Orbital speed variation as a planet orbits the Sun

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tempack
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Summary:: At what distance from the Sun will the speed of the planet be equal to the average orbital speed?

I'm not sure where to place this question, please move it in the right thread.

[Mentor Note -- thread moved from the technical forums, so no Homework Template is shown]

At what distance "r" from the sun will the speed of the planet be equal to the average orbital speed? If it is possible, I need a formula, where the Perimeter of the ellipsoid is not involved as a parameter. If you have a link with more details, then even better.

Great thnx
 
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tempack said:
Summary:: At what distance from the Sun will the speed of the planet be equal to the average orbital speed?

I'm not sure where to place this question, please move it in the right thread.At what distance "r" from the sun will the speed of the planet be equal to the average orbital speed? If it is possible, I need a formula, where the Perimeter of the ellipsoid is not involved as a parameter. If you have a link with more details, then even better.

Great thnx
Can't you just get that from Kepler's laws and conservation of energy?

Except, of course, the perimeter of an ellipse is not so easy to calculate. What's the relevance of average speed?
 
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tempack said:
At what distance "r" from the sun will the speed of the planet be equal to the average orbital speed?
Just to be clear, you mean its own mean orbital speed? Wikipedia quotes (without derivation, unfortunately, but with a reference) a series for the mean orbital speed and an exact expression for orbital speed as a function of radius.
 
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The total energy of the orbiting system is ##E = -\frac{GMm}{2a}## (for a quick way in, look at the Virial theorem), and we have$$ -\frac{GMm}{2a} = \frac{1}{2}mv^2 - \frac{GMm}{r}$$That gives you the Vis-viva equation$$v = \sqrt{GM \left (\frac{2}{r} - \frac{1}{a} \right)}$$I don't know how to calculate the perimeter of an ellipse but if someone more clever knows how to do that then you can just divide by ##\frac{2\pi}{\sqrt{GM}}a^{\frac{3}{2}}##.

Maybe there is a better way that avoids having to approximate the perimeter :wink:
 
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Sorry for my english, I'm using google translator.
I understand so:
Mean orbital speed Vm=P/T, where P - perimeter, T - Perioud. For my problem I need Vm=Vi, where Vi - instantaneous orbital speed

I need an alternative way to get "r", by ex from Kepler's law.

PS: There I'm nub...
 
Sorry, "Perimeter" = ellipse circumference
"Perioud"= orbital perioud
@etotheipi, thnx for your answer, but this method I know. I need an other way