# How do you explain Kepler's third law in general terms without complex math?

I understand the first law-elliptical orbits, and the second-equal area in same time, but I need help with the third one. Note that I am not in an AP course or taking calculus at the moment so simple quadratics/cubics/CP level explanations would suffice.

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You don't need complex maths, it can modelled using real numbers only. – gerrit Mar 21 '13 at 8:54

In 1619, almost a century before Newton published his groundbreaking Principia Mathematica, Johannes Kepler made a revolutionary contribution to observational astronomy. He noticed that the square of the orbital period $(P^2)$ (the time that it takes for a planet to go around the sun) of a planet's orbit is directly proportional to the cube of the length of the semi-major axis (for almost all planetary orbits, the average distance from the sun) of that orbit, ($a^3$). In the right units, taking $a$ to be in AU (Astronomical Units) and $P$ to be in years, this relationship can be expressed:

$$P^2 = a^3.$$

Kepler's third law captures an empirical trend. It makes no claims about the nature of gravitation, or the fundamental physical forces that govern the motions of the celestial bodies—it represents a mathematical pattern that Kepler found in data. Looking for trends like these is still a big part of observational astronomy today. Kepler's 3rd law says that orbital periods and orbital distances are related—an important discovery, because, until this point, it was not widely understood or accepted that the behavior of natural phenomena could be expressed in mathematical language.

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Thank you for the background/conceptual answer. Now I can tie it with the answer from above. – brolyspikes Mar 21 '13 at 4:52

Alternatively you can use dimensional analysis. You want a time $[T]$ and the only things you have that matter are

• $G$ with dimensions $[M^{-1}] [L^3] [T^{-2}]$
• $M$ mass of the central body with dimensions $[M]$
• $a$ semi-major axis with dimensions $[L]$ (if you're not convinced that the semi-major axis is the distance that matters just think of this as some generalized average "size of the orbit" - the scaling will be the same)

If you put these together arbitrarily you get

$$G^i M^j a^k,$$

with dimension $[M^{j-i}] [L^{3i+k}] [T^{-2i}]$. Asserting that you get a time $[T^1]$ gives:

$$\begin{array}{rcl} j-i &=& 0 \\ 3i+k &=& 0 \\ -2i &=& 1 \end{array}$$

Solving these gives $i=j=-1/2$ and $k=-3i=3/2$, thus

$$T \propto \sqrt{\frac{a^3}{GM}}.$$

You can't work out the proportionality constant by this method, but this is enough already to give you Kepler's third law.

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If you don't want to solve it for a general ellipse , you can try for a circle. (special case of ellipse ). Now for centripetal force equating to gravitational force= $$mv^2 /R = GmM /R^2$$ (we use constant gravitational force )

and for total time in one rotation. $$2\pi R / v =T$$ (we use perimeter of a circle)

Now, use first to get velocity and put it into second equation $$T = 2\pi R^{3/2} /(GM)^{1/2}$$

ie. $T$ proportional to $R^{3/2}$ .

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Thanks a bunch, the math is simpler to read now. – brolyspikes Mar 21 '13 at 4:51
Physics is so beautiful , we can solve anything without including dirty maths – ABC Mar 21 '13 at 5:02
exploringnet, just a reminder that you shouldn't be telling people to upvote or accept your answers. You can remind a new user that if they liked your answer, the way to indicate that is by upvoting and/or accepting it, but don't be pushy about it. ;-) – David Z Mar 21 '13 at 6:13