Kepler's 3rd Law: Solving Questions & Math

[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 gravitational 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 12000132.94Also when do I use p^2=a^3/(M1+M2) ?

 

[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.

 

[SteamKing]

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?

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

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 gravitational 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 12000132.94Also when do I use p^2=a^3/(M1+M2) ?

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.

 

[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

 

[pmacias]

Hi there,

First of all, great job on studying astronomy independently and working through Kepler's 3rd law! It's a fundamental law in understanding the motion of planets and other celestial objects.

To answer your first question, yes, M in Kepler's 3rd law refers to the masses of the two objects in solar masses. This is because the law was originally formulated using the masses of the Sun and planets, which are typically measured in solar masses.

Now, let's address your calculations. It looks like you have the correct equation for Kepler's 3rd law, which relates the orbital period (p) to the semi-major axis (a) and the masses of the two objects (M1 and M2). However, it seems like you may have made a few mistakes in your calculations.

First, when plugging in the values for a, M1, M2, and G, make sure to use parentheses to keep the values in the correct order. So your equation should look like this:

p^2 = 4π^2/G * (M1 + M2) * a^3

Next, when solving for p, you need to take the square root of the entire right side of the equation. So your calculation should look like this:

p = √(4π^2/G * (M1 + M2) * a^3)

Then, plug in the values and use parentheses to keep everything in the correct order. It should look like this:

p = √(4π^2/6.67*10^-11 * (0.31+0.00095) * 1^3)

Finally, when you take the square root, make sure you are taking the square root of the entire value, not just the number in front of the square root. So your final calculation should look like this:

p = √(39.47/11.44031908×10^12)

p = 1.95 years

I'm not sure where the number 10.31065 came from in your notes, but it seems like you may have made a mistake in your calculations there. Make sure to double check your math and use parentheses to keep the values in the correct order.

In terms of when to use p^2=a^3/(M1+M2), this is just another way of writing Kepler's 3rd law. In this form, you use the