# Prove gravitational constant $G=1$ in geometric units

I've seen that often geometric units are defined by setting $G=c=1$, however, I have been working with a different definition, namely, to multiply time in seconds by $c$ so that it's measured in meters, and multiply the mass in kilograms by $\frac{G}{c^2}$ so that it's measured in meters.

It should then follow that the law of universal gravitation expressed in geometric units is rewritten as $$F=\frac{Mm}{r^2}$$ But I can't seem to get this expresion. Consider the original one $$F=G\frac{Mm}{r^2}$$ Since force is measured in $kg\cdot m\cdot s^{-2}$, we need to multiply the left side by $\frac{G}{c^2}\frac{1}{c^2}$ and we need to multiply the right side by $\frac{G^2}{c^4}$ since there is two masses. The equation the results $$F\frac{G}{c^4}=G\frac{Mm}{r^2}\frac{G^2}{c^4}\iff F=G^2\frac{Mm}{r^2}$$ As you can see, I don't get the desired result, that is, that $G$ dissapears (or is equal to $1$).

EDIT: I just noticed the mistake and posted it as an answer to this question.

• If $G=1$, what makes you think Force is still in $\frac{kg \, m}{s^2}$? Apr 20 '17 at 16:05
• I started from the original (SI units) formula, and I then multiply by the needed coefficients to try to get $G=1$ as a consequence (since it's not my starting hypothesis/definition). Apr 20 '17 at 16:08
• Why have you multiplied LHS by $G^2/c^4$ and RHS by $G/c^4$ since you used $GMm/r^2$ ? Because you should have multiplied $Mm/r^2$ with $G^2/c^4$ ... Apr 20 '17 at 16:09
• That's what I did though, I multiplied $\frac{Mm}{r^2}$ by $\frac{G^2}{c^4}$. Apr 20 '17 at 16:15
• Hi @F.Webber, it would be nice for you to post the answer below, in the answers section, and not in the question itself. Thanks! Apr 20 '17 at 16:54

In order to introduce the geometric units we have to introduce the coefficients stated above, namely multiply all masses by $\frac{G}{c^2}$ and all times by $c$.
However, we also need to introduce the inverse of them so as to not change the expressions value, and we can then proceed to change units. Precisely, if we use overlines for geometric units (the others being in SI units) we have $$F=G\frac{Mm}{r^2}\iff\left(F\frac{G}{c^2}\frac{1}{c^2}\right)\frac{c^2}{G}c^2=G\frac{M\frac{G}{c^2}m\frac{G}{c^2}}{r^2}\left(\frac{c^2}{G}\right)^2 \iff\overline{F}\frac{c^4}{G}=\frac{c^4}{G}\frac{\overline{M}\overline{m}}{r^2}\iff\overline{F}=\frac{\overline{M}\overline{m}}{r^2}$$ And we get the desired formula, equivalently that $\overline{G}=1$.