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Say we built two large scale portals and sent one to Venus and one to Mars much like in this question but with portals large enough that mass flow rates are not an issue seeing as nobody wants to wait around for 108 billion years for some more habitable planets.

The idea is to dump some of Venus' atmosphere on Mars to get it up to a reasonable temperature, thus leaving us with two semi-terraformed planets rather than just one. (the rest of Venus' excess atmosphere gets dumped out in space and so isn't part of our energy problem)

Now in one of the answers, Demi pointed out that naturally, gas would flow towards Venus despite a 15'000:1 difference in ground-level atmospheric pressure as Venus sits much deeper in the Sun's gravity well. Hence our problem, we need to relocate some of Venus' atmosphere to Mars but just how big of a power station are we going to need...

What I'd like to know is: assuming that we built a whole fleet of air-compressors to force Venus' atmosphere to Mars through our portal (ideally before the end of the millennium), just how much energy would we realistically need to achieve this? (assuming that the portals only represent a small fraction of the total energy expenditure)

Edit: these portals do conserve energy and momentum, if they didn't then all we'd need to worry about is the atmospheric pressure difference. But, when we start adding gravity wells (gravitational potential energy) and orbital velocity (kinetic energy) then the end result becomes... too much for my brain to get a grip on. Something bad presumably if the current answers are anything to go by.

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    $\begingroup$ xkcd.com/681 $\endgroup$
    – Raditz_35
    Jun 7, 2017 at 11:26
  • $\begingroup$ Pressure is the least of your concern... you need to match Mars orbital velocity starting from Venus one. $\endgroup$
    – L.Dutch ♦
    Jun 7, 2017 at 11:29
  • $\begingroup$ @L.Dutch Ah, I forgot about that. Well, I suppose that's going to blow out the energy budget quite a bit. $\endgroup$
    – Samwise
    Jun 7, 2017 at 11:34
  • $\begingroup$ It probably makes sense to say that those portals represents some real tech or way to move the air and are for simplicity of of the question. And they do conserve energy and momentum of that 2 body system. And energy of creating portals is representation of efficiency of the pumping system. If it is the case, because otherwise Q is not interesting, and properties of those portals are unknown and might be any. $\endgroup$
    – MolbOrg
    Jun 7, 2017 at 12:07
  • $\begingroup$ @L.Dutch OP has "Portals". That alone removes orbital characteristics from the list of considerations. $\endgroup$
    – KareemElashmawy
    Jun 7, 2017 at 16:57

4 Answers 4

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I would set the portal on Jupiter and add the requirement that extracted atmosphere be subjected to fractional distillation. That would allow taking the gases needed in the ratios desired to actually perform terraforming.

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You won't need to spend power to get Venus's atmosphere into the portal, since there's such a pressure differential (90 atmospheres to 0.006) Bernoulli's Principle gives us a speed of:

$$ \text{Air speed}=\sqrt{2*\frac{\text{Venus Surface Pressure(9,000,000)}-\text{Mars Surface Pressure(negligible)}}{\text{Density of air on Venus surface(67)}}}=518\textrm{ m/s}$$

You will not have any problems getting the air through. You'll be getting 34,706 kg per second per square meter of portal (518m/s * 67kg/m^3)

However, (and this is what I'm assuming the question is about) there's still the difference in energy between Venus and Mars that you have to worry about.

Energy from orbital speed

Venus's minimum orbital speed is 34.78 km/s, and Mars' maximum is 26.5km/s. The air goes from about 605 MJ/kg to 351MJ/kg. Your portal is going to have to absorb 244 MJ/kg from orbital speed. Assuming the area of the portal is 1 square meter, you will be dissipating around 8.22 TW. I don't know what your portal is made of but it's probably going to melt from absorbed energy, except for the next part.

Energy from orbital altitude

But there's also the energy cost needed for the change in altitude (from the sun.) At Venus, the air has a gravitional energy of -12.2GJ/kg. At Mars, the air has a gravitational energy of -6.41GJ/kg. To get the air to Mars, you need to expend 5.8GJ/kg. Or, with the above portal (1 m^2, 518m/s low rate) that's 201TW to operate. Now, the waste heat of operation will melt it.

Conclusion

You're going to be spending about 5.6 gigajoules per kilogram of air sent. If open a portal with an area of one square meter, the air will rush through at 518 m/s. With the density of Venus's atmosphere, this means your portal will require 192TW to operate. I don't know what it's made of but the waste heat will probably melt it.

