91Ó°ÊÓ

The mass of the sphere A with coordinates is$m_A=40kg,\left(0,50cm\right)$

The mass of the sphere B with coordinates is$m_B=35kg,\left(0,0cm\right)$

The mass of the sphere C with coordinates is$m_c=200kg,\left(-80cm,0\right)$

The mass of the sphere D with coordinates is$m_D=50kg,\left(40cm,0\right)$

The energy that an item has or acquires when its location changes as a result of being in a gravitational field is known as gravitational potential energy.

We can use the concept of gravitational potential energy to find the energy of the given system.

Formula:

$U=-\frac{GMm}{r}$

Where ,

G is the potential energy

U is the gravitational constant ()

M is the mass of earth

m is the mass of body

r is the distance of body from the center of earth

Gravitational potential energy of the three particle system:

The distance between the sphere C and D sphere is$r_{CD}=1.2m$

The total gravitational potential energy of the system is

$$\begin{gathered} U=-\frac{G{M}_B{M}_C}{r_C}-\frac{G{M}_B{M}_D}{r_D}-\frac{G{M}_C{M}_D}{r_{CD}} \\ U=-\frac{\left(6.67×{10}^{-11}\frac{{\text{Nm}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)×35\text{kg}×200\text{kg}}{0.80\text{m}}-\frac{\left(6.67×{10}^{-11}\frac{{\text{Nm}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)×35\text{kg}×50\text{kg}}{0.40\text{m}} \\           -\frac{\left(6.67×{10}^{-11}\frac{{\text{Nm}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)×200\text{kg}×50\text{kg}}{1.2\text{m}} \end{gathered}$$

$U=-1.43×{10}^{-6}J$

If sphere A is put back, compare the potential energy of the four particle system:

The gravitational potential energy is a negative term. When we put the sphere, it lowers the total potential energy of the system.

When we remove the sphere A it increases the total energy of the system; hence, it is positive work done.

When we replace the sphere A, it decreases the total energy of the system; hence, it is negative work done. The system becomes more negative.