Solution 6
Concept:
To catch up to station B, the spaceship from A moves to a lower orbit. In a lower orbit, the orbital velocity is higher, and the angular velocity is significantly higher. This allows the ship to “gain angle” relative to the outer orbit.
Calculations:
Let the initial orbit be at radius $R$ and velocity $v$. Angular velocity $\omega = \frac{v}{R}$.
The spaceship shifts to a lower orbit $R – h$ (where $h \ll R$).
The relationship between angular velocity $\omega$ and radius $r$ is given by Kepler’s Law ($\omega^2 \propto r^{-3}$):
$$ \omega = \sqrt{\frac{GM}{r^3}} $$Differentiating to find the change in $\omega$ for a small change in $r$ ($dr = -h$):
$$ \frac{d\omega}{dr} = \sqrt{GM} \left( -\frac{3}{2} r^{-5/2} \right) = -\frac{3}{2} \frac{1}{r} \sqrt{\frac{GM}{r^3}} = -\frac{3\omega}{2r} $$The increase in angular velocity for the lower orbit is:
$$ \Delta \omega = \left| \frac{d\omega}{dr} \Delta r \right| = \frac{3\omega}{2R} h = \frac{3(v/R)}{2R} h = \frac{3vh}{2R^2} $$Station A is behind Station B by a linear distance $s$. In terms of angle, $\Delta \theta_{req} = s/R$.
The spaceship must gain this extra angle $\Delta \theta_{req}$ over the time duration $\Delta t$.
Solving for the separation $h$:
$$ s = \frac{3vh \Delta t}{2R} $$ $$ h = \frac{2sR}{3v \Delta t} $$Answer: Separation $h = \frac{2sR}{3v \Delta t}$