Potential and kinetic energy on spherical surface 

A small particle of mass $m$ is atop of a semi-sphere as shown in the figure. A little push was given to the particle. Prove that the particle will leave the spherical surface at a height of (2/3)$R$.

Now, I've managed to get this result (i.e, (2/3)$R$) by equating the potential energy with the kinetic energy.But to be honest, it was a mere hunch. My question is why is that true? 
 A: A mass moving in a circle has centripetal acceleration $v^2/r$ directed toward the center of the circle. You can get $v$ from potential energy. 
The mass here has two forces on it. Gravity is constant and down. The reaction force of the surface (assuming no friction) is normal to the surface. When the sum of these two forces becomes less than centripetal force, the mass will leave the surface. 
A: There follows my try to decompose the solution into a minimal
amount of calculation and apart from that only geometrical considerations.
The centripetal acceleration is $a_c=\frac{v^2}R$. It is directed towards the center.
We define the $z$ coordinate as starting at the top and pointing vertically downwards (see the following Figure).

The conservation of energy says $\frac{v^2}2 = g z$. Therefore, we have $2g z= a_c R$. That means $a_c\sim z$.
Furthermore, if the mass was constrained to the sphere then we would have at the equator $a_c(z=R)=2g$.
Now, you can draw a diagram that relates the centripetal acceleration to the angular position of the mass.
(Do not worry. In our consideration we do not need any specific numeric value for the angle.)

At the equator the mass would have normal acceleration of
$a_c=2g$. this is expressed by the big circle.  Depending on $\varphi$
the centripetal acceleration is shrinking proportional to the distance
of the top line to the dashed one. To get the direction of the
centripetal acceleration right we have drawn in the small circle.
The mass detaches from the surface when the normal component of the
gravitational acceleration (marked with $g$. equals the centripetal
acceleration (line from the center of the small circle to the inner
intersection with the small circle). These lines form two edges of the
triangle for the decomposition of the gravitational acceleration into
the normal and the tangential component. The triangle composed of the
blue and the red line and the blue arc has also one edge with length
equal to the centripetal acceleration, the angles are also
equal. Therefore they are congruent and the red line has length $g$.
Finally, we can just use the intercept theorem for the $2g$-diameter
the $3g$ stretch composed of the skew $2g$ diameter and the red line
and the top and the dashed line to obtain that the mass detaches at
$\frac 23$ height of the sphere.
