Let $f: \mathbb{S}^2 \longrightarrow \mathbb{R}$ be a positive differantiable function. Define $$ S(f) = \{f(p)p \in \mathbb{R}^3 \ | \ p \in \mathbb{S}^2\}. $$ It can be shown that $S(f)$ is a compact surface diffeomorphic to the sphere $\mathbb{S}^2$, with the diffeomorphism given by $$ \phi(p) = f(p)p, \quad p \in \mathbb{S}^2. $$

Now, if $S$ is a compact connected surface, we say that $S$ is star-shaped with respect to the origin if $0 \notin S$ and if one cannot draw a straight line from $0$ that is tangent to $S$.

Our aim is to prove that $S$ is star-shaped with respect to the origin if and only if it is one of the surfaces $S(f)$.

One implication is easy:

Take $S(f)$. Of course $0 \notin S(f)$ because $f > 0$. Now, $S(f)$ is not star-shaped, that is, suppose we could draw a line between $0$ and $f(p)p$ tangent to $S(f)$ at $f(p)p$. Then we would have that $p \in T_{\phi(p)}S(f) = (d\phi)_p(T_p\mathbb{S}^2)$. Then we have that for some $v \in T_p\mathbb{S}^2$ $$ p = (d\phi)_p(v) = (df)_p(v)p + f(p)v. $$ Taking the scalar product with $v$, we have that $v = 0$, which is a contradiction. Hence $S(f)$ is indeed star-shaped.

My question is: How to prove the converse implication?