FIGURE CP13.71 shows a particle of mass m at distance 𝓍 from the center of a very thin cylinder of mass M and length L. The particle is outside the cylinder, so 𝓍 > L/2 . (a) Calculate the gravitational potential energy of these two masses.

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1. Identify the formula for gravitational potential energy between two point masses, which is given by U = -G \frac{m_1 m_2}{r}, where G is the gravitational constant, m_1 and m_2 are the masses, and r is the distance between the centers of the two masses.
2. Recognize that the cylinder can be approximated as a line of mass, and the problem can be approached by integrating the contributions to the gravitational potential energy from each infinitesimal mass element dm along the length of the cylinder.
3. Set up the integral for the gravitational potential energy. Let dm = \frac{M}{L} dx, where M is the total mass of the cylinder and L is its length. The variable x will vary from -L/2 to L/2 along the cylinder.
4. Express the distance from the particle to an element dm of the cylinder as a function of x. Since the particle is at a distance x_0 from the center of the cylinder, the distance to a point at x along the cylinder is r = \sqrt{x_0^2 + x^2}.
5. Integrate the expression for the potential energy U = -G \int_{-L/2}^{L/2} \frac{m dm}{r} = -Gm \int_{-L/2}^{L/2} \frac{M/L}{\sqrt{x_0^2 + x^2}} dx, where x_0 is the distance from the center of the cylinder to the particle. This integral will give the total gravitational potential energy between the particle and the cylinder.