Post History

[NASA's *Fundamentals of Electric Propulsion: Ion and Hall Thrusters* by Goebel and Katz, JPL, March 2008](https://descanso.jpl.nasa.gov/SciTechBook/series1/Goebel__cmprsd_opt.pdf) discusses the issue of current ion drive beam focus limits (along with many other matters relevant to ion drives).

Section 5.3.3, page 207:

> Another significant grid issue is alignment of the grid apertures. The ion trajectories shown in Fig. 5-6 assumed perfect alignment of the screen and accel grid apertures, and the resultant trajectories are then axi-symmetric along the aperture centerline.

Figure 5-6(b), page 203, shows a well-aligned ion drive, with optimal perveance. In that case, the beam appears to diverge at a rate of approximately $\frac{1 \times 10^{-4}}{5 \times 10^{-3}}$. Therefore, assuming a point exhaust for simplicity, at a distance of 100 km the exhaust will have spread over a radius of approximately $100 ~\text{km} \times \frac{1 \times 10^{-4}}{5 \times 10^{-3}} = 2 ~\text{km}$.

Assuming that the beam's lack of focus is symmetric around the engine's long axis, we can estimate the area covered by particles ejected from the engine as a circle of this radius.

The area of a circle of radius 2 km is a little less than 13 km<sup>2</sup>.

So, at 100 km distance, the total force of the engine is spread out over approximately 13 km<sup>2</sup>.

Now here's the tricky part. Current ion drives are in the single to tens of kilowatts class (NASA's [HiPEP](https://www.grc.nasa.gov/www/ion/present/hipep.htm "high-power electric propulsion program") "is focused on the development of a 20-50 kW class ion thruster" with a specific impulse in the range 6000-9000 seconds), but you are postulating a 500 MW ion drive. That's quite a lot of scaling up, but I'll make it easy for myself by assuming that we can meet current state of the art at such a massively scaled up power level, and that the specific impulse will be similar to NASA's HiPEP drive. You can substitute your own values below if you wish.

[Wikipedia indicates](https://en.wikipedia.org/wiki/Ion_drive#General_working_principle) that the thrust force of an ion drive, in Newton, can be calculated as $$ F = 2\cfrac{\eta P}{g I_{sp}} $$ where $\eta$ is the engine efficiency, $P$ is the power in watts, $g$ is the gravitational acceleration (9.81 m/s<sup>2</sup> on Earth), and $I_{sp}$ is the specific impulse in seconds. Assuming a 100% efficiency (worst case for the rearward object), and a specific impulse of 10,000 seconds, that gives $$ F = 2 \cfrac{500\,000\,000 ~\text{W}}{9.81 ~\text{m/s}^2 \times 10\,000 ~\text{s}} \approx 10\,200 ~\text{N} $$

10,200 N is approximately 1,040 kgf, but again, this will be spread over an area of about 13 km<sup>2</sup>.

The force exerted by the engine on the object 100 km rearward will thus be approximately $$ \cfrac{1\,040 ~\text{kgf}}{\pi 2\,000^2 ~\text{m}^2} \approx 0.00008276 \cfrac{\text{kgf}}{\text{m}^2} $$ as compared to Earth sea-level atmospheric pressure of (remember that Pa = N/m<sup>2</sup>, and again g = 9.81 m/s<sup>2</sup>) $$ 101\,325 \cfrac{\text{N}}{\text{m}^2} \approx \cfrac{10\,329 ~\text{kgf}}{\text{m}^2} $$

This is a difference of approximately $\frac{10\,329}{0.00008276} \approx 1.25 \times 10^8$ times between the atmospheric pressure from inside the rearward spacecraft, and the engine exhaust pressure from outside it.

Even if they are running a low-pressure atmosphere (say, approximately 20% of Earth sea level pressure but pure oxygen, as the early US space missions in the 1960s did), I really don't think the captain of the rearward ship needs to worry about any significant damage. If the captain of the rearward ship does worry, all that's needed is to back off a little further.