Is Earth is doomed in Project Hail Mary ?
Let's ignore that happy ending and do some science :)
So I've been pondering Earth in the Hail Mary's absence and specifically what we can derive from canon to determine a climatic model for what happens. In order to figure this out I went through the book with the finest comb possible to pick out all the instances of related information that Weir has stuck in there.
This is Part 1 of 3. First we will explore the assumptions established in the canon text, then we will model those scenarios utilizing a couple different simulators, and finally, we will wrap up the viability of these scenarios (and what it might look like to actually survive them!)
So here's what I have from that:
(please feel free to comment any other book-specific information I missed! I will not be using any movie or author notes outside the book to keep things easily citable. Thanks!)
1) The sun's luminosity is decreasing. Specifically, we have the numbers 0.01% for the start of the novel, then 1% over 9y and 5% in 20y.
I shrugged. “So? Where are we in the solar cycle?”
She [Marissa] shook her head. “It’s not the eleven-year cycle. It’s something else. JAXA accounted for the cycle. There’s still a downward trend. They say the sun is 0.01 percent less bright than it should be.”
(...)
“It’s right,” she said. “The sun’s output will drop a full percent over the next nine years. In twenty years that figure will be five percent. This is bad. It’s really bad.”
- Chapter 2, Project Hail Mary
p.s. love marissa. wish we could've seen her in the movie 💔
2) A year after this, the ArcLight probes are launched. These are noted to be outside the typical preferred launch window for a Venus-Earth alignment, hence the extra time.
Browne cleared his throat. “We received confirmation about ninety minutes ago that ArcLight successfully inserted into orbit around Venus. Now we’re just waiting for that first batch of data.”
It had been a heck of a year since the JAXA announcement about the Petrova problem.
- Chapter 2, Project Hail Mary
3) However! All stars affected by Astrophage will only lose up to 10% luminosity per the text.
"They only get to about ten percent dimmer before they stop dimming. We don't know why. It's not obvious to the naked eye, but"
"But if our sun dims by ten percent, we're all dead,"I said.
"Pretty much."
- Chapter 5, Project Hail Mary
4) The Sahara Paving - 2 trillion m² of paved blackpanels
“But it would be ridiculously slow,” I said. “If you had a one-square-meter box and ideal weather conditions…say, one thousand watts per square meter of solar energy…”
“It’s about half a microgram per day,” he said. “Give or take.”
“That’s a far cry from ‘a thousand kilograms’ per day.”
He smiled. “It’s just a matter of how many square meters you make of it.”
“You’d need two trillion square meters to get a thousand kilograms per day.”
“The Sahara Desert is nine trillion square meters.”
My jaw dropped open.
“That went by fast,” said Stratt. “Explain.”
“Well,” I said. “He wants to pave a chunk of the Sahara Desert with blackpanels. Like…a quarter of the entire Sahara Desert!”
“It’d be the biggest thing ever made by humanity,” he said. “It’d be starkly visible from space.”
I glared at him. “And it would destroy the ecology of Africa and probably Europe.”
“Not as much as the coming ice age will.”
- Chapter 13, Project Hail Mary
5) Bombing Antarctica extends Dr. LeClerc's halved population prediction from 19y to 27y. LeClerc is using a 1500 calorie model to determine his population numbers here. Some effects are discussed.
Stratt pinched her chin. “Nineteen years isn’t enough time. It’ll take thirteen years for the Hail Mary to get to Tau Ceti, and another thirteen for any results or data to come back. We need at least twenty-six years. Twenty-seven would be better.”
(...)
He looked out to sea. “And I’m ordering a nuclear strike on Antarctica. Two hundred and forty-one nuclear weapons, courtesy of the United States, buried fifty meters deep along a fissure at three-kilometer intervals. All going off at the same time.”
(...)
He checked his tablet. “The shelf will cleave at the line of explosions and slowly work its way into the sea and melt. Sea levels will rise about a centimeter over the next month, the ocean temperature will drop a degree—which is a disaster of its own but never mind that for now. Enormous quantities of methane will be released into the atmosphere. And now, methane is our friend. Methane is our best friend. And not just because it’ll keep us warm for a while.”
“Oh?”
“Methane breaks down in the atmosphere after ten years. We can knock chunks of Antarctica into the sea every few years to moderate the methane levels. And if Hail Mary finds a solution, we just have to wait ten years for the methane to go away. You can’t do that with carbon dioxide.”
- Chapter 14, Project Hail Mary
+1) Temperatures could drop 10 to 15 degrees <30y.
I stood and paced slowly in front of the class. "We don't know. But if it breeds like algae does, at about that same speed, climatologists are saying Earth's temperature could drop ten to fifteen degrees."
(...)
"Climatologists think it'|l happen within the next thirty years," I said.
Chapter 4, Project Hail Mary
The Challenge
So! This last piece of lore established by the text is the one we will be challenging with these models. Will the average temperature actually drop ten to fifteen degrees? Or will they go even lower? If this theory is proven incorrect, we'll also model out how we could've potentially ended up with 10-15 + how Earth will fare.
