Particle sliding on a sphere

We put the circular orbit of the particle on a straight line and convert the motion to a 1-dimensional rectilinear motion as follows : The arc length, the natural parameter $\:s(t)\:$ is the distance travelled on the straight line till time $\:t\:$. The speed $\:v(t)\:$ on the straight line is the magnitude of the tangent to the circle velocity. Now, on the straight line the particle is moving like under the influence of the tangent force which is $\:f_{t}=mg\sin(\theta)\:$ so under a variable acceleration $\:a_{t}=g\sin(\theta)\:$. But $\:\theta=s/R\:$ so the differential equation of motion is

\begin{equation} \dfrac{\mathrm{d}^{2} s}{ \mathrm{d}t^{2} }-g\sin\left(\dfrac{s}{R}\right)=0, \qquad \left[\dfrac{ \mathrm{d} s }{\mathrm{d} t}\right]_{t=0}=0, \qquad s(0)=0 \tag{01} \end{equation}

since the particle starts at rest on the origin.

On the other hand the condition for the particle to leave the sphere is the normal force to be zero \begin{equation} N=mg\cos(\theta)-ma_{c}=mg\cos(\theta)- \dfrac{mv^{2}}{R}=0 \tag{02} \end{equation} that is \begin{equation} \boxed { \bbox[#FFFF88,8px]{\:\:\left(\dfrac{ \mathrm{d} s }{\mathrm{d} t}\right)^{2}-gR\cos\left(\dfrac{s}{R}\right)=0 \:\:}} \tag{03} \end{equation}

Now, we must solve (01) to find at which point the condition (03) is satisfied. But it'll proved to be not necessary. So, multiplying (01) by $ \dfrac{ \mathrm{d} s }{\mathrm{d} t} $ we have \begin{equation} \dfrac{ \mathrm{d} s }{\mathrm{d} t}\dfrac{\mathrm{d}^{2} s}{ \mathrm{d}t^{2} }-g\dfrac{ \mathrm{d} s }{\mathrm{d} t}\sin\left(\dfrac{s}{R}\right)=0 \tag{04} \end{equation} or \begin{equation} \dfrac{ \mathrm{d} s }{\mathrm{d} t} \dfrac{ \mathrm{d} }{\mathrm{d} t} \left(\dfrac{ \mathrm{d} s }{\mathrm{d} t}\right) +\dfrac{ \mathrm{d} }{\mathrm{d} t}\left[gR\cos\left(\dfrac{s}{R}\right)\right]=0 \tag{05} \end{equation} that is \begin{equation} \dfrac{ \mathrm{d} }{\mathrm{d} t} \Biggl[ \left(\dfrac{ \mathrm{d} s }{\mathrm{d} t}\right)^{2}+2gR\cos\left(\dfrac{s}{R}\right) \Biggr]=0 \tag{06} \end{equation}

This means that we have found a constant of integration of (01) and more explicitly using the initial conditions

\begin{equation} \Biggl[ \left(\dfrac{ \mathrm{d} s }{\mathrm{d} t}\right)^{2}+2g\cos\left(\dfrac{s}{R}\right) \Biggr]=\text{constant}=\Biggl[ \left(\dfrac{ \mathrm{d} s }{\mathrm{d} t}\right)^{2}+2gR\cos\left(\dfrac{s}{R}\right) \Biggr]_{t=0}=2gR \tag{07} \end{equation} or \begin{equation} \boxed { \bbox[#FFFF88,8px]{\:\: \left(\dfrac{ \mathrm{d} s }{\mathrm{d} t}\right)^{2}+2gR\cos\left(\dfrac{s}{R}\right) =2gR \:\:}} \tag{08} \end{equation}

Substructing equations (08) and (03) side by side we have finally

\begin{equation} \cos\left(\theta\right)=\cos\left(\dfrac{s}{R}\right) =\dfrac{2}{3} \tag{09} \end{equation}

Notes :

1. The differential equation of motion (01) is identical to that in the Dvij's answer but with respect to $\:s(t)=\theta(t)R\:$ instead of $\:\theta(t)\:$.

2. I find the constant of integration (07) of equation (01) motivated by the fact that there exists a constant : the energy. I inserted the energy conservation through the back door.