Swing-by maneuver

The momentum conservation for the probe with mass $m$ and the planet with mass $M$ reads $\begin{aligned} m v + M V &= m v' + M V' \\ \Rightarrow \quad V' &= V + \gamma ( v - v') \end{aligned}$ with the mass ratio $\gamma = m/M \ll 1$ . The energy conservation reads $\begin{aligned} & & \frac{1}{2} m v^2 + \frac{1}{2} M V^2 &= \frac{1}{2} m v'^2 + \frac{1}{2} M V'^2 \\ \Rightarrow & & \gamma v^2 + V^2 &= \gamma v'^2 + (V + \gamma ( v - v'))^2 \\ & & &= \gamma v'^2 + V^2 + 2 \gamma (v - v') V + \gamma^2 (v - v')^2 \\ & & &\approx \gamma v'^2 + V^2 + 2 \gamma (v - v') V \\ \Rightarrow & & v^2 &= v'^2 + 2 (v - v') V \\ \Rightarrow & & v + v' &= 2 V \end{aligned}$ (Here, we neglected the term proportional to $\gamma^2$ .) The insertion of the numerical values provides the result $\begin{aligned} V &= -2 \pi \frac{a}{T} = -2 \pi \frac{7.8 \cdot 10^{8}}{11.9 \cdot 365 \cdot 24 \cdot 60 \cdot 60} \frac{\text{km}}{\text{s}} \approx -13 \frac{\text{km}}{\text{s}} \\ \Rightarrow \quad \Delta v &= -(v + v') = -2 V \approx 26 \frac{\text{km}}{\text{s}} \end{aligned}$