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 the force of gravitational attraction between the two parts is maximum .
(A) M = 2m .
(B) M = 3m/2 .
(C) M = m/2 .
(D) M = m .

Answer 1

Hint
When a mass M is broken into two parts take one part as m and other part as (M-m). After that write the equation of force of gravitation between two bodies ie. $F = \dfrac{{GMm}}{{{R^2}}}$ and apply the concept of maxima and minima to solve the problem.

Complete step by step answer
As a mass M is broken into two parts, take one part as m and other part as (M-m).
Now, we know that the force of attraction between two bodies as given by the Universal law of gravitation is,
$\Rightarrow F = \dfrac{{GMm}}{{{R^2}}}$
Using the values of masses as mentioned we have,
$\Rightarrow F = \dfrac{{G(M - m)m}}{{{R^2}}}$
$\Rightarrow F = \dfrac{{G(Mm - {m^2})}}{{{R^2}}}$
Now for force to be maximum we know that the differentiation of Force with respect to mass m must be zero.
Therefore, $\dfrac{{dF}}{{dm}} = \dfrac{{G(M - 2m)}}{{{r^2}}} = 0$
So, $(M - 2m) = 0$
$\Rightarrow M = \dfrac{m}{2}$
Hence, $\Rightarrow M = \dfrac{m}{2}$ is the answer, which is option (C).

Note
Here we do not need to double differentiate the function $\dfrac{{{d^2}F}}{{d{m^2}}}$ to know whether the function attains a maximum value or a minimum value. This is because the question itself tells us that it is attaining a maximum value. So, definitely the value of $\dfrac{{{d^2}F}}{{d{m^2}}}$ will automatically come out to be negative. You can try this and check for yourself.