# How to estimate the energy output of artificial suns

## Setting

I'm designing a world where only a thin belt wrapping the equator of a planet is hospitable. It would look something like this:

The backstory, diluted for brevity, is that the central sun was once bright but due to unfortunate events its luminosity was rendered negligible. The planet, previously a water world and tropical paradise, began rapidly cooling. Luckily, the human inhabitants, though unable to save themselves, were in the good graces of aliens who built one or more artificial suns to warm the equator. The artificial suns, aka "lamps", shine "spotlights" of solar-spectrum light tailored for the human ecosystem (minus some of the unnecessary radio and x-rays).

At some point, I'd like the story centered on this world to explore the nature of these "lamps", which in part means getting a decently accurate total energy output (and then justifying it via some means of energy production).

However, there are a few caveats which may complicate things just a little more. My world is not Earth-sized, but instead Mars-sized. Despite this, it still has Earth-like pressure and temperature at sea level, which means a much taller atmosphere (I calculate ~290 vs 100 km for the Karman line equivalent (scale height 24.8 km)). However, I also calculate that the atmospheric mass is roughly the same (3.7e18 vs 5.2e18 kg), which to me suggests that both atmospheres might also have a similar energy content and a linear approximation between the two might be ballpark feasible.
(If you want to see these calculations I can edit them into the question, but if they're not needed I'd rather not sully the question with a wall of math. I found the atmo. mass by first assuming my world's atmosphere is structured like Earth's and scaling the graph of Earth's density gradient to my world's lower gravity. I then took the sum of density times the volume of successively larger spherical shells extending to the Karman line equivalent.)

Planet properties:

• Radius: 3,100 km
• Surface gravity: 3.34 m/s^2
• Mass: 4.8e23 kg

## Approximating energy output

As a first order approximation, I thought to just find the energy Earth receives and scale it down to my world. Earth receives 1,400 W/m^2 solar incidence over a cross-sectional area of $$\pi R_{Earth}^2$$, for a total of 1.8e17 watts. With some assumptions about the area of the hospitable belt around my world, I can multiply that wattage by the fraction of belt area divided by Earth area to get a fraction of that wattage, and the total output of all the lamps.

The area of the belt is $$A_{belt}=4\pi R^{2}\sin\left(\theta/2\right)$$, (derived from subtracting off the area of two spherical caps from the spherical surface area) where $$\theta$$ is derived from the arc length of the belt, $$L$$, as it extends north-south of the equator: $$\theta = \frac{L}{R}$$; essentially, the diameter of the lamp's "spot". I'd like the belt to be relatively thin, stretching no more than ~500 km above and below the equator, for an arc length of 1,000 km. Like portrayed in the picture above, I expect the actually warm zone to be much thinner than this due to the frozen wastes of higher latitudes leaking/mixing in (likely manifesting severe storms along the way). Anyway, I calculate a belt area of 1.9e13 m^2 (~16% of the planet's surface area, or ~4% of Earth's) and subsequently 4% of Earth's incident power, about 6.8e15 watts.

That's a lot of watts. Packed into just one or two spotlights, they would probably fry everything below, right? So, perhaps we should divvy it up. The lamps are in a circular orbit over the equator, pointed constantly "down" at the surface, and do not deviate or turn off. No need to worry about 12-hour day-night shifts, which are likely more trouble than they're worth given the extreme climate anyway, so the energy can be evenly divided among multiple lamps until it's "safe". Normal ground insolation is IIRC ~900 W/m^2, and the "spot size" has a circular area about $$\pi 1000000^2$$ m^2, to an approximation.

$$W_{spot}=900\cdot\pi1000000^{2} \to \frac{W_{total}}{W_{spot}} = 2.4 \ \text{lamps}$$

Oh, well maybe not then.

## Distributing energy output

However, with only 2 lamps in orbit, any arbitrary point on the equator receives illumination only 10% of the time. With an effective breadth of 1,000 km, each lamp's spotlight covers ~5% of the equatorial circumference (19,500 km), leaving 90% of the circumference dark at all times. To get 50% mean coverage would require 10 lamps, their spotlights side-by-side, each outputting about 25% less power, or 6.8e14 watts. It may be favorable to break the power up among more anyway to evenly distribute the energy and not "stir the pot" with localized hot spots, creating hectic storms. This question is not about the severity of weather/climate—I'll ask a separate question about that if all goes well here—but I expect that'll be a problem going forward.

