Kepler’s Second Law from Newton’s Law of Gravitation

Brief Explanation: Today we continue our series of posts regarding the derivation of Kepler’s Laws from Newton’s Law of Gravitation. You can see the derivation of Kepler’s Third Law here. Our problem today will be a little more complicated but it is also quite elegant.

Just realized this demonstration (that I learned in my calculus class) is presented in the book Differential Geometry of Curves and Surfaces by Manfredo Perdigão do Carmo. If you are interested about the mathematical description of curves check out this book!

— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — —

Question: Derive Kepler’s Second Law from Newton’s Law of Gravitation.

— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — —

Again, let’s consult Wikipedia for the statement of the law:

A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

The first thing we’ll have to do is to visualize what they mean by these segments and areas.


So, we have a line joining the Sun and the Planet. If we look at that picture later in time, the planet will have moved relative to the Sun, and we have another vector joining them… And an area defined by these two vectors:

2 copy.png

That area is represented in gray in the picture. The law says that if we take equal times (one month, for example) the area swept by these vectors will be the same. In conclusion, the Earth covers as much area from the 1st of January to the 31st of January as in the 1st of August to the 31st of August. This law seems very intuitive but remember that the orbit of the planets are shaped like an elipse! Given that the difference between the axis of the elipse are small, we can approximate the orbit to a circle.

Now, how do we prove this area swept has a constant rate? The expression ‘rate’ should ring an immediate bell: taking derivatives. One way of doing it is proving the derivative is constant!

How de we calculate the area of a circular section whose angle is increasing? The area of this section is given by:

Area secção circular.png

(click here if you don’t know where this formula comes from)

If we divide the sector in infinitesimal parts, each one with an angle of , the area of the sector is given by:

area secção circular integral.png

All we need is to take the derivative of this area. We’ll deal with that later.

Another thing we need to visualize is the gravitational force exerted by the Sun on the planet. This is the picture we used last time:


The gravitational force is centripetal (directed to the center). This is a very important fact because the acceleration is also centripetal! This happens because:

Segunda Lei Newton Vector.png

Force is a vector (acceleration) multiplied by a scalar, this means that the force and acceleration have the same direction. Because the acceleration only has a centripetal component (the tangential component is zero), our derivation will be much simpler.

— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — —


First, the derivative of the area of the circular sector. If we take the derivative with respect to time of the expression we had above:


(Sorry if the calculus part is quite hard. I used the chain rule and the Fundamental Theorem of Calculus to solve this one)

All we need is to prove that last part is constant! How do we do it? We derive the expression of the acceleration of a general curve. This part is quite tricky and there is no problem if you skip it!

Let’s define a general curve. The position x(t) is given by:


We transformed the curve into polar coordinates. The vectors ui and uj represent the unit vectors.

If we call the stuff inside the parenthesis ur, we get the expression:


Please take a time to understand why these two following expressions are this way. Remember what ur is:


The two components ur and uθ are perpendicular. They are the normal and tangential components of movement.

Now, let’s derive the velocity and after the acceleration. The velocity is given by:


(This is just the chain rule of derivation applied)

Now, the acceleration:


(I’ve simplified these derivations to show only the result. If you want to find out why this is, tell me and I’ll write a post about it or check the book I talked about in the beginning of the post.)

Now, we know the tangential component is zero! This corresponds to the uθ part. If we work this part of the expression we get:


(This is some clever manipulation of the expression, but quite hard in my opinion)

This expression corresponds to the tangencial acceleration, so we can make it equal to zero. Doing this implies the following:


(Remember that the derivative of a constant is zero.)

Now, returning to the rate of the area swept:


So we proved the rate of the area swept by the position vector is constant! Our exercise ends here.

— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — —

Conclusion: I personally find this exercise quite hard, and I’m not sure this is the smartest way of getting to Kepler’s Second Law. If I find another way of proving this I’ll post it here!

Tell me your thoughts about this one!


3 thoughts on “Kepler’s Second Law from Newton’s Law of Gravitation

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s