5
$\begingroup$

In this answer to the question "What would the full hypothetical Mars terraforming roadmap look like ?" there's a link to the article "Zubrin on Terraforming Mars" in Universe Today,
From an interview with Robert Zubrin in that article:

RZ: If one considers the problem of terraforming Mars from the point of view of current technology, the scenario looks like this:

  1. A century to settle Mars and create a substantial local industrial capability and population.
  2. A half century producing fluorocarbon gases (like CF4) to warm the planet by ~10 C.

(Emphasis by me)

Part of the main question is: did he sufficiently take into account the extremely large amount of rock that needs to be excavated and processed to extract the required amount of Fluorine ?

The MSL rover Curiosity detected fluorine containing minerals in Gale crater.

$\endgroup$
3
  • 3
    $\begingroup$ The Case For Mars is probably the most detailed version that Zubrin has published. $\endgroup$ May 30, 2021 at 16:35
  • 1
    $\begingroup$ Note that part 2 follows part 1, developing "substantial local industrial capability". So it seems reasonable to assume that he was indeed taking this into account. $\endgroup$ May 30, 2021 at 17:50
  • $\begingroup$ @Andrew Indeed, thank you. I've added "sufficiently". $\endgroup$
    – Cornelis
    May 30, 2021 at 19:45

1 Answer 1

3
$\begingroup$

I dug out The Case For Mars (1997) and while it gives some very basic outlines of this, it does not go into any great detail. From Chapter 9:

The industrial effort associated with such a power level would be substantial, producing about a trainload of refined material every day and requiring the support of several thousand workers on the Martian surface. Power levels of about 5000 MWe might be needed, which is about as much power as is required today by a large American city such as Chicago. A total project budget of several hundred billion dollars might well be required. Nevertheless, all things considered, such an operation is hardly likely to be beyond the capabilities of the mid-twenty-fist century.

These calculations in the book are based on producing enough CFCs (probably CF4) to raise the global temperature by around 10K, which he calculates as 0.04 microbar of CFCs, needing about 880 tonnes/hour for twenty years, and an ongoing production rate of about a fifth of that to maintain it once it's built up.

He does not specifically discuss the mining infrastructure needed to get the fluorine, but as it's talking about thousands of workers and city-scale power requirements, it's clear he's aware of how substantial a project it would be.

$\endgroup$
15
  • 1
    $\begingroup$ Query: When the authors mentions years, is that Earth years or Martian years? There's a big difference. $\endgroup$
    – Fred
    May 31, 2021 at 16:07
  • 2
    $\begingroup$ @Fred aha, I've just noticed the previous table in this chapter explicitly says "Earth years"; I think we can probably guess that is the time units being used throughout. $\endgroup$ May 31, 2021 at 18:17
  • 1
    $\begingroup$ @Cornelis The model he's using is described in the chapter, but the specific calculation for CFCs isn't given (it's dealt with in about a page, as one of several options). $\endgroup$ May 31, 2021 at 18:20
  • 2
    $\begingroup$ "such an operation is hardly likely to be beyond the capabilities of the mid-twenty-fist century." I just hate handwavium. $\endgroup$
    – RonJohn
    Jun 1, 2021 at 4:57
  • 2
    $\begingroup$ @Cornelis: CF4 is an aggressive greenhouse gas (much more aggressive than CO2). But to get reasonable estimates of warming you need a proper atmospheric model complete with lapse rates for both temperature & pressure: simple-minded single-layer models don't work, at all. Given known lapse rates it is possible to hand-crank such a calculation as Arrhenius did to get quantatively-OK estimates, or to run a very trivialised computer model. That might be good enough on Mars as there's not much water in the atmosphere. $\endgroup$
    – user21103
    Jun 1, 2021 at 9:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.