For now, I'm leaving the orbital altitude of the lamps open to modification, though I'd like it to be within the range 500-5,000 km. With an assumption about the beam angle of the lamp, $$\theta_{lamp}$$, you can calculate the orbital altitude with the equation:

$$z=\frac{R_{P}\sin\left(\frac{1}{2}\frac{L}{R_{P}}\right)}{\tan\left(\frac{1}{2}\theta_{lamp}\right)}.$$

With a beam angle of 45° for example, orbital altitude is ~1,200 km, and the other elements follow.

## Question

With that out of the way, Is my approximation reasonable? Am I neglecting something? How can you estimate the energy output of these artificial suns?

As an aside, I know a Mars sized planet has little chance of developing such a disproportionate atmosphere naturally or retaining it for long. However, if such an atmosphere was made it would persist for many millions of years. Space weather is not that bad.

• What stops your atmosphere from precipitating out near the poles? With only a fraction of normal energy input, I'd assume all the gases would freeze. Commented Feb 6, 2023 at 4:41
• @jdunlop That may be so. The planet is getting something like like 1/5th the power input it had when the central sun was strong, only highly localized near the equator. Unsure how well/unwell that energy distributes into the rest of the atmosphere. (Inputting enough energy to keep the atmosphere from condensing in short order might be a big concern for the lamp builders.)
– BMF
Commented Feb 6, 2023 at 5:03
• OK, tl;dr, did you account for the altitude of the lamps? The lower they are, the less instantaneous power they need to generate. Are they global lights (like a star) or are they focused? If focused, what's the arc distribution of the focus and why wouldn't they spread the love along a larger portion (longitude) of the equator than 10%? Finally, you might be overengineering the math. Our sun hits the Earth with avg 1380 W/m^2. So, pick your number, decide how many m^2 you have, and multiply. Account for altitude if necessary. Job done.
– JBH
Commented Feb 6, 2023 at 5:56
• @JBH Yeah, all that is addressed in the Q. 1) The lamps act like "spotlights": cones of light shining down on the planet. I fixed the spot size to 1000 km diameter and allowed beam angle and subsequent orbital altitude to be variable. 2) What you described in your last few sentences is basically what I did. The question is whether that approach is actually good enough to get the job done. User jdunlop seems to have already found a hole in that plan (or at least an additional variable that needs accounting).
– BMF
Commented Feb 6, 2023 at 8:49
• There are useful calculations in two answers in this question: worldbuilding.stackexchange.com/questions/237946/…
– user86462
Commented Feb 7, 2023 at 13:51

## 1 Answer

It sounds like you are talking about fusion candles. As for the energy being emitted, it would be tailored (as you mentioned) to release the desired light wavelengths. The higher energy spectrum (ionizing radiation) is what causes harm to lifeforms, DNA, and breaks down molecular bonds. Other forms of radiation tend to cause damage via heat.

The fusion candles would likely be “tuned” to release light in the visible spectrum and reaching down into the infrared. It would stop before reaching microwave and radio radiation frequencies, but those are low power and not really a huge drain. This significantly reduces the power needed to provide light to your planet. https://en.wikipedia.org/wiki/Electromagnetic_spectrum This breaks down the energy per photon. Gamma and Hard X-rays take the most power to produce (1.24 MeV and 124 keV, respectively). Dropping into the visible and near infrared reduces power requirements to 1.24 eV and 124 meV. These are several orders of magnitude apart, so by avoiding the higher-powered radiation bands, the fusion candles can be quite energy efficient and safe. I count seven reductions in x10 power to reach the visible spectrum.

This takes your 6.8e15 watts down to 6.8e8 watts, or 6,800 Megawatts. ITER is estimated to be able to produce 500 megawatts. It would take 13.6 ITER-sized reactors to produce the power needed. As the gaps between the lights would likely be further apart than desired, you could simply double the number of candles and halve the energy production capabilities.

While this might not seem like a lot of electricity, it is important to understand that the average home only uses 10% of its electricity for lighting. We are also assuming the fusion reaction is producing the photons of the wavelength we want and nothing outside that range. By default, nuclear fusion produces gamma-rays, which we do not want. The conversion of higher energy radiation to lower energy radiation could potentially utilize future tech. Perhaps a brief explanation about how an electrical field converts the wavelengths into the desired output.

The other elephant in the room is the evaporation and loss of water which would occur over time. The light from the fusion candles would cause water to evaporate. Some of that water would condense and fall as snow in the cold regions of the planet. There is the potential that the water along the equator would decrease in volume as more and more became trapped in the polar regions. Ice would need to enter the equatorial belt to thaw, either from glacial creep (land) or icebergs (ocean).

• Over 40% of the sun's energy is already in the visible spectrum, under 10% is beyond ultraviolet. Ignoring everything but the visible spectrum will drop your power requirement by about half, not by 99.9999%. Commented Sep 1, 2023 at 18:30