Kepler's Third Law

**PiRsq** · Nov15-03, 02:43 PM

How did Kepler come about to conclude that his third law is
Radius³/Period²? How did he derive this equation?

**jamesrc** · Nov15-03, 02:50 PM

I believe this law was the result of data mining through mountains of observational data collected by Tycho Brahe; I don't think he derived it at all. It was later found to be consistent with Newton's laws.

**Norman** · Nov15-03, 06:45 PM

Actually,

Kepler spent most of his life trying to derive his relationship between the period of orbit and the radius of orbit along with the rest of his laws. He never was able to and was very confounded and frustrated with it. Once Newton came along, he was able to derive Kepler's laws, which was one of the biggest tests of Newton's Laws, giving him a lot of credibility.
Cheers

selfAdjoint · Nov15-03, 07:46 PM

Kepler never knew the formula for centrifugal force (or centripetal if you prefer). This was announced by Huygens in the 1670's I believe. Several people, including Newton, combined the Kepler third law in the case of perfectly circular orbits with the centrifugal force law (i.e. the force needed to keep a planet from flying off at a tangent), and derived the inverse square law for this special case.

excercise for the viewer. Can you do what they did?

**PiRsq** · Nov15-03, 08:28 PM

Im having trouble with understanding mathematics of the equation. How can it be Radius³? Are there any books on the history of the equation?

selfAdjoint · Nov16-03, 12:22 PM

We say R³ but Kepler expressed it in proportions, which can have any relationships and don't necessarily force a gemetric image. The modern view is that the exponents on factors in a product aren't "physical", only the reduced dimensions of the whole product.

**PiRsq** · Nov16-03, 01:01 PM

So does R³ have any relationship with the fact that the planets are orbiting in 3 dimension?

**enigma** · Nov16-03, 03:19 PM

I can't explain how Kepler did it first, but I can attempt to go through a derivation. Hopefully it will be clear: I don't have the skill with the new latex tools to do graphics yet.

Given:

$LaTeX Code: \\mu = G*M$ where G is the gravitational constant and M is the mass of the attracting body.

p is the semilatus rectum, the distance from the attracting focus to the ellipse perpendicular to the direction to periapsis, and is equal to

a is the semimajor axis,

b is the semiminor axis and is equal to $LaTeX Code: \\sqrt{ap}$

e is the eccentricity,

P is the period,

$LaTeX Code: \\vec{r}$ is the position vector from the attracting body

$LaTeX Code: \\vec{v}$ is the velocity vector of the satellite

$LaTeX Code: \\nu$ is the true anomaly, the angle from the closest point (periapsis) to the position vector
$LaTeX Code: \\vec{h}$ is the angular momentum, is constant for the orbit, and is equal to: $LaTeX Code: \\vec{r} \\times \\vec{v}$ and $LaTeX Code: \\sqrt{\\mu p}$

The first steps aren't going to make sense without a picture. They take the angular momentum vector, and geometrically rearrange the terms getting:

$LaTeX Code: h = r^2 \\dot{\\nu}$

or $LaTeX Code: h = \\frac{r^2d\\nu}{dt}$ (1)

Looking at the differentially small area swept out by the r vector as the satellite moves through a differentially small distance, you get

$LaTeX Code: dA = \\frac{1}{2}r^2d\\nu$ (2)

plugging (1) into (2) and rearranging gives

$LaTeX Code: dt = \\frac{2}{h}dA$

Integrating that over a complete revolution,

$LaTeX Code: 2*\\pi$ radians of $LaTeX Code: \\nu$

gives

$LaTeX Code: P = \\frac{2\\pi a b}{h}$

where pi*a*b is the area of the ellipse.

From the geometry of the ellipse,

$LaTeX Code: b = \\sqrt{a^2(1-e^2)} = \\sqrt{ap}$

Combining that with the definitions of h, gives:

$LaTeX Code: P = 2 \\pi \\sqrt{\\frac{a^3}{\\mu}}$

rearranging once more gives:

$LaTeX Code: \\frac{a^3}{P^2} = \\frac{\\mu}{2\\pi} = constant$

If you want to look through where I got all that from with pictures:

Vallado. Fundamentals of Astrodynamics and Applications pages 24-30

...fingers crossed that the Latex worked...
EDIT: not too bad... three edits

**Janus** · Nov16-03, 03:49 PM

Originally posted by PiRsq
So does R³ have any relationship with the fact that the planets are orbiting in 3 dimension?

No. What we are talking about is a porportional realtionship between the Period of an orbit and its radius.

