Ball rolling down a ball, does it make sense that my result depends on the radius? So I have this system below. The upper ball of radius $r$ and mass $m$ is placed right at the top of the lower ball and starts rolling down this ball, which is fixed to the ground, with radius $R$. And the question is, at which angle $\theta$ of the lower ball will the upper ball separate from the lower ball? ($\theta$ is defined from the point of contact between the two balls at $t=0$ and the point contact at any $t$).

The result I got is that $\theta$ is that with this cosine:
$$
\cos \theta = \frac{4}{6 + \frac{R^2}{r^2}}
$$
which depends on $r$ and $R$.
Then, I searched for the solution of this problem (since it's a problem that can be found in Goldstein, so it probably had been solved before) and the result that whoever solved it had got was:
$$
\cos \theta = \frac{1}{2} \quad \theta = \frac{\pi}{3}
$$
which obviously doesn't depend on $r$ and $R$, which I feel is a more complete result, and therefore it seems the right one.
But I don't think it is, because of its procedure, which I think is wrong, but I don't want to upload it (yet) unless it's necessary, because I want to discuss intuitively the result.
The thing is that (for me at least) it's very difficult to say intuitively if the point at which the balls will separate should depend on the radius. Let's consider the upper ball puntual such that $r=0$. Then $\cos \theta = 0$ and $\theta = \frac{\pi}{2}$, which makes quite a lot of sense I think. Meanwhile in the other result, since it doesn't depend on $r$ it will leave the other ball at $\theta = \frac{\pi}{3}$, and I can't judge intuitively if this makes sense.
As I said, it's difficult for me to see it intuitively, that's why I rely more in the procedure, and obviously I think mine is the correct one. So my question is, can anybody tell me intuitively (with arguments) if the result should depend on the ratio of the radius or not?
 A: The upper ball gets separated from the lower ball, when its linear speed exceeds the speed supported by the centripetal force. 
So, intuitively, the sooner the upper ball will gain high enough linear speed, the sooner it will take off. Also, intuitively, we can predict that, given the same linear speed of the upper ball COM and the same radius of the COM's trajectory, $R+r$, the separation will take place at the same angle, regardless of the ratio between $R$ and $r$. 
So, the question comes down to the following: given the same $R+r$, will the ratio between $R$ and $r$ affect the linear acceleration of the upper ball? We can show that it does not.
As the upper ball is rolling down, its potential energy is converted to the linear kinetic energy and rotational kinetic energy. The linear kinetic energy indicates the linear speed. So, if we show that the ratio between the linear kinetic energy and the rotational kinetic energy does not depend on the ratio between $R$ and $r$, we'll prove that the take-off angle also does not depend on the ratio between $R$ and $r$.
$E_{lin}=\frac 1 2 mv^2$
$E_{rot}=\frac 1 2 I\omega^2$ 
Substituting the moment of inertia, $I$, by $\frac 2 5 mr^2$ and the angular speed, $\omega$, by $\frac v r$ (see more accurate formula in Mr.Nobody's answer, leaving this as is to preserve the integrity of the post), we get $E_{rot}=\frac 2 5 mv^2$ and $\frac {E_{lin}} {E_{rot}} = \frac 5 4$.
So, the ratio between $E_{lin}$ and $E_{rot}$ is a constant and, therefore, does not depend on the ratio between $R$ and $r$. This means that, given the same $R+r$, the rate of growth of the linear kinetic energy and the linear speed of the upper ball and, therefore, the take-off angle also do not depend on the ratio between $R$ and $r$. Update: this would be correct if we accepted the approximation $\omega=\frac v r$. See Mr.Nobody's answer for a more accurate formula.  
A: I'm answering my own question, but it's actually a comment on V.F.'s answer, which was very helpful, but since it's too long I have to post it like this.
First of all, let me rename the radius of the upper ball with $a$ instead of $r$, you'll see later why. Now, to compare $E_{lin}$ and $E_{rot}$ you need to convert $w$ into $v$, true, but it's not as easy as doing $\omega=\frac{v}{r}$. You have a constraint that relates $v$ and $\dot \theta$, and another that relates $\dot \theta$ and $\dot \phi$, which corresponds to the rotation of the upper and is also our $\omega$. The constraints are:
$$
v = (a+R)\dot \theta \quad \quad R \dot \theta = a \dot \phi
$$
so joining these two together:
$$
v = \frac{(a+R)a}{R}\dot \phi = \frac{(a+R)a}{R} \omega
$$
Now we can compare the energies (where I used $I = \frac{1}{2}ma^2$ which is different, but only a constant changes so it doesn't matter):
$$
\frac{E_{lin}}{E_{rot}} = \frac{\frac{1}{2} m ((a+R)\frac{a}{R})^2 w^2}{\frac{1}{4} m a^2 w^2} = 2 \frac{(R + a)^2}{R^2}
$$
So we see that it does change depending on $R$ and $a$. But this doesn't match with the equation I got for the cosine of $\theta$. But that's a mistake I made in my notation. I used $a$ as the radius of the upper ball, and $r=R+a$ and then forgot about it and thought that $r$ was the radius of the upper ball (which was how it was presented, that's why I thought it was the radius, but since I had changed the notation...). So actually the rate of change matches with the energy ratio.
