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Chapter 4: Problem 27

*The potential energy of a particle of mass \(m\) is \(V(r)=k / r+c / 3 r^{3}\), where \(k<0\) and \(c\) is a small constant. (The gravitational potential energy in the equatorial plane of the Earth has approximately this form, because of its flattened shape - see Chapter 6.) Find the angular velocity \(\omega\) in a circular orbit of radius \(a\), and the angular frequency \(\omega^{\prime}\) of small radial oscillations about this circular orbit. Hence show that a nearly circular orbit is approximately an ellipse whose axes precess at an angular velocity \(\Omega \approx\left(c /|k| a^{2}\right) \omega\).

Short Answer

Expert verified
Question: Given a particle moving in a central force field, where the potential energy function is given by \(V(r) = \frac{k}{r} + \frac{c}{3r^3}\), find the angular velocity of the precessing ellipse when the particle undergoes small radial oscillations around a circular orbit. Answer: The angular velocity of the precessing ellipse can be found using the following expression: \(\Omega \approx \left(\frac{c}{|k| a^2}\right)\sqrt{\frac{1}{ma}\left(\frac{-k}{a^2} + \frac{c}{a^4}\right)}\), where a is the radius of the circular orbit.

Step by step solution

01

Determine the force acting on the particle

We need to find the force acting on the particle. The force can be obtained from the gradient of the potential energy function, V(r), with respect to the radial coordinate r: \(F(r) = -\frac{dV(r)}{dr} = -\frac{d}{dr}\left(\frac{k}{r} + \frac{c}{3r^3}\right)\) Solving the derivative, we find the force: \(F(r) = -\frac{-k}{r^2} + \frac{c}{r^4}\)
02

Write the centripetal force equation for a circular orbit

For a circular orbit, the centripetal force required to maintain the orbit is equal to the force acting on the particle. Centripetal force is given by: \(F_{c}(a) = m\omega^2a\) So, we have: \(m\omega^2a = \frac{-k}{a^2} + \frac{c}{a^4}\)
03

Solve for the angular velocity, ω

To find the angular velocity ω, we need to solve the above equation for ω. We can do this by rewriting the equation as: \(\omega^2 = \frac{1}{ma}\left(\frac{-k}{a^2} + \frac{c}{a^4}\right)\) Now we can take the square root of both sides: \(\omega = \sqrt{\frac{1}{ma}\left(\frac{-k}{a^2} + \frac{c}{a^4}\right)}\)
04

Determine the effective potential for radial motion

To find the angular frequency ω', we need to first determine the effective potential energy for the radial motion. The effective potential is: \(V_{eff}(r) = V(r) + \frac{1}{2}m\omega^2r^2\)
05

Write the equation for the angular frequency, ω', of the small radial oscillations

The equation for angular frequency ω' can be obtained by finding the second derivative of Veff(r) evaluated at r=a: \(\omega'^2 = \frac{1}{m}\frac{d^2V_{eff}(r)}{dr^2}\bigg|_{r=a}\)
06

Solve for the precessing angular velocity, Ω

Finally, we need to find the angular velocity of the precession Ω. From the given hint, we can write the equation as: \(\Omega \approx \left(\frac{c}{|k| a^2}\right)\omega\) Substituting the expression for ω obtained in Step 3: \(\Omega \approx \left(\frac{c}{|k| a^2}\right)\sqrt{\frac{1}{ma}\left(\frac{-k}{a^2} + \frac{c}{a^4}\right)}\) Thus, we have found the expression for the angular velocity of the precessing ellipse in terms of the given constants and the radius of the circular orbit.

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