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A mass $M$ is split into two parts $m_0$ and $M-m_0$. These two masses are then separated by a distance $D$. If the gravitational force between the parts is maximum, then the ratio $\frac{m_0}{M}$ is
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The correct answer is:
0.5
According to given situation,
Again, $m_0$ and $M-m_0$ are kept at a distance $D$ gravitational force,
$F=\frac{G m_0\left(M-m_0\right)}{D^2}$
$F$ will be maximum, when $\frac{d F}{d m_0}=0$
$\begin{array}{ll}
\Rightarrow & \frac{d}{d m_0} \frac{G m_0\left(M-m_0\right)}{D^2}=0 \\
\Rightarrow & \frac{d}{d m_0}\left(\frac{G m_0 M-G m_0^2}{D^2}\right)=0 \\
\Rightarrow & \frac{G M-2 G m_0}{D^2}=0 \\
\Rightarrow & G M-2 G m_0=0 \Rightarrow M-2 m_0=0 \\
\Rightarrow & 2 m_0=M \Rightarrow \frac{m_0}{M}=\frac{1}{2}=0.5
\end{array}$
Again, $m_0$ and $M-m_0$ are kept at a distance $D$ gravitational force,
$F=\frac{G m_0\left(M-m_0\right)}{D^2}$
$F$ will be maximum, when $\frac{d F}{d m_0}=0$
$\begin{array}{ll}
\Rightarrow & \frac{d}{d m_0} \frac{G m_0\left(M-m_0\right)}{D^2}=0 \\
\Rightarrow & \frac{d}{d m_0}\left(\frac{G m_0 M-G m_0^2}{D^2}\right)=0 \\
\Rightarrow & \frac{G M-2 G m_0}{D^2}=0 \\
\Rightarrow & G M-2 G m_0=0 \Rightarrow M-2 m_0=0 \\
\Rightarrow & 2 m_0=M \Rightarrow \frac{m_0}{M}=\frac{1}{2}=0.5
\end{array}$
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