Kepler's 2nd.law and calculation

[shounakbhatta]

Hello,

While dealing with Kepler's 2nd.law, when calculating the area of the triangle which sweeps out:


Area of a triangle = 1/2 base x height

Now, we know dθ = dx/r (where dx = arc length, r=radius)

So, dx= rdθ

So, area of a triangle = 1/2 base(rdθ) x height(h)

Is this correct?

Thanks

 

[mfb]

If dx is defined to be something like an "arc length in tangential direction", yes. It is not the actual arc length if the motion is not circular.

 

[K^2]

Keep in mind that Kepler's Second Law is just another way of saying that angular momentum is conserved. In other words, $\omega r^2$ is a constant. Keeping in mind that for a small section h = r, this is in agreement with what you wrote.

 

[shounakbhatta]

Thank you very much for letting me clear the confusion.

 

[Philip Wood]

Hope I'm not going to regenerate the confusion, but for an elliptical orbit there are just two places where the triangle area is given by \frac{1}{2} r\ ds = \frac{1}{2} rv\ dt, and those are the two ends of the major axis. [I'm using ds as infinitesimal portion of arc length, r as distance from focus and v as speed of orbiting body.]

At every other point, r and ds are not perpendicular to each other, so a sine or cosine factor has to be inserted.