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area of sector of circle/ellipse (that the planet sweeps out): (1/2)(r^2)O

O is theta!

dA/dO = (1/2) (r^2)

dA= (1/2) (r^2) dO

dA/dt = (1/2) (r^2) dO/dt

It can't be that simple???? Can it??

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- Thread starter Master J
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- #1

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area of sector of circle/ellipse (that the planet sweeps out): (1/2)(r^2)O

O is theta!

dA/dO = (1/2) (r^2)

dA= (1/2) (r^2) dO

dA/dt = (1/2) (r^2) dO/dt

It can't be that simple???? Can it??

- #2

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nasu

Gold Member

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Is my proof of Kepler's 2nd law correct?

dA= (1/2) (r^2) dO

dA/dt = (1/2) (r^2) dO/dt

It can't be that simple???? Can it??

Is not. The derivative in respect to time is incomplete. The radius is not a constant unless is circular motion and then the problem is trivial anyway.

And the first three equations are a little bit redundant.

You have dA=1/2r^2*d(Theta) to start with. You cannot write the area itself this way (as a triangular segment) but only an infinitely small segment of area, dA.

Then you can take the the time derivative but both r and theta are time dependent.

Kepler's 2-nd law is a consequence of Newton's laws in the case of a central force. The above formula - for areal speed - is valid for any kind of motion, with any force. So it cannot give Kepler's law unless you introduce the specific force.

It's much easier to start with conservation of angular momentum - a consequence of central force motion.

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