Astronomical Constant in Astronomical units? I'm doing a computer simulation of the solar system and I'm having trouble working with big numbers (implementation specific problem). So what would be the Newtonian gravitational constant $G$ in relation with the Earth mass instead of kilograms and astronomical units instead of meters?
 A: From Kepler's third law you can find that
$$
\frac{GM_\odot}{4\pi^2} = 1 \frac{\text{AU}^3}{\text{year}^2}
$$
where $M_\odot$ is the mass of the sun.  For a solar system simulation these units will be more convenient than Earth masses.
A: This is a typical "unit conversion" problem. Write $G$ in SI units:
$$G=6.6738\times10^{-11} \frac{\text{m}^3}{\text{kg}\cdot\text{s}^2}.$$
Now find out how many kilograms are in an Earth mass, and how many meters are in an astronomical unit. Also consider converting seconds to some other more convenient measure of time so that $G$ comes close to unity. (Thanks, Davidmh.)
All of this should help you convert units. See this page for further help.
A: In practice, one does not use G in astronomical calculations in this way.  
GM is known more exactly than G or M.  The mass of the earth, for example, is GM=398.6005 E12 m³/s², and is this value as far back as i have been able to trace it.
This is the value that NASA uses when they put man on the moon, so to speak.  
