Calculating G using Kepler's Third Law: The Relationship Between k, a, and P

[Benzoate]

Homework Statement

Using astronomical units as the unit of length , years as the time , and the mass of the Sun as the unit of mass , the value for k in kepler's third law is 1. In these units , what is the value of Newton's constant of gravitation G?

Homework Equations

P^2 =k*a^3
P^2= 4*pi^2*a^3/(G(m(1)+m(2))
Possibly k=4*pi^2/G*M(sun)
P is the period

The Attempt at a Solution

since k = 1 and k =4*pi^2/G*M(sun), all I have to do is switch variables. k=4*pi^2/(G*M(sun)) => 4*pi^2/(M(sun))=G , but I found that method to be a problem because the G I calculated isn't equal the universal Gravitational constant we are all familar with.

 

[learningphysics]

So taking the mass of the sun as 1 solar mass... your answer should be $$G = 4{\pi}^2 (years)^{-2}(AU)^3(solar masses)^{-1}$$

To check that this is the same G you need to convert units...

 

[D H]

The universal gravitational constant has units. The value with which you are familiar, $$6.6730\times10^{-11}$$ is in units of $$m^3/kg/s^2$$. The value differs when you change the system of units. For example, G is $$7.616\times10^{-5}$$ in the ever-popular furlongs-fortnights-stones system. You are asked to specify G in the AU-Msun-year system. The answer is not $$6.6730\times10^{-11}$$.

 

[Benzoate]

The universal gravitational constant has units. The value with which you are familiar, $$6.6730\times10^{-11}$$ is in units of $$m^3/kg/s^2$$. The value differs when you change the system of units. For example, G is $$7.616\times10^{-5}$$ in the ever-popular furlongs-fortnights-stones system. You are asked to specify G in the AU-Msun-year system. The answer is not $$6.6730\times10^{-11}$$.

Are my equations correct? I just need to convert my units to SI units?

 

[learningphysics]

Are my equations correct? I just need to convert my units to SI units?

Yes, I think the equations are right. just convert the units.

 

[D H]

Your instructor did not ask for the value of G in SI (MKS) units. He asked for it in AU-solar mass-year units. Learning physics gave you the answer in these units post #2.

You only need to convert to SI if you want to verify that your answer is correct. If you do the conversion right, you will get [itex]6.672\times10^{-11}\text{m}^3/\text{sec}^2/\text{kg}[/tex], which compares very favorably with the published value of [itex]6.673\times10^{-11}\text{m}^3/\text{sec}^2/\text{kg}[/tex].