Angular and orbital speed at perihelion

  • #1
pobro44
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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
246347

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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 :).
 
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Answers and Replies

  • #2
kuruman
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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.
 
  • #3
pobro44
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Thank you kuruman, I also derived a general expression
246375

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)
246378

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.:wideeyed:
 

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  • #4
kuruman
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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?
 

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