Orthogonality between geodesics and normal spheres in a Riemannian manifold In do Carmo's Riemannian Geometry, right after proving Gauss' Lemma and defining normal neighborhoods and normal balls, the author writes:

By Gauss' Lemma, the boundary of a normal ball $B_\varepsilon(p)$ is a hypersurface (a submanifold of codimension $1$) in $M$ orthogonal to the geodesics leaving $p$, which is denoted by $S_\varepsilon (p)$ and called the normal (or geodesic sphere).

As I understand, the orthogonality refers to ortogonality between the tangent vector of the geodesic and the tangent plane to the said hypersurface. Is it correct?
Anyway, how to prove the claim about the sphere being a submanifold and the claim about orthogonality?
Thanks in advance.
 A: In short, the boundary of the normal ball is a hypersurface of codimension one because it just the image of a sphere in $T_pM$ under the diffeomorphism $\exp_p$ defined on some neighbourhood of zero vector. The orthogonality is pretty straightforward from Gauss' Lemma. Here's the details.
Let $V \subset T_pM$ is a normal neighbourhood of $p$ and $B_{\varepsilon}(p)= \text{exp}_p \big(B_{\varepsilon}(0)\big)$ be a normal ball. We know that the sphere $S_{\varepsilon}(0) = \{ v \in V \mid \|v\|=\varepsilon \}$ is a hypersurface of codimension one of $V$ since it is a regular level set of the smooth function $r : V \to \mathbb{R}$ defined as $r(x)=\|x\|$. More precisely $S_{\varepsilon}(0) = r^{-1}(\varepsilon)$. Since $\text{exp}_p : V \to U\subseteq M$ is a diffeomorphism, we have the normal sphere
$$
S_{\varepsilon}(p) := \partial B_{\varepsilon}(p) = \text{exp}_p \partial B_{\varepsilon}(0) =  \text{exp}_p (S_{\varepsilon}(0)) = \text{exp}_p \circ r^{-1}(\varepsilon) = (r \circ \exp_p^{-1})^{-1}(\varepsilon).
$$
Therefore $S_{\varepsilon}(p) \subseteq U$ is a regular level set of the smooth map $R := r \circ \exp_p^{-1} : U \to \mathbb{R}$. This establish that $S_{\varepsilon}(p)$ is a hypersurface of codimension one.
To show the orthogonality claim, let $\gamma(t) =\exp_p(tv)$ is a geodesic emanating from $p$ and passing through $S_{\varepsilon}(p)$, say $\|v\| = \varepsilon$ and $t$ defined in some open interval contain $[0,1]$, so $q:=\gamma(1) \in S_{\varepsilon}(p)$. For any $w \in T_q S_{\varepsilon} = \text{ker } dR_q$, we have
$$
0=dR_q(w) =dr_v \circ d(\exp_p^{-1})_q(w) = dr_v(w_0),
$$
with $ w_0:= d(\exp_p^{-1})_q(w) \in T_vV \approx T_v(T_pM) \approx T_pM$.
This means that
$$
w_0 \in \text{ker }dr_v = \text{tangent space of }S_{\varepsilon}(0) \text{ at }v.
$$
Therefore,
\begin{align}
\langle \gamma'(1),w\rangle &= \langle d(\exp_p)_v(v),d(\exp_p)_v(w_0) \rangle \\
&=\langle v,w_0 \rangle, \qquad \text{by Gauss' Lemma} \\
&= 0.
\end{align}
