# Kepler's Laws, mysterious equation?

Homework Helper
I got a question that depicted the elliptical orbit of a comet over two intervals of the orbit, covering sectors, with the sun at a foci. It is given that the ratio of the "radii of the sectors" are 12:60 = 1:5. It is given the sectors have equal area, what is the ratio of the speeds of the comet at the aphelion to the perihelion.

First of all, yes it would be interesting to know how to solve that. I can't see any direct application of Kepler's laws that would help. Equal areas, so equal time taken, so the ratio of the velocities will just be the ratio of the arc lengths covered, how do we find that out?

Also, on the solutions page, it automatically starts with some equation, " Area = Rv", and using that equation it easily works out the ratio to be 1:5. I am really interested to find out how they got that equation? Is that exact or just an approximation? Is it even correct?

Thanks for helping out guys.

## Answers and Replies

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tiny-tim
Homework Helper
Also, on the solutions page, it automatically starts with some equation, " Area = Rv", and using that equation it easily works out the ratio to be 1:5. I am really interested to find out how they got that equation? Is that exact or just an approximation? Is it even correct?
Hi Gib Z!

I haven't worked out the picture of the main question …

but the equation "Area/time = Rv" is true for an infinitesimal sector at perihelion and aphelion …

area/time = r(r + ∆r)∆θ/∆t ~ r²dθ/dt = rv.

area/time = r(r + ∆r)∆θ/∆t ~ r²dθ/dt = rv.
I'm not sure your equation is correct here. The orbit of the comet is not a circle. Hence in general we cannot say that rdθ/dt = v. What I understand is that in general.

$$\vec{v}= \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}$$

However if the comet is at the aphelion or perihelion, your equation is true.

tiny-tim