Making a Left Turn

For the sake of variety, here's how I did it ( $m$ = probe mass, $M$ = total ion mass, $v$ = probe speed, $V$ = ion speed):

$\large{F = \frac{m v^2}{R} = \dot{M} V \implies \dot{M} = \frac{m v^2}{R V}}$

The time taken for the quarter circle is:

$\large{t_{qc} = \frac{(\pi/2)R}{v} = \frac{\pi R}{2v}}$

The total expended ion mass is:

$\large{M = \frac{m v^2}{R V} \frac{\pi R}{2v} = \frac{\pi m v }{2V}}$

The total ion kinetic energy is:

$\large{2E = M V^2 = \frac{\pi m v V }{2}}$

The probe kinetic energy is:

$\large{2T = m v^2}$

The ratio is therefore:

$\large{\frac{E}{T} = \frac{\pi m v V }{2 m v^2} = \frac{\pi V }{2 v} = \frac{\pi}{2} \frac{5}{2} \approx \boxed{3.927}}$

During a incremental change in direction $d\theta$ , the change in momentum is $mv d\theta$ . Let $M$ be the mass of ions ejected during this time. Then for momentum to be conserved, we have

$mv d\theta=MV$

The energy required to propel this mass $M$ to velocity $V$ is $\frac{1}{2}M{V}^{2}$ , so we have

$\frac{1}{2}mvV d\theta=\frac{1}{2}M{V}^{2}$

We integrate over $\theta$ to get

$E=\frac{1}{2}mvV\Delta \theta$

so that

$\dfrac{E}{T}=\dfrac{V}{v}\Delta \theta=\dfrac{V}{v}\dfrac{\pi}{2}=3.927...$

Moral: Making a left turn in outer space is a costly proposition

The force on the rocket is, on one hand, the needed centripetal force for a circle of radius $R$ (not specified here!); on the other hand, it is equal to the rate at which momentum is given to the ions. Thus $F = m\frac{v^2}R = \mu V,$ where $\mu$ is the mass rate at which the ions are ejected.

The total kinetic energy given to the ejected ions is $E = \tfrac12\mu V^2 t,$ where $t$ is the time needed for the quarter turn. This is easily found to be $t = \tfrac12\pi R/v$ . Thus $E = \tfrac12\mu V^2\cdot \tfrac12\pi \frac Rv = \frac\pi 4 \left(m\frac{v^2}R\right) V\frac R v = \frac\pi 4 mvV.$ Finally, using $V = \frac 5 2 v$ , we find $E = \frac{5\pi}4\cdot \tfrac12mv^2\ \ \ \ \therefore\ \ \ \ \frac E T = \frac{5\pi}4 \approx \boxed{3.927}.$

The assumption here is that the ions are ejected at their maximum speed. It costs less energy (but more time) to eject them at slower speed.

The initial momentum p has spinned 90 degrees, the arc represents for how and how much momentum is changed, the radial length is p, the arc's length is p(3.14/2), (because p=mv, e=mv^2/2), E/T = (arc V)/(radius v) = 3.14(5/4)