(The other questions about this were not helpful).
Given a point $p\in S^2=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=a^2\}$ and a vector $v\in T_pS^2$, I want to find a curve $\sigma$ of the great circle which goes through $p$ with velocity $v$.
My idea was to start with the curve $\alpha(t)=(\cos t,\sin t,0)$ and then apply rotations to it so that the plane defined by the resulting great circle has normal vector equal to $p\times v$. Hence, if we write $p\times v = N(\sin \theta_N \cos \phi_N), \sin \theta_N \sin \phi_N, \cos \theta_N$, we had the curve $\gamma(t)=R_z(\phi_N)R_y(\theta_N)\alpha(t)$.
I am not sure that this is correct, and the computations are tiresome. IS there other way to do it?
(Context: I want to know the parametric geodesic $\gamma$ that goes through $p$ with velocity $v$ so that I can determine the exponential map $\exp_p(v)=\gamma(1)$ )
I will consider the unit sphere ($a=1$).
Let $w:=\tfrac{v}{\|v\|}$.
As $p \perp w$, the curve defined by
$$\tag{1}s(t)=\cos(t)p+\sin(t)w$$
goes, at $t=0$, through point $p$ ($s(0)=\cos(0)p+\sin(0)w=p$).
belongs to the sphere and describes a great circle on it (consider the orthonormal basis $\{p,w,p \times w\}$; this curve is the unit circle in the "XOY" plane of this basis).
has the following velocity at $t=0$: $s'(t)|_{t=0}=-\underbrace{\sin(t)p|_{t=0}}_{=0}+\underbrace{\cos(t)w|_{t=0}}_{=1w}=w$, instead of $v$.
We will obtain an initial speed vector equal to $v$ by replacing (1) by:
$$\tag{2}\sigma(t)=\cos(t\|v\|)p+\sin(t\|v\|)w$$
(proof: differentiation of (2) gives $\sigma'(t)|_{t=0}=-\underbrace{\sin(t\|v\|)\|v\|p|_{t=0}}_{=0}+\underbrace{\cos(t\|v\|)\|v\|w|_{t=0}}_{=1v \ \text{by def. of w}}$)