Question

A particle of mass m moves under an attractive central force F(r) = -Kr4 with angular momentum L. For what energy will the motion be circular, and what is the radius of the circle? Find the frequency of radial oscillations in the particle is given a small radial impulse.

Answer 1

Answer:

Angular velocity is same as frequency of oscillation in this case.

ω = $\sqrt{\frac{7K}{m} }$ x $[\frac{L^{2}}{mK}]^{3/14}$

Explanation:

- write the equation F(r) = -K$r^{4}$ with angular momentum L

- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.

- Write the energy of the orbit in relative to r = 0, and solve for "E".

- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.

- Solve for effective potential

- ω = $\sqrt{\frac{7K}{m} }$ x $[\frac{L^{2}}{mK}]^{3/14}$