Solution 7

Method: Planet Frame of Reference

Let $\vec{V}_p$ be the velocity of the planet and $\vec{V}_s$ be the velocity of the spaceship in the laboratory frame.
Given: $|\vec{V}_p| = |\vec{V}_s| = v_0$ and the angle between them is $120^\circ$.

1. Relative Velocity (Input)

We switch to the planet’s frame of reference. The initial relative velocity of the ship is $\vec{u}_{in} = \vec{V}_s – \vec{V}_p$.

Using the law of cosines:

$$ u = |\vec{u}_{in}| = \sqrt{v_0^2 + v_0^2 – 2v_0 v_0 \cos(120^\circ)} $$ $$ u = \sqrt{2v_0^2 + v_0^2} = \sqrt{3}v_0 $$

2. The Gravitational Scattering

In the planet’s frame, the spaceship follows a **hyperbolic trajectory**.

• The **Vertex** is the point of closest approach.
• The **Planet** is located at the **Focus** of the hyperbola.
• For an attractive force (gravity), the Focus lies **inside** the curvature of the path.
• Conservation of energy ensures speed is constant in this frame: $|\vec{u}_{out}| = |\vec{u}_{in}| = \sqrt{3}v_0$.

3. Maximizing Reverse Speed

To maximize the speed in the reverse direction in the lab frame ($\vec{V}_{final}$), the spaceship must exit such that $\vec{V}_{final}$ is antiparallel to $\vec{V}_s$. Let $\hat{u}_s$ be the unit vector of $\vec{V}_s$. We require $\vec{V}_{final} = -v_{max} \hat{u}_s$.

Using vector addition ($\vec{V}_{final} = \vec{V}_p + \vec{u}_{out}$):

$$ -v_{max} \hat{u}_s = \vec{V}_p + \vec{u}_{out} \implies \vec{u}_{out} = -(v_{max} \hat{u}_s + \vec{V}_p) $$

Squaring both sides:

$$ |\vec{u}_{out}|^2 = v_{max}^2 + V_p^2 + 2v_{max}(\vec{V}_p \cdot \hat{u}_s) $$

Substituting values ($u^2 = 3v_0^2$, $V_p = v_0$, $\theta = 120^\circ$):

$$ 3v_0^2 = v_{max}^2 + v_0^2 + 2v_{max}(v_0)(-1/2) $$ $$ 3v_0^2 = v_{max}^2 + v_0^2 – v_{max} v_0 $$

Solving the quadratic $v_{max}^2 – v_0 v_{max} – 2v_0^2 = 0$:

$$ (v_{max} – 2v_0)(v_{max} + v_0) = 0 $$