# Stars more affected by gravity of stars further away, need help understanding argument why

In the first chapter of Binney and Tremaine's "Galactic Dynamics" there is a statement that I don't yet understand. The first chapter of the book is linked here, straight from the publisher. The statement is in the second paragraph of section 1.2, which begins on PDF page 34 (print page 33).

The argument is that, in a galaxy, a star is more affected by the gravitational force from stars further away than stars closer by. Since force is proportional to $r^{-2}$, and (assuming uniform density) the number of stars seen by an observer at a radius $r$ in a given angle is proportional to $r^2$. However, there is an extra factor of $r$ that comes in (the radius of each "octave", as they've defined octave) and I am not sure the physical reason why this $r$ comes in. With this extra $r$, the gravitational force is proportional to $r^{-2} \times r^2 \times r = r$.

What is the physical reason for that extra factor of $r$? Why does the length of the octave come in?

## 1 Answer

That's a very confusing paragraph. They use the term 'octave' specifically for a doubling of distance (r) --- that's not important. The argument is simply that the force per star falls off as $r^{-2}$, but the number of stars actually increases as $r^3$. So as you increase distance, the gravity increases by $r$.

We could write this more precisely as for a constant number density of stars $n$, $$F \propto \int n \frac{1}{r^2} dV = \int n \frac{1}{r^2} r^2 dr \propto r$$

• That last step only makes sense if you integrate from $0$ to $r$. But we're integrating over similar sized shells (at least same size in radius) at different radii. If the shell size is $\Delta r$, then the integral gives a value of $\Delta r$ for every shell, independent of its radius. At least that's how I understand it. – NeutronStar Apr 9 '16 at 18:42
• @Joshua, yes - you're right, its just whatever with dr you look at. but that's why they specified an 'octave' --- i.e. a length proportional to r (in this case 2r). As long as that's true, this proportionality holds. In astro, since most things are power-laws, you assume that variations are on the order of the initial values, i.e. $\delta x \sim x$. – DilithiumMatrix Apr 9 '16 at 19:15
• I'm still having a difficult time understanding. What are so magical about octaves that it's okay that they give a different result than the one I mention with $\Delta r$? – NeutronStar Apr 9 '16 at 19:23
• @Joshua they don't give a conflicting result. The import thing is that the force scales roughly linearly with distance, it doesn't matter if that's $r$ or $\Delta r$. I was trying to additional point out, that if the integral bounds are something like $\alpha r$ to $\beta r$ (for some constants $\alpha$ and $\beta$) then the specific form of the result $F \propto r$ holds. This is an order of magnitude type argument. – DilithiumMatrix Apr 10 '16 at 1:13