Angular and orbital speed at perihelion

[pobro44]

Homework Statement

I want to derive how are angular and orbital speeds related in perihelion of eliptical orbit

Relevant Equations

Angular momentum of reduced body in polar coordinates

Hello to all good people of physics forums. I just wanted to ask, whether the angular and linear (orbital) speed in perihelion of eliptical orbit are related the same way as in circular orbit (v = rw). If we take a look at the angular momentum (in polar coordinates) of reduced body moving in eliptical orbit

if we equate the underlined factors and solve for v_theta, we get v_theta = rw. As I understand v_theta is a component of velocity perpendicular to position vector, and generaly is not equal to orbital velocity, but it is in perihelion/aphelion.

Is this reasoning correct? Thank you for taking the time to read and respond :).

 

[kuruman]

It's simpler than that. At perihelion/aphelion the radial component of the velocity vector is zero by definition. So the velocity is perpendicular to the radial direction at these two points.

 

[pobro44]

Thank you kuruman, I also derived a general expression

where v is the orbital (tangential) velocity, and theta is the angle between velocity and position vector.

I derived it by using the magnitude of angular momentum of reduced body (one body problem)

and equating that expression with general one for angular momentum L = μrvsinθ and solving for ω. In polar coordinates, vsinθ is actually speed in theta hat direction, and in perihelion/aphelion theta is 90 degrees and angular speed becomes ω = v/r. However, my professor claimed this was wrong, and if I remember correctly, that I can't equate those two expression for angular momentum, but I can't figure out why.

 

[kuruman]

The starting point for considerations of this kind is that angular momentum is conserved because the orbiting mass is moving under the influence of a central force which can exert no torque. This means that at all points on the orbit $\vec L =\vec r \times \vec p=const.$ The magnitude of the cross product is maximum when the linear momentum vector $\vec p$ is perpendicular to the position vector $\vec r$. This occurs at perihelion and aphelion. The magnitude of the constant angular momentum is commonly calculated at either one of these points, $|\vec L|=r_a~p_a=r_p~p_p$ where subscripts "a" and "p" stand respectively for aphelion and perihelion. If you want to bring in $\omega$ through $|\vec L|=\mu \omega r^2$, you would have to write $\omega_a r_a^2=\omega_p r_p^2$. The expression $\omega=v/r$ is better written as $\omega_a=v_a/r_a$ at aphelion or $\omega_p=v_p/r_p$ at perihelion. At other points on the orbit a sine will be required.

I cannot speak for your professor, but I think his/her objection is that writing $\omega=v/r$ is meaningless and misleading because $\omega$ varies along the orbit and it is equal to the ratio of the speed to the distance only at aphelion and perihelion with the understanding that $\omega_a \neq \omega_p$. So how useful is this expression?