To show the math:

Force of gravity is

$LaTeX Code: F_{g}= \\frac {GMm}{R^2}$

M is the mass of the sun in this case, and m the mass of the planet. The R² just means that the force falls off by the square of the distance. (double the distance and the force decreases to a quarter of its strength.)
The centripetal force needed to hold the planet in a circular path is

$LaTeX Code: F_{c} = \\frac{mv^2}{R}$

For a circular orbit, these two forces are equal, so:

$LaTeX Code: \\frac {GMm}{R^2} = \\frac{mv^2}{R}$

m apears in both numerators, so they cancel out, and R appears in both demoninators, canceling out one R of R² which gives you:

$LaTeX Code: \\frac {GMm}{R} = v^2$

Taking the squareroot of each side gives you the orbital velocity of the Planet:

$LaTeX Code: \\sqrt{ \\frac {GMm}{R}} = v$

In one orbit the planet will travel the circumference of a circle with a radius of R or:

$LaTeX Code: C = 2\\pi R$

The time or period of this orbit is equal to distance/velocity or:

$LaTeX Code: P = \\frac{2\\pi R}{\\sqrt{ \\frac {GMm}{R}}}$

Rearranged:

$LaTeX Code: P= 2\\pi \\sqrt{ \\frac{R^3}{GM}}$

Now let's say that you want to compare the periods of two different orbits at different values of R, then:

$LaTeX Code: \\frac{P_{1}}{P_{2}}<BR>= \\frac{ 2\\pi \\sqrt{ \\frac{R_{1}^3}{GM}}}<BR>{ 2\\pi \\sqrt{ \\frac{R_{2}^3}{GM}}}$

$LaTeX Code: 2\\pi$ cancels out:

$LaTeX Code: \\frac{P_{1}}{P_{2}}= \\frac{ \\sqrt{ \\frac{R_{1}^3}{GM}}}{ \\sqrt{ \\frac{R_{2}^3}{GM}}}$

The squareroots combine:

$LaTeX Code: \\frac{P_{1}}{P_{2}}=\\sqrt{ \\frac{ \\frac{R_{1}^3}{GM}}{ \\frac{R_{2}^3}{GM}}}$

The

s drop out:

$LaTeX Code: \\frac{P_{1}}{P_{2}}=\\sqrt{ \\frac{ R_{1}^3}{ R_{2}^3}}$

Square both sides:

$LaTeX Code: \\left(\\frac{P_{1}}{P_{2}}\\right)^2={ \\frac{ R_{1}^3}{ R_{2}^3}$

or
$LaTeX Code: \\frac{P_{1}^2}{P_{2}^2}={ \\frac{ R_{1}^3}{ R_{2}^3}$

Kepler's Third law.

**PiRsq** · Nov17-03, 10:25 PM

Wow those are beautiful!! Just beautiful how they fit into place! Thx guys for helping me understand.

**zeronem** · Dec5-03, 10:41 PM

I was wondering if you could derive Keplers third law

$LaTeX Code: \\frac{P_{1}^2}{P_{2}^2}={ \\frac{ R_{1}^3}{ R_{2}^3}$

by the the equations of the ellipse described in Cartesian Coordinates?

$LaTeX Code: \\frac{x^2}{a^2} + { \\frac{y^2}{b^2} = 1}$

By understanding Keplers 3rd law deals with 2 ellipses, one inner ellipse and one outer ellipse. Perhaps we could combine 2 ellipse equations described in Cartesian Coordinate to derive Keplers 3rd law.

The only thing is that, the ellipse equations in cartestian coordinates deal with two foci. Although, I dont know if that will even make a difference. Is there a possibility to derive Kepler's Third law from the Ellipse equations that describe the ellipse in Cartesian Coordinates? I just want to know this answer.

Ive actually started trying to do such derivation. But ofcrouse the ellipse equations that describe the ellipse in Cartesian Coordinates will have to be modified for its fit to be derived into Keplers 3rd law. Just tell me if there is a possibility to do such derivation before I waste alot of time trying. I've found that in order to acquire time correctly among the ellipse equations is by the understanding the velocity of the elliptical orbiting object. We would have to some how bring the velocity in through Angular Momentum, $LaTeX Code: \\ p=mvr$

selfAdjoint · Dec6-03, 10:33 AM

You can't get Kepler's third law just from the shape of an ellipse (or two ellipses). You have to involve time somehow, because it's a relation between "time taken" and "radius of orbit".

It should be possible to derive Kepler's third law from his first two, since those two plus centripetal force are sufficient to prove the inverse square central force law, and the third law follows from inverse square and centripetal force.

**zeronem** · Dec6-03, 12:11 PM

Yes, but if I can use Angular momentum - p=mvr somehow, I thus can find the Velocity. Then by Velocity I can find the time. The only thing is, I would have to possibly add something to the Ellipse Equations, or modify the equations.

EDIT MESSAGE: 2 Cartesian ellipse's have 2 focus. Perhaps I can develop the idea of only using the focus of the inner cartesian ellipse. Other then that, I would have to ofcrouse apply angular momentum for the 2 Elliptical bodies orbiting the Focus. I could also study the ellipse eccentricity and by this understand the Gravitational pull that causes the 2 elliptical orbiting bodies to orbit the Focus of a given mass. Then by this, I could determine the Mass of the Orbiting bodies. Therefore I could maybe develop a Cartesian version of Keplers Third Law. I could get the time of revolution with the Velocity in Angular Momentum, p=mvr. Also I could somehow determine the Mass of the Focus which is causing the 2 bodies to orbit elliptical by the gravitation pull; which I stated earlier I could possibly find the Gravitational force of the Focus opening the door to determining the Mass of the Focus. I would need to also find a Mathematical way to get the distances of the outer elliptical bodie from the Focus at any given point of the Outer Ellipse.

**ObsessiveMathsFreak** · Dec10-03, 08:59 AM

Kudos to anyone who can remember Kepplers

law.