Question

Mass $$m$$ is divided into two parts $$Xm$$ and $$\left(1-X\right)m$$. For a given separation the value of $$X$$ for which the gravitational force of attraction between the two pieces becomes maximum is:
A. $${\displaystyle \frac{1}{2}}$$
B. $${\displaystyle \frac{3}{5}}$$
C. $$1$$
D. $$2$$

Answer 1

Hint: In this question, the mass $$m$$ is divided into two parts $$Xm$$ and $$\left(1-X\right)m$$ , and we need to find the value of $$X$$ for which the gravitational force of attraction between the two pieces becomes maximum. To solve this, find the gravitational force of attraction between the two masses. For maximum value $${\displaystyle \frac{dF}{dX}}=0$$. Differentiate force with respect to $$X$$ and equate it to zero. Solve the equation, to find out the value of $$X$$

Formula used:
The magnitude of Gravitational force $$F$$ between two particles $$m_1$$ and $$m_2$$ placed at a distance $$r$$ is given by,
$$F={\displaystyle \frac{G{m}_1{m}_2}{r^2}}$$
Where $$G$$is the universal constant called the Gravitational constant.
$$G=6.67\times {10}^{-11}\text{ N - m/k}{\text{g}}^2$$

Complete step by step answer:
Mass $$m$$ is divided into two parts $$Xm$$ and $$\left(1-X\right)m$$.
Let the distance between them be $$R$$ meters
Therefore, Gravitational force between them is given by $$F={\displaystyle \frac{G{m}_1{m}_2}{r^2}}$$
Substituting the values in the formula we get,
$$F={\displaystyle \frac{GXm\left(1-X\right)m}{R^2}}$$
$$\Rightarrow F={\displaystyle \frac{GX\left(1-X\right)m^2}{R^2}}$$
The gravitational force for a given distance will be maximum when $$X\left(1-X\right)$$ will be maximum
Thus, for maxima, $${\displaystyle \frac{dF}{dX}}=0$$
Differentiating the gravitational force between the two masses $$F$$ with respect to $$X$$ we get,
$${\displaystyle \frac{dF}{dX}}={\displaystyle \frac{G{m}^2}{R^2}}{\displaystyle \frac{d\left(X\left(1-X\right)\right)}{dX}}$$
$$\Rightarrow {\displaystyle \frac{dF}{dX}}={\displaystyle \frac{G{m}^2}{R^2}}{\displaystyle \frac{d\left(X-X^2\right)}{dX}}$$
$$\Rightarrow {\displaystyle \frac{dF}{dX}}={\displaystyle \frac{G{m}^2}{R^2}}\left(1-2X\right)$$
For maxima, $${\displaystyle \frac{dF}{dX}}=0$$
$$\Rightarrow {\displaystyle \frac{dF}{dX}}={\displaystyle \frac{G{m}^2}{R^2}}\left(1-2X\right)=0$$
$$\Rightarrow \left(1-2X\right)=0$$
On solving we get,
$$X={\displaystyle \frac{1}{2}}$$
The gravitational force between the masses has a maximum value at $$X={\displaystyle \frac{1}{2}}$$
The mass $$m$$ should be divided into $${\displaystyle \frac{m}{2}}$$ and $${\displaystyle \frac{m}{2}}$$ for maximum gravitational force.
Hence the correct option is option (A).

Note:
Unlike the electrostatic force, Gravitational force is independent of the medium between the particles. It is conservative in nature. It expresses the force between two-point masses (of negligible volume). However, for external points of spherical bodies, the whole mass can be assumed to be concentrated at its center of mass.