[proofplan]
We prove the two conservation laws by differentiating the energy and angular momentum along the $C^2$ trajectory. The central-force equation makes the force parallel to $q(t)$, so the energy derivative cancels by the chain rule and the torque $q(t) \times F(t)$ vanishes. The planar statement follows because the conserved angular momentum is orthogonal to both $q(t)$ and $\dot q(t)$. Finally, for a scattering orbit, taking the asymptotic limits and using $V(r) \to 0$ gives the energy and impact-parameter formulas.
[/proofplan]
[step:Differentiate the energy and cancel the kinetic and potential contributions]
Define the force map along the trajectory
\begin{align*}
F_q: \mathbb{R} \to \mathbb{R}^3, \qquad F_q(t) = -V'(|q(t)|)\frac{q(t)}{|q(t)|}.
\end{align*}
Since $q(t) \in \mathbb{R}^3_0$ for every $t$, the map $t \mapsto |q(t)|$ is $C^1$, and
\begin{align*}
\frac{d}{dt}|q(t)| = \frac{q(t)\cdot \dot q(t)}{|q(t)|}.
\end{align*}
Using the product rule for the kinetic term and the chain rule for the potential term, we obtain
\begin{align*}
\frac{d}{dt}E_q(t) = m\ddot q(t)\cdot \dot q(t) + V'(|q(t)|)\frac{q(t)\cdot \dot q(t)}{|q(t)|}.
\end{align*}
The equation of motion gives
\begin{align*}
m\ddot q(t)\cdot \dot q(t) = -V'(|q(t)|)\frac{q(t)\cdot \dot q(t)}{|q(t)|}.
\end{align*}
Substitution yields
\begin{align*}
\frac{d}{dt}E_q(t) = 0.
\end{align*}
Therefore $E_q$ is constant on $\mathbb{R}$.
[guided]
The energy has two pieces, and the point of differentiating it is to see that the work done by the force is exactly balanced by the change in potential energy. Define
\begin{align*}
F_q: \mathbb{R} \to \mathbb{R}^3, \qquad F_q(t) = -V'(|q(t)|)\frac{q(t)}{|q(t)|}.
\end{align*}
This is well-defined because the trajectory avoids the origin, so $|q(t)| > 0$ for every $t$.
We first compute the derivative of the radial coordinate. Since $q \in C^2(\mathbb{R};\mathbb{R}^3_0)$, the function $t \mapsto |q(t)|$ is $C^1$, and differentiating $|q(t)|^2 = q(t)\cdot q(t)$ gives
\begin{align*}
2|q(t)|\frac{d}{dt}|q(t)| = 2q(t)\cdot \dot q(t).
\end{align*}
Because $|q(t)| > 0$, division by $2|q(t)|$ gives
\begin{align*}
\frac{d}{dt}|q(t)| = \frac{q(t)\cdot \dot q(t)}{|q(t)|}.
\end{align*}
Now differentiate the energy function
\begin{align*}
E_q: \mathbb{R} \to \mathbb{R}, \qquad E_q(t) = \frac{m}{2}|\dot q(t)|^2 + V(|q(t)|).
\end{align*}
The product rule applied to $|\dot q(t)|^2 = \dot q(t)\cdot \dot q(t)$ gives the derivative $m\ddot q(t)\cdot \dot q(t)$ of the kinetic energy. The chain rule applied to $V(|q(t)|)$ gives
\begin{align*}
\frac{d}{dt}V(|q(t)|) = V'(|q(t)|)\frac{q(t)\cdot \dot q(t)}{|q(t)|}.
\end{align*}
Hence
\begin{align*}
\frac{d}{dt}E_q(t) = m\ddot q(t)\cdot \dot q(t) + V'(|q(t)|)\frac{q(t)\cdot \dot q(t)}{|q(t)|}.
\end{align*}
The central-force equation states that
\begin{align*}
m\ddot q(t) = -V'(|q(t)|)\frac{q(t)}{|q(t)|}.
\end{align*}
Taking the Euclidean [inner product](/page/Inner%20Product) of both sides with $\dot q(t)$ gives
\begin{align*}
m\ddot q(t)\cdot \dot q(t) = -V'(|q(t)|)\frac{q(t)\cdot \dot q(t)}{|q(t)|}.
\end{align*}
This is exactly the negative of the potential-energy derivative. Substituting into the derivative of $E_q$ yields
\begin{align*}
\frac{d}{dt}E_q(t) = 0.
\end{align*}
Thus the energy function is constant on $\mathbb{R}$.
[/guided]
[/step]
[step:Differentiate the angular momentum and use that the torque vanishes]
By the product rule for the cross product,
\begin{align*}
\frac{d}{dt}L_q(t) = m\dot q(t)\times \dot q(t) + m q(t)\times \ddot q(t).
\end{align*}
The first term is zero because a vector crossed with itself is zero. For the second term, the equation of motion gives
\begin{align*}
m q(t)\times \ddot q(t) = q(t)\times \left(-V'(|q(t)|)\frac{q(t)}{|q(t)|}\right).
\end{align*}
Since the vector on the right is a scalar multiple of $q(t)$, the cross product vanishes. Therefore
\begin{align*}
\frac{d}{dt}L_q(t) = 0.
\end{align*}
Thus $L_q$ is constant on $\mathbb{R}$.
[/step]
[step:Identify the fixed plane containing the trajectory]
Let $L \in \mathbb{R}^3$ denote the constant value of $L_q$. For every $t \in \mathbb{R}$,
\begin{align*}
q(t)\cdot L = m q(t)\cdot \bigl(q(t)\times \dot q(t)\bigr) = 0.
