# Orbital motion with varying radius

The model of a orbital motion in many Physics textbooks and online resources (e.g., Wikipedia on circular motion) assumes that the orbit is a circle, i.e. the radius is constant and the speed is constant, when deriving the many relationships such as $v = R \omega$ and $a_{\text{centripetal}} = R\omega^2 = \frac{v^2}{R}$.

Would the many relationships hold when the orbit's radius is no longer constant but varies with time, for example?

I argue that the many relationships would not hold because the model from which the relationships are derived does not take into account the change in the orbit's radius.

If my argument is wrong, what is the logical basis to say that the relationships still hold even when the model does not take into account the change in the orbit's radius?

If my argument is right, any pointer to a better model that takes into account the change in the orbit's radius?

• The reason introductory textbook only look at the circular case is the more difficult math needed for a complete answer to the non-circular case. We usually teach the full version in an upper-division course that assume the students have had both calculus and differential equations. Some of the relationship from the circular case still hold absolutely (equal area AKA the conservation of angular momentum), other apply only locally and still others are must be modified. – dmckee --- ex-moderator kitten Apr 6 '14 at 14:15
• @dmckee Perhaps you have a name for the full version that I can google? – Tadeus Prastowo Apr 6 '14 at 15:30
• Any text called "Classical Mechanics" or "Classical Dynamics" should have it. I use Marion&Thornton at the upper-division level and learned from Goldstein in grad-school. – dmckee --- ex-moderator kitten Apr 6 '14 at 15:38
• So, is it true that I cannot conclude the following relationship between $F_\text{tangential}$ and the varying radius $\frac{dR}{dt}$ of a circular motion depicted in !a circular motion with varying radius because the model assumes a constant radius R, and therefore, I need more advanced analytical tools? $F_\text{tangential} = m a_\text{tangential} = m R \alpha = m R \frac{d\omega}{dt} = m R \frac{\omega_f-\omega_i}{dt} = \frac{m R \omega_f - m R \omega_i}{dt} = \frac{m R_f \omega - m R_i \omega}{dt} = m \omega \frac{R_f - R_i}{dt} = m \omega \frac{dR}{dt}$ – Tadeus Prastowo Apr 6 '14 at 16:41
• You are trying to work the problem the hard way. Neglecting external forces on the body, the energy and angular momentum of the system are conserved and applying these principles is often the easiest way to understand specific questions about orbits. If that won't cut it you generally have to perform the full solution. – dmckee --- ex-moderator kitten Apr 6 '14 at 17:01

The two relations that you give, are strictly valid only for circulation motion. You give an example of circular motion with variable radius, but, that is not circular motion anymore. Also, how would you define the centre of the motion, and thus angular velocity or cetripetal accelaration?

If you have a well-defined case, you can always derive similar relations. You then need to start from the basic principles that are always for classical mechanics, such as conservation of momentum, and conservation of angular momentum. In circular motion, there is only one force, directed towards the centre of the circle. For your case, you will have to make a free body diagram, and choose a smart origin.

• I am trying to analyze the experiment described by Tom Young in this article. Empirically I came up with this depiction of the motion with a varying radius !A circular motion with a varying radius. I was about to use the relationships in my Physics textbook about circular motion. But, I realized that the model used in the textbook was different. I was thinking to extend the model to cater to varying radius but found no logical basis to do it. Should I go with more advanced mechanical analysis tools? – Tadeus Prastowo Apr 6 '14 at 15:58

Take for example the equation for the velocity:

$$v = r\omega$$

In circular motion the velocity is always normal to the radius. In non-circular motion, e.g. an elliptical or hyperbolic orbit, the velocity is not normal to the line joining the object to the focus of the orbit. However the velocity can be expressed as a vector sum of a normal, $r_\perp$, and radial, $r_\parallel$ component. The radial component is obviously unrelated to the angular velocity because it can take any value and the angular velocity due to it would still be zero. Therefore only the normal component affects the angular velocity and we define:

$$v = r_\perp\omega$$

Likewise for acceleration, which can be split into normal and radial components.

It depend on which "many relationships" you are referring to. A better model would be a Kepler orbit with an eccentricity ($\epsilon$, a measure of how elliptical an orbit is) smaller than one. This model describes a two body problem. But the downside of this model is that it can not explicitly be expressed as a function of time. Namely you would have to solve Kepler's equation: $$M=E+\epsilon\sin{E}$$

• For the motion depicted here !A circular motion with a varying radius, I argue that the relationships given about circular motion in basic Physics textbooks cannot describe the transition from the circle of radius R' to the circle of radius R and then back again. That is, can I say the following? $F_\text{tangential} = m a_\text{tangential} = m \frac{dv}{dt} = m \frac{R d\omega}{dt} = m R \frac{\omega_f-\omega_i}{dt} = \frac{m R \omega_f - m R \omega_i}{dt} = \frac{m R_f \omega - m R_i \omega}{dt} = m \omega \frac{R_f - R_i}{dt} = m \omega \frac{dR}{dt}$ – Tadeus Prastowo Apr 6 '14 at 16:16
• Sorry, I wrote the first three terms incorrectly, the equation should be the following one (can I really conclude that $F_\text{tangential} = m\omega\frac{dR}{dt}$)? $F_\text{tangential} = m a_\text{tangential} = m R \alpha = m R \frac{d\omega}{dt} = m R \frac{\omega_f-\omega_i}{dt} = \frac{m R \omega_f - m R \omega_i}{dt} = \frac{m R_f \omega - m R_i \omega}{dt} = m \omega \frac{R_f - R_i}{dt} = m \omega \frac{dR}{dt}$ – Tadeus Prastowo Apr 6 '14 at 16:26