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TomLikesPhysics
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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.
 2 years ago
 2 years ago
TomLikesPhysics Group Title
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.
 2 years ago
 2 years ago

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TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Here is what I did so far.
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

ivanmlerner Group TitleBest ResponseYou've already chosen the best response.0
What did you get when you integrated?
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
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?
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
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).
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
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?
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
I'm curious how you derived your differential equation. Is that the equation that is given to you for Kepler's second law?
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
here is a picture of it
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
However, I am still clueless how the angle depends on time. Is it the integral of omega (the angular velocity)?
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
I'm not sure. Alpha is going to be a function of r(t) and r(t+dt).
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
alpha=omega*t
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
That one should be helpfull...
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
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)
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
Alpha, as an angle, should be the radial angle if it's omega*t. dw:1352577895185:dw
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
Because the angular velocity is defined as: \[\omega =\frac{d \theta}{dt} \rightarrow \omega\ dt = d \theta\]
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
but that gives me:
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
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:\[dA=\frac{1}{2}r^2d\theta\]\[dA=\frac{1}{2}r^2\omega dt\]
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
but I want to find T as a function of r. How does this help?
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits_{A(0)}^{A(T)}dA=\int\limits_{0}^{T}\frac{1}{2}r\omega\ dt\]\[A(T)=\frac{1}{2}r\omega T\]\[\pi r^2=\frac{1}{2}r\omega T\]
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
Sorry, sorry. I missed an r term. One second.
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
like that?
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
Yep. Looks okay to me.
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
It seems that we only used mathematics but not Kepler´s second law.
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
Well, Kepler's second law just states that the area swept out by the radius vector sweeps out equal areas in equal time intervals.
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
... ah so we used that to claim that omega is constant!
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
But than we did not use any stuff of kepler. The formula for A follows from the unit circle and the rest was mathematics.
 2 years ago

Xishem Group TitleBest ResponseYou've already chosen the best response.1
It's not that. It's that we made a simplification based on our understanding of the problem.
 2 years ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Where did we simplify?
 2 years ago
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