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  • $\begingroup$ That seems to pretty much make any attempt to move any appreciable amount of planet further out in the solar system pretty pointless as basically unless I tap a whole star, either I'm just not going to have the energy or I'm going to char grill the planet in question. I guess I need to look elsewhere for my new martian atmosphere. $\endgroup$
    – Samwise
    Jun 7, 2017 at 23:29
  • $\begingroup$ a few sentences on how you assume the flow trough portal and thus powers required would significantly improve the answer. $\endgroup$
    – MolbOrg
    Jun 8, 2017 at 13:52
  • $\begingroup$ @Samwise if you want to open the can of worms that is breaking the conservation of energy, knock yourself out. $\endgroup$
    – ltmauve
    Jun 8, 2017 at 21:39
  • $\begingroup$ @MolbOrg Added a line after the flow speed calculation that explains it. $\endgroup$
    – ltmauve
    Jun 8, 2017 at 21:44
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Suppose we can set aside the gravity thing, given we're already working with "portals" which connect two distant points as if they were contiguous. (This makes sense, I guess, because if it were otherwise, opening one end of a portal anywhere but very close to the other end would cause major disruptions.)

The Venusian atmosphere weighs $4.8 × 10^{20}$ kg and almost all of it is CO₂, the handiest of greenhouse gases. If you just want to increase the temperature on Mars, you need only a bit of CO₂ for a greenhouse effect, so you can import maybe only one hundred-thousandth of the Venusian atmosphere. That would give you a bit more than double the CO₂ content of Earth's atmosphere. If relative pressures are the only concern, I think the problem will be to stop the outgoing flow once you've opened the portals.

But of course you'd need a much denser atmosphere to keep the heat, and it would be convenient to increase the atmospheric pressure on Mars so that people can walk around outside without a spacesuit (only with breathable air tanks). So it might be better to import much more of the Venusian atmosphere. At the surface of Venus the pressure is 93 bar. Mars has 42% of Venus' gravity, so you would need more atmosphere, proportionally, to achieve the same surface pressure. If your target was Earth sea level pressure (~1 bar), you should need

$$ {{4.8 \times 10^{20} kg} \over{93 \times 0.42}} = 1.2 \times 10^{19} kg$$

This is an extremely crude calculation, since things are not that neatly lineal. Also, I have no idea whether Mars will be able to keep all that gas over time.

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  • $\begingroup$ Methane and water vapor are even better green house gasses. Just saying... $\endgroup$
    – ventsyv
    Jun 7, 2017 at 14:30
  • $\begingroup$ @ventsyv Yes, but at least water vapor tends to not be particularly vapory at the temperatures commonly seen on Mars, which could be a problem... $\endgroup$
    – user
    Jun 7, 2017 at 14:47
  • $\begingroup$ Note that Mars' atmosphere is also almost exclusively CO$_2$, clocking in at 96% CO$_2$ according to Wikipedia. Not to be outdone, though, Venus clocks in at 96.5% CO$_2$. $\endgroup$
    – user
    Jun 7, 2017 at 14:50
  • $\begingroup$ @ventsyv Methane could be obtained in large amounts by placing a portal on Titan, but it wouldn't be a good idea because any attempt to then add oxygen to the atmosphere would provoke a combustion. $\endgroup$
    – pablodf76
    Jun 7, 2017 at 15:25
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    $\begingroup$ Might actually be a good idea. Burning CH4 will give you CO2 and water, so you might want to use CH4 to bring the temperature up quicker, then burn it off. But we are way off topic here... $\endgroup$
    – ventsyv
    Jun 7, 2017 at 15:31
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Easiest to do some rough computation is to use the Gravity Well metaphor; in all the computations that follow the numbers (as pointed out by @MolbOrg) virtual meters of 9.81 m/s^2 equivalents. This means we turn the variable gravity field in a fixed equivalent and do all computations "as-if" on surface of Earth. Due to Conservation of Energy these factors can be added trivially.

In Sun's gravity well Venus is at 124.9Mm and mars is at 59.3Mm, so we have a difference of 124.9-59.3=65.6.

Venus own gravity well is 5407km while Mars is only 1274km, so You have to add another 5.407-1.274Mm = ~4.25Mm for a grand total of ~70Mm.

This means You need to lift the atmosphere do be transferred as-if lifting it 70000km. Energy expenditure is quite high.

You would be much better off using Jupiter or Saturn as source and gaining energy in the process.

Otherwise You can dump part of Venus atmosphere very near the Sun so gaining energy; choosing appropriately the place where to dump your excess "air" you can compensate exactly the energy needed to pump what needs to go to Mars.

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  • $\begingroup$ "In Sun's gravity well Venus is at 124.9Mm and mars is at 59.3Mm" - wrong numbers if you mean orbits and distances of planets from the sun $\endgroup$
    – MolbOrg
    Jun 8, 2017 at 23:09
  • $\begingroup$ @MolbOrg: no, I'm meaning depth of gravity well. For the number you cite is how deep the planets are in sun's gravity well, then there is the gravity well of the planet itself: you need to climb out of Venus's and drop down into Mars's (see: spiralwishingwells.com/guide/Gravity_Wells_Mirenberg.pdf; numbers were actually taken from: explainxkcd.com/wiki/index.php/681:_Gravity_Wells) $\endgroup$
    – ZioByte
    Jun 8, 2017 at 23:16
  • $\begingroup$ ok, those are virtual meters of 9.81 m/s^2 equivalents. You probably should clarify that in the answer. $\endgroup$
    – MolbOrg
    Jun 8, 2017 at 23:48

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