Lesson 34: Kepler's Three Laws of Planetary Motion

Kepler took the data that Brahe had spent his life collecting and used it (especially the information on Mars) to create three laws that apply to any object that is orbiting something else.

Kepler’s First Law

Kepler’s First Law went against the major assumption that scientists of the time had about orbits… in fact it is probably against the image of orbits that you have!

“The path of any object in an orbit follows the shape of an ellipse,
with the orbited body at one of the foci.”

So what does all that mean?

Figure 1

Remember that the object being orbited sits at one focus, and the other object follows the path of the ellipse.

Figure 2

The diagrams I have drawn here are exaggerated quite a bit to show the elliptical shape and focus clearly.

Newton was able to show that his laws of gravitation gave the same results as Kepler's. In fact, Newton was able to take things farther with his strong math background to show that the shape of the orbits were conic sections (for those of you that have seen that stuff in math). We'll be looking more at Newton's contributions in a lesson coming up.

Kepler’s Second Law

Kepler’s Second Law is based on the speed of the object as it orbits.

Kepler didn’t talk about speed when he wrote out his second law. Instead, he looked at a mathematical detail that pops out because we are talking about ellipses.

“An imaginary line from the sun to the planet
sweeps out equal areas in equal times.”

If we were to look at the area the Earth sweeps out in a 15 day period, first when close to the sun (Figure 3) and then when far away (Figure 4 ), we would get diagrams that look like this.

Figure 3

Figure 4

Notice how in Figure 3 we have a stubby, fat, (basically) triangular area that was swept out by the line, but in Figure 4 we have a tall, thin, (basically) triangular area swept out.

Kepler’s Third Law

The big mathematical accomplishment for Kepler is in his Third Law, where he relates the radius of an orbit to it’s period of orbit (the time it takes to complete one orbit).

“The square of the period of orbit,
divided by the cube of the radius of the orbit,
is equal to a constant (Kepler’s Constant) for that one object being orbited.”

The formula looks like this...

T = period (in any unit)
r = radius (in any unit)
K = Kepler's Constant

There's a few weird things about this formula compared to many other physics formulas:

Figure 5

Example 1: Using the values for Kepler’s Constant that were calculated above for Earth, Mars, and Jupiter orbiting the sun…

a) determine the average Kepler’s Constant for anything orbiting our sun.

b) Neptune has an average orbit of 4.5e12 m from the sun. Determine how long it takes to complete one orbit.

a) The average of the three values is 3.99e-29.

b) T2 = Kr3 = (3.99e-29) (4.5e12m)3 = 6.0e4 days = 165 years!

It takes Neptune about 165 years to go once around the sun!!!

You can also write out this formula a couple of other different ways. They are just ways of rearranging things.

e -> values for Earth
m -> values for Mars

Example 2: If the orbit of Mars is 1.52 times greater than the orbit of Earth, determine how much time it takes Mars to complete one orbit.

If I was able to look up the orbit of Earth in a book, this would be a fast question. All I would do is multiply that number by 1.52 to get the orbit of Mars, and then I would have both the radii I need. But what if I was doing this question on an exam and I didn’t know the value for Earth’s orbit.

Note: Do not use the “Radius of Earth” number on your data sheet. That is the distance from the centre of the Earth to the surface of the Earth, not the size of Earth’s orbit.

Here’s what I do know…

rm = 1.52 re

Te = 365 days

Let’s see if we can substitute that into the formula…

Tm = 684 days