Kepler’s Third Law from Newtons Law of Gravitation

Brief Explanation: This will start a series of posts regarding the derivation of Kepler’s Laws for planetary movement that I’ve been wanting to make for a while now. I’ve been reading Professor Feynman’s lectures and I’ve been wondering how to derive these laws.

This first exercise, deriving the Third Law, is actually the easier one. Nevertheless, it deals with some important concepts that we need to know.

On these proofs I’ll be taking a more mathematical formalism: I actually prefer this notation on comparison to the one physicists normally apply. But that should not make a difference at all for a reader familiarized with these notations.

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Question: Derive Kepler’s Third Law from the Universal Law of Gravitation developed by Newton.

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Comprehending: So, what does Kepler’s Third Law say? According to Wikipedia:

The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

Mathematically speaking, this means that:


If we let Τ be the orbital period and r the semi-major axis of the orbit. Given that the orbit of the Earth is approximately circular, we can say that r is the radius of this orbit. I actually want to address this question in a different post.

Now, what about the Newton’s Law of Gravitation? It can be written in the form:

Força Gravítica.png

Where G is Newton’s Constant, m1 and m2 the different masses and r their distance. So far, so good.

Another useful notion will be Newton’s Second Law, which states that the resulting force equals mass times acceleration. This can be written as:

Segunda Lei Newton.png

(Two notes here: I chose not to represent the forces as vectors because we don’t need their directions to solve this problem. But remember, a force always has a magnitude and a direction. Second, F stands for the resulting force (the sum of all forces)! This is a mistake I see a lot of people doing when approaching basic Classical Physics).

Now, all we need is to picture the problem. Imagine a planet going around a star, what forces act on the planet? The answer is quite simple:


As you can see in the picture, the only force that acts on the planet is the gravitational force. This means the gravitational force is the resulting force! This will be very helpful when solving the problem.

Another thing: what type of movement are we dealing with here? Given that we approximated the orbit to a circle, we have circular motion. This implies that the resulting force is centripetal, meaning the gravitational force is centripetal. Remember the following notions regarding centripetal acceleration:

aceleração centripeta.png

Where ω is the angular velocity and r the radius of curvature. These are some basic concepts regarding circular motion, allowing us to get an expression relating the centripetal acceleration with the period of the rotation.

We now have all we need, so let’s advance to solving the problem.

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Solving: All we need is to relate the concepts we analyzed previously. Given that the gravitational force is the resulting force, we can relate these two equations:


Where m is the mass of the planet and the mass of the star. Solving for the acceleration:

3.pngNow remember, this acceleration is centripetal! If we apply the equality we developed previously regarding centripetal acceleration and develop the expression:


And there we have it! Just for the looks of it, if we define a constant of proportionality k:


Then we have:


This means that:


We finished it! We were able to derive Kepler’s Second Law from Newton’s Law of Gravitation!

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Conclusion: This one was quite easy and straightforward, but I can assure you the next one wont be! I’ll post the derivation of Kepler’s Second Law in the next days!


2 thoughts on “Kepler’s Third Law from Newtons Law of Gravitation

  1. Pingback: Kepler’s Second Law from Newton’s Law of Gravitation – Matiphy

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