Where's the invisible matter?

We first look at the expression for tangential velocity of an object in circular motion. This is: $v = \sqrt{\frac{F*r}{m}}$ Where v is the tangential velocity, F is the radial force, r is the radius of the orbit, and m is the mass of the orbiting object. We see that if v is to remain constant, as stated in the problem, we must have: $F*r = \mbox{constant}$ $\Rightarrow F = \frac{\mbox{constant}}{r} \quad (1)$ . But, from Newton's Law of Gravitation, we also know: $F = \frac{M*C}{r^2} \quad (2)$ , where C is some constant and M is the total mass within the sphere of the object's orbit; that is, within a sphere with radius r. We omit from the equation the mass of the orbiting objects (in this case, the stars) because the problem statement specifies that their masses are negligible. Now, we want to find the required form of the density distribution function p(r), so we reformulate M in terms of $p(r): M = \frac{4*\pi}{3}*r^3*p(r) \quad (3)$ using the formula for the volume of a sphere. Upon substitution of (3) into (2), we find $F = \frac{\frac{4*\pi}{3}*r^3*p(r)*constant}{r^2} = \frac{4*\pi*r*p(r)}{3} \quad (4)$ . Now, if we look back to equation (1), we remember than we need F to vary as $\frac{1}{r}$ . From equation (4), we see that this condition is satisfied when p(r) varies as $\frac{1}{r^2}$ . In this case, we can write $\frac{p(10R)}{p(R)} = \frac{1}{100*R^2}*\frac{R^2}{1} = \frac{1}{100}$ And hence the desired ratio is 0.01.

The stars are moving in circular orbits and so the gravitational acceleration of each star must be equal to the centripetal acceleration,

$\frac {GM} {r^2} = \frac {v^2} {r}$

where G is Newton's gravitational constant, M is the total mass of stuff inside the star's orbit, and v is the speed of the star. Since v is a constant, we know that $M=constant \times r$ . However, M is also a function of the density $\rho(r)$ . Since the distribution is spherically symmetric,

$M(r)=\int_0^r \rho(r') 4 \pi (r')^2 dr'=constant \times r$ .

Taking the derivative of both sides with respect to r yields $4 \pi r^2 \rho(r)=constant$ , which implies $\rho(r)$ falls off as $1/r^2$ . Therefore $\rho(R)/\rho(10R)=R^2/(10R)^2)=0.01$ .

This is actually one way in which astronomers have concluded that dark matter must exist in our universe.