Question about Keplers third law.

[bloodhound]

I just have to do a simple question using his third law.

 

Can someone give me the sets of units that are compatible with the equation..

 

cos my data is given in days and meters.. and most of the stuff of the web says it should be in years and AUs

 

whats the most convinient to do it?

[swansont]

How are you writing the law? If you write it as

 

(T₁/T₂)² = (a₁/a₂)³

 

then you use the same units for numerator and denominator.

 

You can always convert to other units.

[bloodhound]

its just the standard

[math]T^2=(\frac{4\pi^2}{Gm})a^3[/math]

[Martin]

its just the standard

[math]T^2=(\frac{4\pi^2}{Gm})a^3[/math]

 

If you use that form of the law' date=' then you are asking to do it in metric units because in the books the only value of the constant G you are going to find is G expressed in metric units. If you have G in metric then the other stuff should all be in metric, or it gets complicated.

 

Kepler orig. stated the law more like way Swanson put it:

 

How are you writing the law? If you write it as

 

(T₁/T₂)² = (a₁/a₂)³

 

then you use the same units for numerator and denominator.

 

You can always convert to other units.

 

Bloodhound look at Swanson version-----if T_2 is a year, and if a_2 is the AU (that is T_2 is the orbital period of one planet, the earth, and a_2 is the semimajoraxis or orbital "radius" of that planet)

 

then a_1/a_2 is the semimajor of the OTHER planet expressed in AU

 

and T_1/T_2 is the period of the OTHER expressed in years

 

so the whole thing becomes ridiculously simple

 

 

--------

from what you call the standard version, you can PROVE the Swanson form of it

 

just write the standard version twice, once sub-1 and once sub-2,

and then divide one equation by the other

and the G shit and pi shit will cancel out and you will have the clean

form that Swanson wrote

 

------------

and that is essentially same as what Kepler wrote in 1618 in his book

Harmonice Mundi

 

because in those days there was no constant G

[Martin]

In his 1618 book, Kepler originally expressed the law using the Latin word "sesquipotentia" which means "three-halves power"

 

for any two planets, the ratio of the periods is equal to the 3/2 power of the ratio of their distances from the sun (their semimajors or approx. avg. dist from sun)

 

4^3/2 = 8

 

so if planet-one is 4 times farther out than planet-two

then its orbitperiod is 8 times longer

 

9^3/2 = 27

 

so if planet-one is 9 times farther out than planet-two

then its orbitperiod is 27 times longer

 

and it actually works (except for slight differences due to different masses of the planets, which you can almost overlook)

 

it actually works----imagine being the first person to notice that

[bloodhound]

bloody hell, didnt need so much info, but thanks anyway mate.