Question

A meteor of mass $M$ breaks up into two parts. The mass of one part is $m$. For a given separation $r$ the mutual gravitational force between the two parts will be maximum if

• A. $m=\left(M/2\right)$
• B. $m=\left(M/3\right)$
• C. $m=\frac{M}{\sqrt{2}}$
• D. $m=\frac{M}{2\sqrt{2}}$

Answer 1

Answer: (A)

$m=\left(M/2\right)$

Force of gravitation between 2 masses $M_1$ and $M_2$ is given by
$F=\frac{G{M}_1{M}_2}{r^2}$
Here, gravitational force between mass $m$ and mass $M-m$ is
$F=\frac{Gm(M-m)}{r^2}$
Differentiate $F$, in order to find the maxima
$\frac{dF}{dm}=\frac{G(M-2m)}{r^2}=0$
Thus, $M=2m$ or $m=\frac{M}{2}$