91Ó°ÊÓ

1. The figurefor a threeparticle system.
2. A particle of mass m is to be moved from an infinite distance to one of the three possible locations a, b and c.

When an object is present in a gravitational field, it has or can gain gravitational potential energy, which is energy that results from a change in position. Using the relation between change in potential energy and work done, we can find the work done by the gravitational force on the moving particle when it is at points a, b and c due to the fixed particles.

Formula:

Formulae:

The gravitational potential energy between two bodies of masses M and m separated by distance r is, $U=\frac{-GMm}{r}$ …(¾±)

The work done by the system due to energy change, $W=-∆U$ …(¾±¾±)

Using equation (i) in equation (ii), we can get the work done of the net gravitational force

as follows:

$W=\frac{GMm}{r}$ …(¾±¾±¾±)

Thus, the work done by the gravitational force on the moving particle when it is at point a, due to the fixed particles can be given using equation (iii) as:

$$\begin{gathered} W=\frac{Gm\left(2m\right)}{d}+\frac{Gm\left(m\right)}{3d} \\ =\frac{2G{m}^2}{d}+\frac{G{m}^2}{3d} \\ =\frac{7G{m}^2}{3d} \end{gathered}$$

Thus, the work done by the gravitational force on the moving particle when it is at point b, due to the fixed particles is given using equation (iii) as:

$$\begin{gathered} W=\frac{Gm\left(2m\right)}{d}+\frac{Gm\left(m\right)}{3d} \\ =\frac{2G{m}^2}{d}+\frac{G{m}^2}{3d} \\ =\frac{3G{m}^2}{d} \end{gathered}$$

The work done by the gravitational force on the moving particle when it is at point c, due to the fixed particles is given using equation (iii) as:

$$\begin{gathered} W=\frac{Gm\left(2m\right)}{d}+\frac{Gm\left(m\right)}{3d} \\ =\frac{2G{m}^2}{d}+\frac{G{m}^2}{3d} \\ =\frac{5G{m}^2}{3d} \end{gathered}$$

Therefore, the ranking of the three possible locations according to the work done by the net gravitational force on the moving particle due to the fixed particles, greatest first is $b>a>c$.