The figure shows four particles, each of mass 17.5 g, that form a square with an edge length of d = 0.600 m. If d is reduced to 0.150 m, what is the change in the gravitational potential energy of the four-particle system? P

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### Problem Description:

The figure shows four particles, each of mass 17.5 g, that form a square with an edge length of \( d = 0.600 \, \text{m} \). If \( d \) is reduced to 0.150 m, what is the change in the gravitational potential energy of the four-particle system?

### Diagram Explanation:

#### Square Formation of Particles:
- The diagram represents four particles situated at the corners of a square.
- Each side of the square is labeled as \( d \), indicating its edge length.
- Initially, the edge length \( d \) is 0.600 m.

### Calculation of Gravitational Potential Energy:

Gravitational potential energy \( U \) between two particles of masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by:

\[ U = -\frac{G m_1 m_2}{r} \]

Where:
- \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)),
- \( m_1 \) and \( m_2 \) are the masses of the particles,
- \( r \) is the distance between the particles.

### Explanation of Changes in Potential Energy:

#### Initial Setup:
- Four particles form a square with \( d = 0.600 \, \text{m} \).
- Distance between adjacent particles is \( d \).
- Distance between diagonally opposite particles (using Pythagoras' theorem) is \( \sqrt{d^2 + d^2} = \sqrt{2}d = 0.849 \, \text{m} \).

#### Modified Setup:
- \( d \) is reduced to 0.150 m.
- Distance between adjacent particles is \( d = 0.150 \, \text{m} \).
- Distance between diagonally opposite particles is \( \sqrt{2}d = 0.213 \, \text{m} \).

For both setups:
- Calculate gravitational potential energy for each pair and sum them up.
- Compare the total gravitational potential energy before and after the reduction in \( d \).

### Conclusion:

The change in gravitational potential energy of the four-particle system can be determined by analyzing the difference between the initial and
Transcribed Image Text:### Problem Description: The figure shows four particles, each of mass 17.5 g, that form a square with an edge length of \( d = 0.600 \, \text{m} \). If \( d \) is reduced to 0.150 m, what is the change in the gravitational potential energy of the four-particle system? ### Diagram Explanation: #### Square Formation of Particles: - The diagram represents four particles situated at the corners of a square. - Each side of the square is labeled as \( d \), indicating its edge length. - Initially, the edge length \( d \) is 0.600 m. ### Calculation of Gravitational Potential Energy: Gravitational potential energy \( U \) between two particles of masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by: \[ U = -\frac{G m_1 m_2}{r} \] Where: - \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)), - \( m_1 \) and \( m_2 \) are the masses of the particles, - \( r \) is the distance between the particles. ### Explanation of Changes in Potential Energy: #### Initial Setup: - Four particles form a square with \( d = 0.600 \, \text{m} \). - Distance between adjacent particles is \( d \). - Distance between diagonally opposite particles (using Pythagoras' theorem) is \( \sqrt{d^2 + d^2} = \sqrt{2}d = 0.849 \, \text{m} \). #### Modified Setup: - \( d \) is reduced to 0.150 m. - Distance between adjacent particles is \( d = 0.150 \, \text{m} \). - Distance between diagonally opposite particles is \( \sqrt{2}d = 0.213 \, \text{m} \). For both setups: - Calculate gravitational potential energy for each pair and sum them up. - Compare the total gravitational potential energy before and after the reduction in \( d \). ### Conclusion: The change in gravitational potential energy of the four-particle system can be determined by analyzing the difference between the initial and
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