I'm sorry for my mistake in presenting the problem. But I don't want to correct, since I see it's better to see the mistake and see where I did wrong, but if necessary I'll change it.
And I'm also a bit startled, since apart from me and what I get, everything tells me that it doesn't depend on the radius. I trust more in my equations than everything else, but if somebody could contrast it, it would be better.
A: $\let\th=\theta \let\om=\omega \def\IP{I_{\rm P}}
\def\half{{\textstyle{1 \over 2}}} \def\cF{{\cal F}} \def\cM{{\cal M}} \def\Fr{F_{\rm r}} \def\Rr{R_{\rm r}} \def\ac{a_{\rm c}} 
\def\dth{\dot\th} \def\thd{\th_{\rm d}}$
IMHO all solutions I've read are faulty. (I couldn't find the problem in Goldstein. Where exactly is it located?) To be true, I do not even understand them.
If I'm right, the answer is 
$$\cos\th = {10 \over 17}.$$
But in order to better discuss limiting cases, I prefer to generalize a little, assuming that the barycentric moment of inertia of the moving ball $\cM$ isn't ${2 \over 5}mr^2$. I will use the generic symbol $I$.
Edit. From here on I've entirely rewritten my proof, mainly because I've found an easy way to explain why the detachment angle depends on $r$ and $I$ - actually, on the adimensional ratio $I/(mr^2)$ - and not on $R$.
Let G be $\cM$'s center, P the point of contact between $\cM$ and the fixed ball $\cF$, $\om$ the angular velocity of $\cM$ (there is only one in a rigid motion), $K$ the total kinetic energy of $\cM$. $K$ is given by König's theorem
$$K = \half m\,v^2 + \half I\,\om^2.\tag1$$
The speed $v$ can be given two forms:
$$v = \om\,r = (R + r)\,\dth.$$
First form obtains because P is momentarily at rest, so that $\cM$'s
motion is a rotation around P with angular velocity $\om$. Second form
results from G's trajectory being a circumference of radius $R+r$,
traveled with angular speed $\dth$. Then
$$K = \half m\,v^2\!\left(\!1 + {I \over m\,r^2}\!\right)\!.\tag2$$
From energy conservation
$$K = mgh \tag3$$
where 
$$h = (R+r)\,(1-\cos\th) \tag4$$ 
is the length G has descended when P's colatitude is $\th$.
Let's study when $\cM$ detaches from $\cF$. This happens when the radial component $\Fr$ of the force $\vec F$ of $\cF$ over $\cM$, initially outward, becomes zero. To compute $F$ the c.o.m. theorem can be used. The resultant of forces acting on $\cM$ is 
$$\vec R = m \vec g + \vec F$$
and its radial outward component is
$$\Rr = \Fr - m g \cos\th.\tag5$$
$\Rr$ is connected to centripetal acceleration $\ac$ of G by
$$\Rr = -m\,\ac$$
(look at sign!). 
Computing $\ac$ requires caution. The right expression is
$$\ac = {v^2 \over R + r}.$$
It would be wrong to write $\ac=\om^2 r$, since for a displacement to second order in time, as is required to get acceleration, motion of G is not a rotation around P: this only holds to the first order, to compute velocities.
Substituting into (5)
$$\Fr = m\,g\,\cos\th - {m\,v^2 \over R + r}$$
and $\Fr=0$ implies
$$g\,\cos\thd =  {v^2 \over R + r}\tag6$$
where $\thd$ is the angle of detachment.
From (2), (3), (4)
$$v^2 = {2\,m\,g\,r^2 \over I + m\,r^2}\,(R+r)\,(1-\cos\thd)$$
and (6) gives
$$\cos\thd = {2\,m\,r^2 \over I + m\,r^2}\,(1-\cos\thd).$$
Solving for $\cos\thd$:
$$\cos\thd = {2\,m\,r^2 \over I + 3\,m\,r^2}.\tag7$$
Eq. (7) is our general solution. If $I={2 \over 5}mr^2$ then 
$$\cos\th = {10 \over 17}.$$

Limit cases 
1) $I=0$. This means that $\cM$ reduces to a point, moving on a sphere of radius $R+r$. Then (7) gives $$\cos\thd = {2 \over 3}.$$
2) $R=0$. There is nothing to change in eq. (7), as it doesn't contain $R$. But it's easy to solve this case separately, thus giving a check of the general solution.
$R=0$ means that $\cM$ has a fixed point P. But this constraint is unilateral: the radial component $\Fr$ of the force exerted by the constraint on the ball must be non-negative. Detachment happens when $\Fr=0$. Then at detachment the only radial force acting on $\cM$ is gravity's component, and it must equal $\ac$, the centripetal acceleration of G:
$$g\,\cos\thd = \ac.$$
The ball's motion is a pure rotation around P, and G's centripetal acceleration is 
$$\ac = v^2\!/r$$
where $v$ is still given by energy conservation
$$v^2 = {2\,m\,g\,r^3 \over I + m\,r^2}\,(1-\cos\thd).$$
Then we find
$$g\,\cos\thd = {2\,m\,g\,r^2 \over I + m\,r^2}\,(1-\cos\thd)$$
and solving for $\thd$ we get again eq. (7).
