I was thinking about stellar engines. One type of stellar engine is the Caplan thruster, which concentrates the star's light onto the star's surface to create beams of solar wind, which are collected by Bussard ramjets to propel the star.

I would like to know how efficient and fast seven Caplan thrusters would be at pushing the Sun. People have talked about building one, but how much less efficient would each one be (due to the angle of the six 'outside' thrusters) if we built seven? I'm thinking of an arrangement with six thrusters at the corners of a hexagon plus one in the middle, all pointing their hydrogen beams at the same spot. I would like to know this, but I don't even know the thrust or size of one thruster, let alone how much this arrangement would decrease efficiency.

Sorry for the order in which I gave out details in this question.

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    $\begingroup$ Welcome to Space Exploration SE ! Could you add some link(s) that explain what Caplan thrusters and stellar engines are ? $\endgroup$
    – Cornelis
    Commented Oct 3, 2020 at 11:15

1 Answer 1


Quick summary of what a "Caplan" thruster is:

  1. A Dyson swarm collecting sunlight, shooting it back at the Sun to stir up mass as solar winds.
  2. Some system to collect that solar wind.
  3. Fusion reactors, using the helium from the collected solar wind.
  4. A fusion product jet of oxygen-14 pointed into space.
  5. A hydrogen jet pointed back at the Sun.

Caplan's paper.

You propose to split 4) and 5) into multiple parts instead of just one.

The Sun is a 1.4 million km wide target. It doesn't matter if you don't shoot at the middle of it, the momentum will be transferred anyway.

Which means that unless the distance between the jets is greater than 1.4 million km, there's no loss in efficiency.

If the distance between is jets is greater than 1.4 million km, your formation is simple trigonometry. If a jet is tilted at an angle of $\theta$, only $\cos{\theta}$ of its acceleration is usable thrust.

For seven equal jets that's an efficiency of:

$$\eta = \frac{6\cos{\theta} + 1}{7}$$

There's no reason to believe this arrangement would provide any more thrust than Caplan's original setup. In fact, his setup doesn't care much about the geometry of it at all, it's a rough best-case estimate. The acceleration should therefore still be about $10^{-9} m/s^2$ as described in his paper.


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