Three masses m\(_{1}\) = 200 kg, m\(_{2}\) = 300 kg and m\(_{3}\) = 400 kg are kept at the vertices of on equilateral triangle of side 20 m. If the masses are shifted to new configuration such that they are at the vertices of an equilateral triangle of 25 m now. Find the work done in this process :

Question

Two identical particles each of mass \( m \) go around a circle of radius \( a \) under the action of their mutual gravitational attraction. The angular speed of each particle will be:

Answer 1

To find the work done in rearranging the masses from the vertices of one equilateral triangle to another, we need to calculate the change in gravitational potential energy of the system.

The gravitational potential energy of a system of three masses placed at the vertices of an equilateral triangle is given by:

U = -\frac{G(m_1m_2 + m_2m_3 + m_3m_1)}{r}

where G is the gravitational constant and r is the side of the equilateral triangle.

Initial potential energy, U_1 with r_1 = 20 \, \text{m}:

U_1 = -\frac{G(200 \times 300 + 300 \times 400 + 400 \times 200)}{20}

Final potential energy, U_2 with r_2 = 25 \, \text{m}:

U_2 = -\frac{G(200 \times 300 + 300 \times 400 + 400 \times 200)}{25}

The work done, W, is the change in potential energy:

W = U_2 - U_1

Substituting the potential energies:

W = -\frac{G \times 290000}{25} + \frac{G \times 290000}{20}

W = G \times 290000 \left(\frac{1}{20} - \frac{1}{25}\right)

Simplifying further:

W = G \times 290000 \times \frac{5}{500} = \frac{G \times 290000}{100}

Given G \approx 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2:

W = \frac{6.67 \times 10^{-11} \times 290000}{100}

W = 1.9354 \times 10^{-7} \, \text{J}

Correct answer: \(1.735 \times 10^{-7}\)J after rounding.

Answer 2