# Proportionality and units

This might be very easy, but I'm not 100% sure how it's done.

Lets say I have this equation:

$$R = R_{0} \cdot \left[1 - \frac{P_{0}R_{0}}{GM_{0}\rho_{0}}\right]^{-1},$$ where I know $P_{0}$, $M_{0}$ and $\rho_{0}$. I'm then told to find $R$, and that $M_{0} \propto R_{0}^{-3}$.

Now, how do I do this ? From the proportionality I can also say $R_{0} \propto 1/\sqrt{M_{0}}$, and then just insert it into the above equation, right ? But let's say that $M_{0} = 10^{28} \text{kg}$, how does that translate into $R_{0}$, with units and all ? I don't end up with $R$ being a length, if $R_{0} = 1/\sqrt{10^{28}\text{kg}}$.

Yeah, I'm just a little bit confused. But I can't seem to get my head around this right now.

• Newtons 2nd law of motion says that the total force $F$ is proportional to the acceleration $a$. So, $F \propto a$. So do $F$ and $a$ have same units? You are forgetting that the proportionality constant itself has units that will make it all fit. – Steeven Oct 21 '14 at 16:32
• What's the question? How to find $R$, or how to reconcile the units between $M_0$ and $R_0$? – garyp Oct 21 '14 at 16:37
• @Steeven gave me what I needed :) Thank you. Very simple :P – Denver Dang Oct 21 '14 at 17:29

It seems that that you're converting from a volume ($R_0^3$) (is the minus sign a typo?) to a mass($M_0$).
The constant of proportionality is a density, presumably the $\rho_0$ that you're given.
In more general terms, the constants that are hidden by the $\propto$ symbol can have units too.
• Well, the minus just means that $M_0$ is inversely proportional to $R^3$. You see: $M_0 \propto R^{-3} \Rightarrow M_0 \propto \frac{1}{R^3}$. – Steeven Oct 21 '14 at 19:33
• Yes, but to get the correct unit I need to have a proportional constant with the unit $\text{kg} \cdot \text{cm}^{3}$. And from what I have been given, it doesn't look like something I can do. – Denver Dang Oct 21 '14 at 20:03