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To melt a planets surface erasing all traces of civilisation, I figured a solar furnace would do the trick. It is more elegant than nuking a planet, not to mention leaves no nuclear fallout. Perfect for the cleansing! Afterward the scorched planet can easily be reinhabited after it has cooled.

Using the reflective surfaces of the spacecrafts, a sufficiently large fleet could form a giant concave mirror. Each of the ships has a diameter of approximately 1 km with a circular shape. The result is a beam of light focused on a single point. Not very efficient for cleaning the surface of a planet so let's make that a line instead. Focusing the beam on the diameter of the planet would end the process in a day. The Odeillo solar furnace measures 48 metres (157 ft) wide and 54 metres (177 ft) tall and can achieve a temperature of 3,500 °C (6,330 °F).

How large of a surface area should this solar furnace be to achieve the forementioned effect?

Take earth as the example. A diameter of 12,742 km (7917.511 miles). Granite has a melting point of about 1215° to 1260° (2219°F to 2300°F). About 20 seconds of exposure is enough to melt it into lava using focused light, though I haven't found the dimensions of the mirrors used. The rotation of earth also needs to be taken into account, clocking at about 460 meters per second (1,000 miles per hour).

I'd crunch the numbers myself if I could. Unfortunately I my math is bad. Do your thing please.

PS: Please don't mock me for being unable to do this 'simple' calculation on my own. Life is hard enough as it is.

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  • $\begingroup$ "Please don't mock ..." - You realise that's an invitation, right? $\endgroup$ Commented Mar 31, 2022 at 19:23
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    $\begingroup$ It seems like you're missing a critical component. How long to you want the surface melting to take. Given long enough and a little bit of handwaving the Odeillo furnace could melt the entire surface. Are you wondering how large a facility you'd need to melt Earth's surface in a day? $\endgroup$
    – sphennings
    Commented Mar 31, 2022 at 19:24
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    $\begingroup$ do you really need to melt the whole planets surface? seems a bit overkill, and there will be no easy repopulation of this lifeless rock. $\endgroup$
    – ths
    Commented Mar 31, 2022 at 23:27
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    $\begingroup$ +1 for the scenario, it reminds me of Niven/Pournelles Footfall.. this is also quite primitive but very effective. I wonder if you could set up an optical defensive shield against it. Like a huge amount of counter-mirrors in orbit, destroying the ships. $\endgroup$
    – Goodies
    Commented Mar 31, 2022 at 23:38

2 Answers 2

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As a rule of thumb, you get about 1 kW of sunlight per meter squared on the earths surface. In space you have about 1.3 kW per meter squared without the atmosphere. So ignoring things like clouds, the precise latitude, time of year, time of day etc. You can say about 30% of the energy doesn't reach the earth.

The temperature rock melts varies a lot, but silicas, and some of the more heat resistant minerals melt in the 1200-1300 C range.

The size of of the Odeillo tower is not as important as knowing the total area of the heliostats that they are using to direct light to the parabolic mirror and the area of the target that is illuminated. In that case they have 63 heliostats that track the sun and total area of about 2835 square meters. Roughly you can say that it is magnifying the sun about 3000 times, and they rate the facility at about 1 megawatt. It seems it can reach temperatures of about 3,000-3500 C, although the details are not that clear.

By the way, there a some theoretical limits as to how much you can concentrate sunlight, and the record is somewhere around 84,000 times. However, that technique isn't very helpful for this application. More generally for a terrestrial solar furnace a magnification of 2000, is pretty reasonable.

Well, you have 1 km diameter space ships that you want to position. The best you can do in directing the light as a beam, (assuming the light from the sun is mostly parallel before it hits the space ship, would be diffraction limited by the angle theta=1.22*wavelength_in _meters/1000meters. This angle is pretty small and is in radians, 6.1xe-10, but if in geosynchronous orbit about 35,760 km then the 1km spot would be 70 meters larger, so that that sounds still about a km. But if as far out at the moon, 363,300 km, then the spot would have a diameter would be 1.72 km in diameter. This would depend on the wavelength with shorter wavelengths spread out more. To make your spots smaller you could put a slight curvature on your 1 km spaceships, but you will still end up being limited by diffraction by the diameter of your space ship.

