What does the 3rd Kepler’s law say?

Hi,
today I want to do a short post about the 3rd Kepler’s law. I kind of really like it because it has very simple explanation but lot of uses at the same time.


The law goes as follow:{\frac  {T_{1}^{2}}{T_{2}^{2}}}={\frac  {a_{1}^{3}}{a_{2}^{3}}}

T stands for time and for semi-major axis of ellipse, that is basicly radius for planets since

What is semi-latus rectum?

their orbit is highly circular. The index and 2 stands for first and second object, basicly you are comparing two objects with each other though they must orbit the same body. This is very useful since you can compare anything in Solar System orbiting Sun with Earth. Why is it useful? Because Earth’s semi-major axis is 1AU and orbit lasts for 1 year which means that this fraction will disappear and you are left only with the object you want to calculate with.

Where did this even came from? The prove for this equation is very simple and basicly stands on the fact that centripetal force equals gravitational force for our orbiting object.

Fg=Fc

We can find the equations for both of these forces and from them finally get to the Kepler’s law:KeplerLaw3

Ok, before you start to freak out, this is completely easy. First line is clear, I have accidentaly indexed Fd instead of Fc because in Czech the force is called “dostředivá”.

Second line shows the forces and their equations, third canceles the mass of the orbiting body and the radius of orbit. Since v=s/t we can write it down as is shown. Also watch out because s is whole orbit so s^2=4π^2

The equation that you have in fifth and six line is also usable equation! It is more general and does not need the second orbiting body but it needs the mass of object. From this equation you can also figure out the mass of Sun which is completely amazing! (You have to watch out for the right units!)

After the small space I have divided the equation by the same one except that it works with some other object orbiting the same star (or planet..), with this step I will get easily rid of all the π, gravitational constant and mass of the center object.

Now we have the original 3rd Kepler’s law!

Dragallur

PS: in the prove we also assumed that r=a which means that planets orbit on circles not ellipses but it is accurate enough

Easy way to get excentricity

Hi,
it has been few days since I looked on some problems from astronomy contest. One of the problem was about excentricity of Pluto.


This has to do with ellipses since according to Kepler’s laws, planets are orbiting on ellipses, shapes just a little different from circles, at least when you consider their equation.

For circle equation is x2+y2=1

For ellipse equation is x2/a+y2/b=1

Sun is always the focus of the elipse, above those two points are -c,0 and c,0. For planets those ellipses are much less excentric which means that in the equation above, “a” and “b” are fairly similar.

In the problem I knew only perihelium and afelium of Pluto.

e = \frac{\varepsilon}{a}=\frac{\sqrt{a^2-b^2}}{a}

“e” is excentricity. “a” is semi-major axis. “b” is semi-minor axis. “ε” is linear excentricity (not really important).

Since the equation goes as the one above you need both semi-major and semi-minor axis to get the result. From knowing afelium and perihelium I easily got semi-major axis. To get semi-minor you must know that the distance from focal point to the top or bottom of ellipse is equal to semi-major axis, from this you can use pythagorean theorem and then all this information insert into the equation. All went right and with perihelium of 29.66 AU and afelium of 49.32 AU the excentricity is 0.246 which is just right, if excentricity is equal to 1 than it is parabola and if greater it is hyperbola.

Dragallur