I am supposed to integrate Kepler´s second Law to find T (period) as a function of R (radius of a circle). However, this is a special case where we talk about a circle and not about an ellipse. See the attachment for details.

- anonymous

- schrodinger

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- anonymous

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- anonymous

Here is what I did so far.

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- anonymous

But I have to no good reason to drop the sin when I integrate.
However I think the end is correct. But I can not justify my steps to get there.

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- anonymous

What did you get when you integrated?

- anonymous

The thing you see on the sol.jepg picture.
On the left side it will be A(T) which is a full circle because we are talking about a circular orbit so when the Period T passes the planet is where it was before and has therefore made a full circle area which is pi*r^2.
On the right side I am kind of clueless how to integrate that stuff properly.

- Xishem

Alright, if we use the equation you're using:
\[dA=\frac{1}{2}r v \sin(\alpha)dt\]Which of those quantities changes in time?

- anonymous

A and the angle alpha. r is constant because we talk about a circular orbit, and therefore the velocity is also constant (or the speed).

- Xishem

Now, I want you to be cautious. In this example, you know that the angle changes linearly in time, but you know exactly how it changes?

- Xishem

I'm curious how you derived your differential equation. Is that the equation that is given to you for Kepler's second law?

- anonymous

Yes, if you look at Kepler´s law and say that dt is very small than you can say that the area made by the radius is a triangle.

- anonymous

here is a picture of it

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- anonymous

However, I am still clueless how the angle depends on time. Is it the integral of omega (the angular velocity)?

- Xishem

I'm not sure. Alpha is going to be a function of r(t) and r(t+dt).

- anonymous

alpha=omega*t

- anonymous

That one should be helpfull...

- anonymous

than I could rewrite 0.5*v*r*sin(alpha) as 0.5*omega*sin(omega*t)
but the integral of that after dt is only
-0.5*cos(omega*t)

- Xishem

Alpha, as an angle, should be the radial angle if it's omega*t.
|dw:1352577895185:dw|

- Xishem

Because the angular velocity is defined as:
\[\omega =\frac{d \theta}{dt} \rightarrow \omega\ dt = d \theta\]

- anonymous

but that gives me:

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- Xishem

I would do it this way.
Since this is a circle, you can define the area of any arc as:
\[A=\frac{1}{2}\theta r^2\]If we look at a small piece of A, then it becomes:\[dA=\frac{1}{2}r^2d\theta\]\[dA=\frac{1}{2}r^2\omega dt\]

- anonymous

but I want to find T as a function of r. How does this help?

- Xishem

\[\int\limits_{A(0)}^{A(T)}dA=\int\limits_{0}^{T}\frac{1}{2}r\omega\ dt\]\[A(T)=\frac{1}{2}r\omega T\]\[\pi r^2=\frac{1}{2}r\omega T\]

- Xishem

Sorry, sorry. I missed an r term. One second.

- anonymous

like that?

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- Xishem

Yep. Looks okay to me.

- anonymous

It seems that we only used mathematics but not Kepler´s second law.

- Xishem

Well, Kepler's second law just states that the area swept out by the radius vector sweeps out equal areas in equal time intervals.

- anonymous

... ah so we used that to claim that omega is constant!

- Xishem

Usually, the more general form of Kepler's law that I use most is:
\[dA=\frac{1}{2}|\vec r \times d\vec r|\]No, not exactly. We could only claim that because it's a circle.

- anonymous

But than we did not use any stuff of kepler. The formula for A follows from the unit circle and the rest was mathematics.

- Xishem

It's not that. It's that we made a simplification based on our understanding of the problem.

- anonymous

Where did we simplify?

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