# Why can you remove the gravitational constant from a computer game simulation?

I've seen in a few gravity simulation games (ie. bouncing balls) the equation:

force = G * m1 * m2 / distance^2


shortened to this by removing the gravitational constant:

force = m1 * m2 / distance^2


I accept that it works fine and saves some calculations, but I'm wondering why it still works? Is the value just too small to matter? What's the physics behind this?

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In simple words: force and distance in games are usually measured in chickens and ducks. –  valdo Sep 5 '11 at 10:56
Gravity is not "removed" there in the sense that attraction is still there. –  Vladimir Kalitvianski Sep 5 '11 at 17:46

$G$ is just a constant of proportionality to get the units right (so that when $m_1$ and $m_2$ are in kilograms and $r$ is in meters you get a force in Newtons rather than wingdingalings or something really weird). Indeed cosmologists like to work in a system of units where $G = c = 1 \text{ (dimensionless)}$, and particle physicists like to work in units where $c = \hbar = 1 \text{ (dimensionless)}$, you can even set all three of these numbers to 1 if you like.
A very similar situation occurs in electrostatics: in SI units, Coulomb's law is $F = (1/4\pi\epsilon_0)(q_1q_2/r^2)$; the funky constant is there because the unit charge (coulomb) is defined in a roundabout way using magnetic force in current-carrying wires. But in Gaussian units, it is $F = q_1q_2/r^2$, with no constant, because the unit charge (esu) is defined in terms of Coulomb's law instead. Look up "geometrized units" on wikipedia for an extreme version of this. –  Stan Liou Sep 5 '11 at 15:19