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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 the ratio of the masses $[m:(M-m)]$ of the two parts is
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Verified Answer
The correct answer is:
1: 1
The gravitation force between two masses,
$$
F=\frac{G m_{1} m_{2}}{r}
$$
here, $\quad m_{1}=m$
and
$$
\begin{array}{l}
m_{2}=(M-m) \\
F=\frac{G m(M-m)}{r^{2}}
\end{array}
$$
$\therefore$
For maximum gravitational force, $\frac{d F}{d m}=0$
$\therefore$
$$
\frac{d}{d m}[m(M-m)]=0
$$
By solving, we get $m=\frac{M}{2}$
So, $\quad \frac{m}{(M-m)}=\frac{1}{1}$
$$
F=\frac{G m_{1} m_{2}}{r}
$$
here, $\quad m_{1}=m$
and
$$
\begin{array}{l}
m_{2}=(M-m) \\
F=\frac{G m(M-m)}{r^{2}}
\end{array}
$$
$\therefore$
For maximum gravitational force, $\frac{d F}{d m}=0$
$\therefore$
$$
\frac{d}{d m}[m(M-m)]=0
$$
By solving, we get $m=\frac{M}{2}$
So, $\quad \frac{m}{(M-m)}=\frac{1}{1}$
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