Related : Artificial star

Instead of creating a literally artificial star, i want to create a satellite that has a fusion plant inside it and lots, LOTS of high power lights outside shining towards earth.


  1. Can it be tidally locked to the rogue planet so that you only need a single side of lights facing the planet ? In other words, how can i avert shining light away from the planet ?
  2. How much power does the fusion plant need to produce (considering 100% efficiency in producing light from it) in order to generate a sun-like illumination of the planet ?
  3. Whats should be the lights peak emission frequency in order to generate a sun-like light able to sustain plants and life ?

In other words, provided we have the materials and dominate fusion power, can we produce a artificial satellite that can mimmick basic sun effects on earth (besides gravity effects), effectively making a rocky rogue planet habitable ?

Consider an earth like atmosphere and a earth like radius, gravity and mass.

  • $\begingroup$ Does the satellite have to have lights? Could it just expose the raw fusion furnace to the planet, sort of like how a star works? Then I think the spectrum could mimic a stars by varying the temperature of the furnace. That would also get you 100% efficiency. $\endgroup$
    – AndyD273
    Commented Apr 24, 2015 at 17:57
  • $\begingroup$ If its possble, then ok. $\endgroup$
    – Jorge Aldo
    Commented Apr 24, 2015 at 18:07
  • $\begingroup$ Why go to all the trouble of putting a fusion reactor up there, and having to bring fuel to it? Just make a big, lightweight mirror (or lots of small mirrors), and use the fusion reactor we already have. $\endgroup$
    – jamesqf
    Commented Apr 24, 2015 at 19:02
  • $\begingroup$ @jamesqf If you're talking about the sun, it's a rogue planet, so no star anywhere close by. $\endgroup$
    – AndyD273
    Commented Apr 24, 2015 at 20:23

2 Answers 2


The absolute minimum amount of power you need is equal to the total top-of-atmosphere insolation of the entire Earth. We can calculate this by multiplying the projected surface area of the planet with the solar constant:

$$ P = E\cdot\pi R^2 = 1.7\times10^{17}~\text{W} $$

Assuming your reactor uses D-T fusion, and does not capture energy from the neutrons produced, your efficcency will be around:

$$ 6.7\times 10^{13}~\text{J}/\text{kg} $$

Meaning that your fuel consumption rate is around:

$$ 2.6~\text{ton}/\text{s} $$

If you want your reactor to look like the Sun, the simplest design is a flat plate maintained at $5778~\text{K}$, the effective temperature of the Sun. In order to figure out what the efficcency is, we need to know how far away the reactor is going to be. Just for simplicity, let's put it at the distance of the Moon. In fact, let's just put it on the lunar surface, since then we don't have the problem of maintaining this thing in orbit. This will work out pretty well since the Moon is already tidally locked to Earth, and the "lunar day" is about equal to the solar day.

From the Moon, the Earth seems to subtend an angle of around $1.9^\circ$. This means that, assuming the plate is an ideal Lambertian radiator the amount of emitted power that lands on the Earth is around:

$$ \frac 1\pi\int\cos\theta \sin\theta\ d\theta\ d\phi=\sin^2 r=0.0275\% $$

This means that we need to multiply our power requirements by $3640\times$:

$$ P = 6.3\times 10^{20}~\text{W} \\ \dot m = 9400~\text{ton}/\text{s} $$

From the Stefan-Boltzmann law we know that our plate emits:

$$ \Phi = \sigma T^4 = 6.3\times 10^7~\text{W}/\text{m}^2 $$

So the surface area of the disk must be at least:

$$ A = \frac P \Phi= 1.0\times 10^7~\text{km}^2 $$

For a dimameter of:

$$ D = \sqrt{\frac{4A}\pi}=3560~\text{km} $$

The diameter of the Moon is around $3475~\text{km}$, so we need to take up the whole Moon with our emitter. This makes intuitive sense: since the Moon is the same angular diameter as the Sun when viewed from Earth, if they had the same brightness then they would appear equally bright.

  • $\begingroup$ @JorgeAldo Note that the power obeys the inverse-square law: if you halve the altitude of your sun-satellite, you can quarter the power. $\endgroup$ Commented Apr 24, 2015 at 20:57
  1. I believe that most satellites, shuttles, and the space station are tidally locked, so that things like communication dishes and cameras can be used. So that's no problem.

  2. I don't know if power is what you really need. If you can make fusion, why not just expose the planet to the raw fusion? That way you're getting the full spectrum. You need light and heat, and converting it from fusion (light and heat) to electricity to light and heat is a pretty lossy process. The extra benefit of this is that any electricity that is produced by the fusion reactor could be beamed down to the planet as energy.

  3. With a fusion reactor you have a mini star. The way I understand it, if you change the temperature, you get a different spectrum. So about 5800 K.

You might need more than one of them. One benefit of this is that you could get rid of timezones. Put the satellites around the planet in a ring (like GPS satellites). Light them up when it's day time and off at night. You wouldn't get sunrises though...

  • 1
    $\begingroup$ Satellites are not large enough to be tidally locked. They use reaction wheels or gravity-gradient torque to stabilize themselves in an Earth-facing configuration. In fact, if they are designed incorrectly they will spontaneously tumble. $\endgroup$ Commented Apr 24, 2015 at 18:17
  • $\begingroup$ @2012rcampion Thanks for the clarification! I knew that they were able to keep themselves aligned toward the earth, but hadn't read up on how they did it. $\endgroup$
    – AndyD273
    Commented Apr 24, 2015 at 18:19
  • $\begingroup$ Some related interesting reading: the Lewis satellite had an incorrect mass distribution that somehow passed review; after about three days of struggling to maintain attitude, the spacecraft turned off its thrusters and went into 'safe mode,' and immediately flipped over, pointing the solar panels away from the Sun. Its batteries were totally dead within hours, and it re-entered the atmosphere around a month later. The gravity-gradient torque is really important! $\endgroup$ Commented Apr 24, 2015 at 19:35

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