Science Time, baby!
Let's start with the central focus of the story: the sun's luminosity! Referencing (1), if we take (9,0.01) & (20,0.05) as points, x-axis as our time in years and y-axis as luminosity percent loss, we can figure out a rough exponential to determine when a luminosity percent loss will achieve that 10% limit.
y = 0.00267988e^(0.146313*x)
y = lumosity percent loss
x = time in years
So plugging in our 10% for the y, we can figure out it will take 24.75 years for the sun to reach maximum luminosity loss. We also have a general model for what the year over year loss will be, which will be helpful later.
(Edit: Hi! Just realized I forgot to account for (0,0.001) in my equation. This doesn't really fit the current exponential I'm using, so we're going to use a separate formula for 0-9 then 9-20. Formula for 0-9 is y = 0.001*e^(0.25585*x). Graph updated!)
Okay, next point (5): methane!! Everyone loves methane in this 'verse, and for good reason: it has a relatively powerful global warming potential of 81.2 (GWP) in comparison to its residence time. GWP as a measurement, btw, is comparing the warming ability of greenhouse gasses to the standard of carbon dioxide so in 20 years a tonne of methane warms the Earth as much as 81.2 tonnes of carbon dioxide. Residence time can vary a lot depending on what source you're looking at, because over the course of this time, the carbon in methane is oxidized (producing carbon dioxide and water vapor) and leaves the atmosphere through various other sinks. This can be best modelled through exponential decay, which can get very complicated very fast.
We're going to say that half of methane will be gone in say 6 years and all of it by 12 years, simplifying this vastly.
Now moving on: how much methane are we adding, exactly? Great question. We don't know :) Even with the specificity in the text, unfortunately we just don't really know how much methane is contained in glaciers and can only utilize what little work has been done on it so far. But in this particular instance, we can figure it out by utilizing some back of the napkin math and the text. So Stratt is asking for 27 years at a minimum (and confirming that we're in the launch year all at once. Thanks Stratt!), when LeClerc previously placed the halved population at 19 years. This means that in a standard model without methane, we will see the average temperature that will halve the population at 19 years post-launch. From there, we'll rewind time and add methane until that temperature doesn't occur until year 27. Now, I'll have to figure out if this will require more than one methane addition (probably), but my general theory here is that the goal will be the minimum emissions possible to achieve this goal.
Amazing! We will then have methane and luminosity... Wait. I'm forgetting something. Oh right. The Sahara!
This is a massive project. Redell rightfully notes that it'll be visible from space, and that affects albedo. What's albedo? Great question! Albedo is a measure of the percentage of sunlight that a surface reflects. You've experienced this if you've ever stuck your hand on a white car versus a black one in hot weather. The white car has a higher albedo percentage, so it absorbs less sunlight thus ends up less hot. In the Astrophage Polar Age, however, we really like having less albedo, not more. So how do we work this out to figure how much paving two trillion square meters would affect the Earth? Back to the napkin we go!
So first we get two trillion into a more usable form. Let's convert to kilometers squared instead, which works out to 2 million km². Excellent. Earth has a surface of 510 million km² of general surface area, divided between 361.13 million km² of ocean and 148.94 million km² of land. So this project would alter 1.34% of all terrestrial surface area and 0.39% of Earth's surface in total. Great job, Bob. Now, just for note, the equator is obviously where the majority of sunlight directly impacts the Earth, so it would likely have an outsized effect in comparison to more extreme latitudes. We can't really account for that without literally losing the plot, but be aware that I'm aware. So we can presume they utilize unoccupied regions of the Sahara for this. So we'll say that the original material present was sand, which has an albedo of around 0.5. Now we don't have an exact equivalent for blackpanels, but we'll use asphalt in a pinch for this exercise, here is a simple module exploring this topic. So that would be an albedo of 0.1 - incredibly low.
Okay so, let's get this show on the road. How is planetary albedo calculated? It is essentially an inventory of the reflectivity of the globe, which we figure out through use of special satellites and a lot of math. A common figure used for planetary albedo is 0.31. Now this one was fairly tricky to work out, but here's how I ended up calculating the potential change in albedo. I utilized this data set from NASA Earth Observations in 720x360, then I altered the "black/9999" cells until we ended up with an average of about 0.31 for the whole set. Yay! Now I need to change 0.39% of that, so 1012 of 259560 cells. My goal is to find cells similar to the sand and replace them with asphalt, so in the 0.5 region. This turned up 957 cells for me which I quickly replaced, and did the additional 55 manually (hahaha, suffering).
So what's the end result? Well. It makes a difference of about 0.002063869 (🥲). I will be utilizing that in the final model, because I'm incredibly petty, but it's negligible in this scenario. In case you were wondering.
Some factors I will be willfully ignoring to preserve my sanity. If you have suggestions for other modifiers we can add, please let me know.
