Question

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.

Answer 1

Here is what I did so far.

Answer 2

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.

Answer 3

What did you get when you integrated?

Answer 4

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.

Answer 5

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?

Answer 6

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).

Answer 7

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?

Answer 8

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

Answer 9

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.

Answer 10

here is a picture of it

Answer 11

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

Answer 12

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

Answer 13

alpha=omega*t

Answer 14

That one should be helpfull...

Answer 15

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)

Answer 16

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

Answer 17

Because the angular velocity is defined as:

\[\omega =\frac{d \theta}{dt} \rightarrow \omega\ dt = d \theta\]

Answer 18

but that gives me:

Answer 19

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:\[dA=\frac{1}{2}r^2d\theta\]\[dA=\frac{1}{2}r^2\omega dt\]

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

like that?

Answer 24

Yep. Looks okay to me.

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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.

Answer 29

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

Answer 30

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

Answer 31

Where did we simplify?