Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I was watching a Fehnman lecture on YouTube, where he used Kepler's second law as an example of something he was explaining.

He was showing geometrically why a line joining a planet and a Sun sweeps out equal areas during equal intervals of time.

I followed his explanation for when the Sun has no pull on the planet. Then, I'm just solving the area of obtuse angles.

His explanation begins here and he loses me when he says that the triangular area swept out when the Sun has a force has the same height.

I understand that both triangles have the same base, but I don't understand how to see that they're the same height.

His explanation was, "...and do they have the same altitude? Sure, because they're included between parallel lines and so they have the same altitude."

That doesn't explain it for me, and so can someone show me how it's done?

share|improve this question
1  
If you take the baseline to be one of the parallel lines, then the corresponding heights of the triangles is just the distance between the two parallel lines. (the "other" parallel line passes through the vertices of the triangles opposite the baseline). Drawing a picture should help. –  David Mitra Feb 14 '12 at 17:07
1  
@David, since the picture is in the video, it didn't appear to help the OP. The key point of confusion must be that in the first half of the argument we too the orbit to be the base of the triangles, and in the second hand we switch to considering the middle one of the sun-planet radii to be the base. –  Henning Makholm Feb 14 '12 at 17:11
    
@HenningMakholm so if the middle of the sun-planet radii is now the base, I can see that the height is the 1 second measure, since the lines are parallel. But then I'd have to show that the 'radius base' is the same as the height of the previous triangle, for the areas to be the same. That's what I don't know. –  Korgan Rivera Feb 14 '12 at 17:22
    
You know what? I just got it. I had to look at it in a different way. I was trying to see how the base of the 3nd triangle was the same as the height of the 1st triangle. But what I really had to do was see that the base of the 3rd triangle was the same as the base of the 2nd triangle. Then I could say that the 3rd triangle had the same area as the 2nd triangle, and I already knew that the 2nd triangle had the same area as the first. Done. Thanks! –  Korgan Rivera Feb 14 '12 at 17:33
add comment

1 Answer

The matter was settled in the comments, by Henning Makholm:

in the first half of the argument we too the orbit to be the base of the triangles, and in the second hand we switch to considering the middle one of the sun-planet radii to be the base.

and by the OP:

I was trying to see how the base of the 3nd triangle was the same as the height of the 1st triangle. But what I really had to do was see that the base of the 3rd triangle was the same as the base of the 2nd triangle. Then I could say that the 3rd triangle had the same area as the 2nd triangle, and I already knew that the 2nd triangle had the same area as the first.

I'll add a calculus version of this argument: placing the Sun into the origin of polar coordinates, we find that the position vector $\mathbf r$ of the planet satisfies $\ddot{\mathbf r}\times \mathbf r=0$ (force collinear to $\mathbf r$). Therefore, $$\frac{d}{dt}(\dot {\mathbf r}\times \mathbf r) = \ddot {\mathbf r}\times \mathbf r+ \dot {\mathbf r}\times \dot{\mathbf r}=0+0=0$$ that is, $\dot {\mathbf r}\times \mathbf r$ is constant. The magnitude of the latter vector is twice the rate at which the area is swept.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.