So to get the energy density concentrating 2000 times, you could use 2000 spaceships and overlap the spots. Then to get the line to scan, you could add them in increments of 2000, each increment adding km to the line. Or less if you want to overlap some for efficiency.

The circumference of the earth is about 40,000 km, so 1/2 that is 20,000 so 20,000 spaceships in lined up in groups of 2000 would be one option. You might think that you could just let the earth rotate underneath the space ship array. Well, then the equator is moving faster than the higher latitudes 40,000/24 hrs is about 1,667 km/hr, so that is a minor problem.... you would be under the spot for about 18 seconds, and might not be quite hot enough to melt the rock at the equator, as you go up in latitude you would spend more time within the spot.

Of course everything up to now neglected the atmosphere, or the oceans, or vegetation. This gets sort of complicated. Initially, assuming turning on one beam from the 2000 spaceships, you will have absorption from the atmosphere, with the atmosphere warming up in the spot, that will create convection. You will also get some changes in the refractive index and a variety of scintillation effects, so the way the light actually reaches the ground will be distorted and changing due to convection and the turbulence. The molten rock of course will also create convection currents and storms. The burning material will be carried up into the atmosphere and change the reflectivity the way the light is absorbed in the atmosphere.

However, water vapor and cloud formation will likely cause the biggest effects. How this might work could be complicated. Usually, one would think that the clouds would be very reflective so much less light would reach the ground. However, there may be enough energy to vaporize water droplets, perhaps punching a hole through the clouds. Then there are secondary effects like greenhouse gas effects...

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Since you're concerned about the math let's make sure the explanation provided shows how to set up the math.

First in the interest of simplicity we're going to ignore atmospheric effects, and assume an ideal energy transfer and perfect positioning of mirrors.

You want to melt the surface of the earth in 1 day. So you need to figure out how much surface area a given size of array can melt in a given time. Then it's just some divisions, and unit conversions to determine how much area you need to melt the whole surface in a day.

So surface_area_of_earth / seconds_in_day / area_melted_per_area_of_mirror_per_second

Judging by one of the videos in one of your links it looks like a one meter lens can melt 10 square centimeters of rock in about 20 seconds. Or .5cm^2 per second for each square meter of mirror.

Wolfram Alpha says that there's roughly 5.1*10^18 square centimeters on the surface of the earth.

Do the math and you'll find you need roughly 1.18x10^14 square meters to melt the surface of the Earth in a day. Or roughly 118,000,000 square kilometers of mirrors. That's roughly seven times the size of Russia. If your mirror ships have a diameter of 1km you'll need almost 150 million ships.

If you wanted to only melt the land you can multiply the figures by .29 (the percentage of the earth's surface that isn't covered in water) and you'll get 34,200,000 km^2 of mirrors, only 2 Russias, or a little less than 43,00,000 ships.

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  • $\begingroup$ Computer-controlled logistics + fuel for station-keeping by 10M ships... sounds like enough energy to use in a more-direct (albeit less cool) fashion. $\endgroup$
    – Tom
    Commented Apr 1, 2022 at 5:13
  • $\begingroup$ The ships are 1km in diameter, not radius. Also, I couldn't reproduce your math. A 1 km^2 mirror can melt 0.00005 km^2/second or 4.32 km^2/day; divide 510 million km^2 of earth by that and you get 118 million square km of mirrors needed. Also since the ships are circular they will overlap some, each contributing a hexagonal area at best (about .65 km^2). That's 181.8 million ships for the whole surface, or 52.7 million ships for land only. Probably need to consider more variables like atmospheric scattering, angle of the reflection, etc. as well. $\endgroup$
    – BoomChuck
    Commented Apr 1, 2022 at 23:44
  • $\begingroup$ @BoomChuck Thanks for pointing out the math issue. I reran the numbers and I got figures similar to yours. However I don't thing that overlapping is a meaningful consideration. There is no requirement for the mirrors to be contiguous, Assuming each space mirror has a diameter of 1 KM as long as each mirror is kept more than .5 KM apart (trivial to do in space) then we will have no overlap to worry about. $\endgroup$
    – sphennings
    Commented Apr 2, 2022 at 0:26

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