\end{align*}
Hence $q(\mathbb{R})$ lies in the fixed plane
\begin{align*}
\Pi_L := \{x \in \mathbb{R}^3 : x\cdot L = 0\}.
\end{align*}
If $L \ne 0$, this is the two-dimensional plane through the origin perpendicular to $L$. If $L = 0$, then $q(t)\times \dot q(t)=0$ for every $t \in \mathbb{R}$. Since $q(t) \ne 0$, define the scalar function $\lambda: \mathbb{R}\to\mathbb{R}$ by
\begin{align*}
\lambda(t):=\frac{q(t)\cdot \dot q(t)}{|q(t)|^2}.
\end{align*}
The identity $q(t)\times \dot q(t)=0$ implies that $q(t)$ and $\dot q(t)$ are linearly dependent, and the displayed definition gives $\dot q(t)=\lambda(t)q(t)$. Define the direction map $u: \mathbb{R}\to S^2$ by $u(t)=q(t)/|q(t)|$, where $S^2:=\{x\in\mathbb{R}^3:|x|=1\}$. Differentiating $u$ gives
\begin{align*}
\dot u(t)=\frac{\dot q(t)}{|q(t)|}-\frac{q(t)(q(t)\cdot \dot q(t))}{|q(t)|^3}.
\end{align*}
Substituting $\dot q(t)=\lambda(t)q(t)$ yields
\begin{align*}
\dot u(t)=\frac{\lambda(t)q(t)}{|q(t)|}-\frac{\lambda(t)q(t)|q(t)|^2}{|q(t)|^3}=0.
\end{align*}
Fix any reference time $t_0\in\mathbb R$. Thus $u$ is constant on $\mathbb{R}$, so $q(t)$ always lies on the fixed line $\mathbb{R}u(t_0)$ through the origin. Hence the motion is contained in any plane through the origin containing that line.
[/step]
[step:Pass to the scattering limits to compute the conserved energy]
Let $E \in \mathbb{R}$ denote the constant value of $E_q$. By the definition of scattering orbit in the theorem statement, there are vectors $a_-,a_+,v_-,v_+\in\mathbb{R}^3$ with $v_-\ne0$ and $v_+\ne0$ such that $q(t)-(a_-+tv_-)\to0$ and $\dot q(t)\to v_-$ as $t\to-\infty$, while $q(t)-(a_++tv_+)\to0$ and $\dot q(t)\to v_+$ as $t\to+\infty$. From the incoming position asymptotic, we have
\begin{align*}
\frac{|q(t)|}{|t|}\to |v_-|
\end{align*}
as $t\to -\infty$, because $q(t)-(a_-+tv_-)\to0$ and $|a_-+tv_-|/|t|\to |v_-|$. Hence $|q(t)|\to\infty$ as $t\to -\infty$. The hypotheses give $\dot q(t) \to v_-$ and $V(r) \to 0$ as $r \to \infty$, so $V(|q(t)|)\to0$. Therefore
\begin{align*}
E = \lim_{t \to -\infty} E_q(t) = \frac{m}{2}|v_-|^2.
\end{align*}
Using $|v_-| = v_\infty$, this becomes
\begin{align*}
E = \frac{m}{2}v_\infty^2.
\end{align*}
From the outgoing position asymptotic, we also have
\begin{align*}
\frac{|q(t)|}{t}\to |v_+|
\end{align*}
as $t\to+\infty$, because $q(t)-(a_++tv_+)\to0$ and $|a_++tv_+|/t\to |v_+|$. Hence $|q(t)|\to\infty$ as $t\to+\infty$. The hypotheses give $\dot q(t)\to v_+$ and $V(r)\to0$ as $r\to\infty$, so $V(|q(t)|)\to0$. Therefore
\begin{align*}
E=\lim_{t\to+\infty}E_q(t)=\frac{m}{2}|v_+|^2.
\end{align*}
Since the same conserved number $E$ is obtained from both limits and $m>0$, it follows that $|v_-|=|v_+|$. Thus the definition $v_\infty:=|v_-|=|v_+|$ is consistent.
[/step]
[step:Evaluate the conserved angular momentum on the incoming asymptotic line]
Let $L \in \mathbb{R}^3$ denote the constant value of $L_q$. We use the incoming scattering asymptotics from the theorem statement. Define the error terms $r_-: (-\infty,0)\to\mathbb{R}^3$ and $w_-: (-\infty,0)\to\mathbb{R}^3$ by
\begin{align*}
r_-(t):=q(t)-(a_-+tv_-),
\end{align*}
\begin{align*}
w_-(t):=\dot q(t)-v_-.
\end{align*}
Then $r_-(t)\to0$, $w_-(t)\to0$, and $t w_-(t)\to0$ as $t\to-\infty$. Bilinearity of the cross product gives
\begin{align*}
q(t)\times \dot q(t)=a_-\times v_-+a_-\times w_-(t)+t v_-\times w_-(t)+r_-(t)\times v_-+r_-(t)\times w_-(t).
\end{align*}
Each error term tends to $0$: this follows from $w_-(t)\to0$, $t w_-(t)\to0$, $r_-(t)\to0$, and boundedness of $w_-$ for all sufficiently negative $t$. Hence
\begin{align*}
L = \lim_{t \to -\infty} m q(t)\times \dot q(t) = m a_- \times v_-.
\end{align*}
Taking Euclidean norms gives
\begin{align*}
|L| = m|a_- \times v_-|.
\end{align*}
By the definition of the incoming impact parameter,
\begin{align*}
b = \frac{|a_- \times v_-|}{|v_-|}.
\end{align*}
Since $|v_-| = v_\infty$, we obtain
\begin{align*}
|L| = m v_\infty b.
\end{align*}
This proves the stated conservation laws, the planar confinement, and the scattering formulas.
[/step]
