Question 9: Gravitation

Question

A mass $M$ is broken into two parts of masses $m_1$ and $m_2$. How are $m_1$ and $m_2$ related, so that force of gravitational attraction between the two parts is maximum?

Accepted Answer

Let $m_1=m$, then $m_2=M-m$
Then, $F=\frac{G m_1 m_2}{r^2}=\frac{G m(M-m)}{r^2}$
$$ =\frac{G}{r^2}\left(M m-m^2ight) $$
Differentiating w.r.t. m , we get
$$ \frac{d F}{d m}=\frac{G}{r^2}(M-2 m) $$
For $F$ to be maximum,
$$ \begin{aligned} \frac{d F}{d m} & =0 \ \frac{G}{r^2}(M-2 m) & =0 \ M-2 m & =0 \ 2 m & =M \ m & =\frac{M}{2} \end{aligned} $$
So, $m_1=\frac{M}{2}$
$$ m_2=M-\frac{M}{2}=\frac{M}{2} $$

Answer

$m_1=\frac{M}{6}, m_2=\frac{5 M}{6}$

Answer

$m_1=\frac{M}{3}, m_2=\frac{2 M}{3}$

Answer

$m_1=\frac{M}{4}, m_2=\frac{3 M}{4}$

Answer

$m_1=\frac{M}{2}, m_2=\frac{M}{2}$

Gravitation

A mass M is broken into two parts of masses m_1 and m_2. How are m_1 and m_2 related, so that force of gravitational attraction between the two parts is maximum?

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