I am working on this article on 'The Slingshot Effect' by R. C. Johnson (http://maths.dur.ac.uk/~dma0rcj/Psling/sling.pdf) and have become stumped on the derivation of the following equation: $$v_{f}^2 = v_{i}^2 + 2 V \{V (1 - \cos\beta) + v_{i}[\cos(\alpha - \beta) - \cos \alpha]\}$$
The question takes two scalene vector triangles as shown below:
where, angle positions for vector triangles:
Using this information, one should end up with the equation given above.
It appears that the law of cosines has been used, however I cannot find the final equation as required.
Any help would be greatly appreciated, Thank you.
The following might not be the “straightforward trigonometry” Johnson had in mind, but it is straightforward. The key to the derivation is that in the planet’s frame, the distant-approach and distant-departure velocities have the same magnitude, i.e., that the two dashed line segments have the same length. The Law of Cosines immediately gives $$V^2 + v_{\text i}^2 - 2 V v_{\text i} \cos\alpha = V^2 + v_{\text f}^2 - 2 V v_{\text f} \cos\alpha'$$ or $$v_{\text f}^2 = v_{\text i}^2 + 2V(v_{\text f}\cos\alpha'-v_{\text i}\cos\alpha)$$ which is promising but still leaves finding an expression for $v_{\text f}\cos\alpha'$, the scalar projection of $\mathbf v_{\text f}$ onto $\mathbf V$, in terms of the other quantities.
I would instead proceed by transforming $\mathbf v_{\text i}$ into $\mathbf v_{\text f}$ directly. The effect of the encounter is to rotate the planet-relative velocity $\mathbf u_{\text i}=\mathbf v_{\text i}-\mathbf V$ through an angle of $\beta$, i.e., $\mathbf v_{\text f} -\mathbf V = R_\beta(\mathbf v_{\text i}-\mathbf V)$. W.l.o.g. choose a solar reference frame in which the planet’s velocity is $V\mathbf i$. Then, with a bit of trigonometric simplification, $$\begin{bmatrix} v_{\text f}\cos\alpha' \\ -v_{\text f} \sin\alpha' \end{bmatrix} = \begin{bmatrix} V \\ 0 \end{bmatrix} + \begin{bmatrix}\cos\beta&-\sin\beta\\\sin\beta&\cos\beta\end{bmatrix} \begin{bmatrix} v_{\text i}\cos\alpha - V \\ -v_{\text i}\sin\alpha \end{bmatrix} = \begin{bmatrix} V(1-\cos\beta)+v_{\text i}\cos(\alpha-\beta) \\ -(V\sin\beta+v_{\text i}\sin(\alpha-\beta)) \end{bmatrix}.$$ The right-hand side is effectively a rotation about $(V,0)$ in the chosen solar frame. This immediately gives you the two identities just below equation (3) in the paper, and taking the square of the norms of both sides produces the equation that you’re trying to derive, again after a bit of trigonometric simplification. The minus signs in the second components of the velocity vectors reflect the fact that I’m using a right-handed coordinate system with $\beta$ positive.