Solution: Gravitational Scattering
Consider a spacecraft of mass $m$ moving with initial speed $v$ at a large distance from a star of mass $M$. The spacecraft follows a hyperbolic trajectory under the gravitational influence of the star. Since $M \gg m$, we assume the star remains stationary at the focus of the hyperbola.
Fig 1: Trajectory of the spacecraft with $90^\circ$ deflection.
Derivation: Eccentricity vs Scattering Angle
1. Polar Equation of the Orbit
The trajectory of a particle under an inverse-square force is described by the polar equation:
$$ \frac{p}{r} = 1 + e \cos \theta $$
Here, $r$ is the distance from the star, $\theta$ is the angle measured from the periapsis (closest approach), and $e$ is the eccentricity.
2. Asymptotes (Distance $\rightarrow \infty$)
Far from the star, $r \rightarrow \infty$, which implies the left side of the equation becomes zero:
$$ 0 = 1 + e \cos \theta_{\infty} \implies \cos \theta_{\infty} = -\frac{1}{e} $$
Here, $\theta_{\infty}$ represents the angle of the asymptotes relative to the axis of symmetry.
3. Geometric Relation to Scattering Angle ($\delta$)
Let $\alpha$ be the acute angle between the asymptote and the axis of symmetry. Since $\theta_{\infty}$ is in the second quadrant:
$$ \cos(180^\circ – \alpha) = -\frac{1}{e} \implies -\cos \alpha = -\frac{1}{e} \implies \cos \alpha = \frac{1}{e} $$
The total scattering angle $\delta$ is related to $\alpha$ by the geometry of the hyperbola:
$$ \alpha = 90^\circ – \frac{\delta}{2} $$
Substituting this into the cosine relation:
$$ \cos\left(90^\circ – \frac{\delta}{2}\right) = \frac{1}{e} $$
Using the identity $\cos(90^\circ – x) = \sin x$, we arrive at the final relation:
$$ \sin\left(\frac{\delta}{2}\right) = \frac{1}{e} $$
Step 1: Calculating Eccentricity
The problem states that the final velocity is perpendicular to the initial velocity, meaning the scattering angle is $\delta = 90^\circ$. Using the derived formula: $$ \sin\left(\frac{90^\circ}{2}\right) = \sin(45^\circ) = \frac{1}{\sqrt{2}} $$ $$ \frac{1}{\sqrt{2}} = \frac{1}{e} \implies e = \sqrt{2} $$
Step 2: Calculating Minimum Distance ($r_{min}$)
The minimum distance of approach $r_{min}$ (periapsis distance) is given by: $$ r_{min} = a(e – 1) $$ From the conservation of energy, the semi-major axis $a$ is related to the initial velocity $v$ at infinity by: $$ \frac{1}{2}mv^2 = \frac{GMm}{2a} \implies a = \frac{GM}{v^2} $$ Substituting $a$ and $e$: $$ r_{min} = \frac{GM}{v^2}(\sqrt{2} – 1) $$
Step 3: Comparing the Two Spacecraft
For the first spacecraft ($r_1, v_1$) and second spacecraft ($r_2, v_2$): $$ r_1 = \frac{GM}{v_1^2}(\sqrt{2} – 1) \quad \text{and} \quad r_2 = \frac{GM}{v_2^2}(\sqrt{2} – 1) $$ Taking the ratio $\frac{r_2}{r_1}$: $$ \frac{r_2}{r_1} = \frac{\frac{GM}{v_2^2}(\sqrt{2} – 1)}{\frac{GM}{v_1^2}(\sqrt{2} – 1)} = \frac{v_1^2}{v_2^2} $$
Conclusion
Solving for $r_2$: