anonymous
  • anonymous
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.
Physics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
Here is what I did so far.
1 Attachment
anonymous
  • 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
  • anonymous
What did you get when you integrated?
anonymous
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
here is a picture of it
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anonymous
  • anonymous
However, I am still clueless how the angle depends on time. Is it the integral of omega (the angular velocity)?
Xishem
  • Xishem
I'm not sure. Alpha is going to be a function of r(t) and r(t+dt).
anonymous
  • anonymous
alpha=omega*t
anonymous
  • anonymous
That one should be helpfull...
anonymous
  • 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
  • Xishem
Alpha, as an angle, should be the radial angle if it's omega*t. |dw:1352577895185:dw|
Xishem
  • Xishem
Because the angular velocity is defined as: \[\omega =\frac{d \theta}{dt} \rightarrow \omega\ dt = d \theta\]
anonymous
  • anonymous
but that gives me:
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Xishem
  • 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
  • anonymous
but I want to find T as a function of r. How does this help?
Xishem
  • 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
  • Xishem
Sorry, sorry. I missed an r term. One second.
anonymous
  • anonymous
like that?
1 Attachment
Xishem
  • Xishem
Yep. Looks okay to me.
anonymous
  • anonymous
It seems that we only used mathematics but not Kepler´s second law.
Xishem
  • 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
  • anonymous
... ah so we used that to claim that omega is constant!
Xishem
  • 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
  • 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
  • Xishem
It's not that. It's that we made a simplification based on our understanding of the problem.
anonymous
  • anonymous
Where did we simplify?

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