G.MATH
CALCULUS
Kepler's Second Law

We start our journey into Kepler’s laws with his second one reason being that this is the easiest of the three to prove. Remember that we are using a purely mathematical argument here with perhaps the only physics concept borrowed is Newton’s universal law of gravitation.
VIDEO LECTURE


MAIN CONCEPTS
The line segment joining a planet to the sun sweeps out equal areas in equal times or
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LESSON
Recall that by Newton’s universal law of gravitation, the particle m moving around the orbit experiences a force, , directed along the line segment from m to M.  is called a central force and since is directed along the line segment it has no  component, or . Thus using the appropriate equations of motion,

Knowing that the second law involves time, we aim to get dt on one side of the equation. It takes a great deal of intuition to see that on multiplying through by r, we get

which is the derivative of a certain expression namely,

Wow! Who would have thought. Just differentiate w.r.t t carefully using chain rule and product rule if you need further convincing. We can how integrate w.r.t t giving us

for some constant h. We assume that h is positive implying is positive which means m is moving around the centre of orbit in a counterclockwise direction. Further point to know, the equation above is vital in proving Kepler’s first law. Remember it.

Now since we are concern with the area swept out by mass m, we let  where  is the area swept out by  from some fixed position. We can now define a small area swept out as

And by integrative this from  to , we get

This represent Kepler’s second law stated as the line segment joining the sun to a planet sweeps out equal areas in equal time intervals, i.e, so long  is the same of both sweeps.

The same diagram has been place here for easy reference.


 

All information presentated, less questions and exercises, is original content of Donny, with slight references to various books.
Video courtesy of YouTube.com service.
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