# Kepler's law questions

1. Oct 11, 2014

### mh8780

Hello I am studying astro independently and have a question on keplers 3rd law and my math.
I set myself with a few givens with Kepler's laws
My first question is, I remember reading M had to be in solar masses, is this true?
For p^2=4π^2/G(M1+M2)/ a^3
Given: a= 1 AU

m1= .31 M☉(Used Gliese 581)
m2= .00095 M☉ (Used Jupiter)
We all know the gravitional constant is 6.67*10^-11
I plug in everything and get

39.47/6.67*10^-11(.31+.00095) *1^3

39.47/6.67*10^-11(.31095) *1^3

39.47/11.44031908×10^12
This is where I always screw up. In my notes I got 10.31065 I don't know how but after this I did:
39.47/10.31065
3.828
√p^2=√3.828
p=1.95 Years

Otherwise I would have got 12000132.94

Also when do I use p^2=a^3/(M1+M2) ?

Last edited: Oct 11, 2014
2. Oct 11, 2014

### TurtleMeister

1) You can use SI units but then the period will not be in years of course.
2) Your equation for Kepler's third law is wrong.

3. Oct 11, 2014

### SteamKing

Staff Emeritus
Not unless you use G expressed in solar masses and 'a' expressed in AUs. Look up the units associated with the value of G in your calculations, and use consistent units accordingly for the other quantities in the formula.

http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6

It's always a good idea to show units in your calculations. It avoids a lot of confusion in figuring out what your calculation means.

4. Oct 14, 2014

### dean barry

Keplers rule disregards the planet masses, the k value being the same for all (negligable mass) planets.
k = p ² / a ³
No good in this case.
You have given both masses and the distance inbetween.
Ive attached the two body data sheet to follow, finding the orbit time for either body (which is the same for both)
can be found.

The derived equation for finding t directly is more involved :

t = square root ( ( 4 * π² * ( M2 / ( M1 + M2 ) ) * d³ ) / ( G * M2 ) )
or
t = square root ( ( 4 * π² * ( M1 / ( M1 + M2 ) ) * d³ ) / ( G * M1 ) )

Using your data, i get :
t = 1.794 years

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