Keplers Laws – n-Body Problem

Astrodynamics, Astrophysics, Engineering, Math, Physics

Things Always Come in 3’s

If two wrongs don’t make a right, try three.

-Laurence J. Peter  

Lot’s of things come in threes, there’s even a whole website devoted to three’s. Is there a reason we’re fixated on groupings of  three? Probably, but that seems like a psychology question and this post is about orbital dynamics. The most common set of three’s in dynamics are Newton’s three laws of motion. These 3 laws govern the movement of objects in the universe (as long as they don’t get too small, too large, too slow, or too fast). Another less famous set of three laws governs the movement of the planets. These are known as Kepler’s 3 laws of planetary motion.

  1. The orbit of a planet is an ellipse with the sun at one of the two focal points
  2. A line segment from the sun to the planet sweeps out equal areas in equal lengths of time
  3. The square of the planets orbital period is proportional to the cube root of the semi major axis of it’s orbit

Kepler’s First Law

The orbit of a planet is an elipse with the sun at one of the two focal points

The Sun is massive. It contains about 99.8% of the mass in our solar system. Because it’s so massive, it provides almost all of the force acting on the planets. This means they are well modeled as 2-body systems. If you’ve read my past posts on 2 body systems (place lings to old posts) you’ll know that in 2-body systems there can be one of three solutions. An elliptic orbit, a parabolic orbit, or a hyperbolic orbit. Only the elliptic orbit is stable and periodic.  This law can be generalized to any two bodies orbit around their combined barycenter. Hint: If you want to prove that for yourself, which you should, take the derivative of angular momentum of a two body system and see what it equals. Here’s a solution so you can check your work.

Kepler’s Second Law

A line segment from the sun to the planet sweeps out equal areas in equal lengths of time

Imagine you had a really long slinky. Take one end and tie it to the center of the sun. Take the other end and tie it to the center of the Earth. If you were to measure the area it swept out over a set period of time, say one hour, wait a half a day and then measure the area it swept out over the same period of time, you would get the same number as your first measurement. If you’re having trouble visualizing the setup the image below, Figure 1, might help.

 

InkedKeplers second law 2.jpg

Figure 1 : Kepler’s second law diagram

Now if the orbit was circular that would make intuitive sense but the orbits of planets are elliptical. This would mean that the planet was adjusting it’s speed as it traveled through it’s orbit. That seems kinda weird at first but lets see what the math says about it.

Note: did you do the derivation from Kepler first law? If you haven’t, you should go and try.

From the post on orbital constants you may remember we began by left crossing the acceleration vector by the radius vector. That led us to the following

\boldsymbol{h}=\boldsymbol{r}\times\boldsymbol{v}

Now try to derive the radius and velocity vectors in polar coordinates. Don’t worry, I’ll wait. Once you’ve derived them and placed them into the above equation, we get

\boldsymbol{h} = r^2\frac{d\theta}{dt}\boldsymbol{i_z}

Because h is only in the z direction we can convert this to the following scalar equation and multiply both sides by 2

2h=2r^2\frac{d\theta}{dt}

The right side of the equation gives us the area that a radius vector covers in a small period of time and the left side is constant. This is a verification of Kepler second law.

Now that we’ve shown it comes from the math why does this happen in reality? Because energy is conserved. As the planet gets closer to the sun it trades its potential energy for kinetic energy. Similar to how a marble would roll down the inside of a bowl, sans friction, picking up speed until it got to the bottom where it would be going the fastest. then the marble would roll up the other side slowing down till it traded the last bit of its kinetic energy for potential energy and repeat the cycle indefinitely

Kepler’s Third Law

The square of the planets orbital period is proportional to the cube root of the semi-major axis of it’s orbit

The wording in keplers third law is a little confusing but can be summed up quite enatly 

Kepler's_third_law

Figure 2 : Our universe showing Kepler third law. http://schools.wikia.com/wiki/File:Kepler%27s_third_law.gif

in the following equation

P^3 = \frac{4\pi^2}{G(m_s+m_p)}a^3

Where P is the planets orbital period, a is it’s semi major axis, Ms and Mp are the masses of the sun and the planet respectively, and G is the universal gravitational constant. The farther out a planet is, the longer it takes for it to revolve around the sun.

Now let’s prove it. We begin with an Kepler second law

r^2 \frac{d\theta}{dt}=h

If we integrate this over one whole orbit, remember the area of an ellipse is πab, and denote the period by P we get

\frac{2\pi a b}{P}=h

By swapping in the definitions and doing some algebra we can obtain the following

P = 2\pi \sqrt{\frac{a^3}{\mu}}

Note: This small proof only works if the mass of the smaller object is much smaller than the mass of the larger object, like with satellites orbiting earth or the earth orbiting the sun.  If the two bodies are of relative sizes we  can use the following equation

P^2 =\frac{2\pi^2}{G(M_1+M_2)}a^3 

Now, P is a constant that we got from integration so that’s a useful term to describe our orbit size. Another constant related to period is the mean motion, n. It’s defined as the following

n = \frac{2\pi}{P}

From this we can write newtons third law of motion as

\mu = n^2a^3

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