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Chapter 12: Problem 1

In one dimension, the magnitude of the gravitational force of attraction between a particle of mass \(m_{1}\) and one of mass \(m_{2}\) is given by $$F_{x}(x)=G \frac{m_{1} m_{2}}{x^{2}}$$ where \(G\) is a constant and \(x\) is the distance between the particles. (a) What is the potential energy function \(U(x) ?\) Assume that \(U(x) \rightarrow 0\) as \(x \rightarrow \infty,(b)\) How much work is required to increase the separation of the particles from \(x=x_{1}\) to \(x=\) \(x_{1}+d ?\)

Short Answer

Expert verified
The potential energy function is \(U(x)=-\frac{Gm_1m_2}{x}\). And the work done to increase the separation of particles from \(x=x_1\) to \(x=x_1+d\) is \(W = U(x_1+d) - U(x_1) = \[-\frac{Gm_1m_2}{x_1+d} + \frac{Gm_1m_2}{x_1}\]\)

Step by step solution

01

Identifying the Potential Energy Function

From the physics of gravitation, we know that the potential energy in a gravity field is given by \(-\frac{Gm_1m_2}{r}\). Since, gravity is always attractive and the force of gravity does work on the particles to decrease the separation. Therefore, the potential energy is negative. Also, it is customary to choose the potential energy to be 0 at infinite separation. So, in this case the potential energy function will be \(U(x)=-\frac{Gm_1m_2}{x}\)
02

Calculating Work Done

To determine the work done, we need to calculate the potential energy difference between two positions. Work is defined as the change in kinetic energy and also we know from the conservation laws of physics that total energy (kinetic + potential) in a closed system is conserved. Hence, work done to move the particles from \(x = x_1\) to \(x = x_1+d\) is: \(W = U(x_1+d) - U(x_1) = \[-\frac{Gm_1m_2}{x_1+d} + \frac{Gm_1m_2}{x_1}\]\)

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Most popular questions from this chapter

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