• A. 1 : 1
• B. 1 : 2
• C. 1 : 3
• D. 1 : 4

Answer 1

The correct option is A 1 : 1

The two pieces have masses as $m$ and $M-m$.

Gravitational force between them $F=\frac{Gm(M-m)}{r^2}$

For this force to be maximum, $\frac{d}{dm}F=0$

$\therefore$ $\frac{d}{dm}\frac{G(mM-m^2)}{r^2}=0$

OR $\frac{d}{dm}(mM-m^2)=0$

We get $M-2m=0$ $⟹m=\frac{M}{2}$

Mass of other piece $M-m=M-\frac{M}{2}=\frac{M}{2}$

$\therefore$ We get $m:(M-m)=\frac{M}{2}:\frac{M}{2}=1:1$