How many cows would you need to drop on Mars to successfully terraform it?

Question

The concept behind this lies in three primary points:

1. It has been theorized that one could increase the heat of Mars' atmosphere to a suitable level by guiding meteors into the atmosphere and crashing them into the surface. Assuming this is possible through pure mass, it should also be possible using cows.

2. Methane release from cows contributes a significant portion of the global warming. This would be extremely beneficial when trying to generate a stable atmosphere on Mars.

3. Fertilizer. Cows contain organic matter, which would assist future farmers when they begin their work on the freshly terraformed planet.

Would it be possible to achieve a useful balance between the mass needed to heat the planet, the methane contained in the dying cows' bodies, and still benefit from having scattered cows fertilizing the planet's surface?

Answer 1

1 Quintillion, or 1x 10¹⁸

This is based on a few assumptions of course, and there is a big caveat on it as well.

Before you start - you need to ensure that you put a magnetic field in place. Without that, all the work you put into bombarding the surface of Mars from orbit with cows is not going to help as the atmosphere and water will be stripped away by solar winds¹ and you run the risk of your cyanobacteria and other extremophile life (see later in this answer) being killed off by large CMEs and other large scale solar eruptions and events.

So; once you have the magnetic field in place, we'll continue.

So; the next step that you want is an ocean of some kind, and of course some organic chemicals either in the ocean or surrounding it. Fortunately cows have both. The average cow can vary in weight, but let's assume that they are around 400Kg - they contain about 60% water, so that translates to approximately 240L of water in the average cow.

Your ocean on Mars doesn't need to be anywhere near the size of that on Earth; for a start, the planet is smaller. Let's say you want an ocean a little smaller than the Indian Ocean, which is approx. 284 million KM³ in volume. To make it easy on us, we'll say we want an ocean of approx. 240 million Km³. A Km³ works out to a Trillion litres of water, or 10¹², meaning that a million of those is a quintillion, and given the number of litres in a cow the number of cows you need is around that figure.

But, getting all that water and those organic compounds to the ground via orbital bombardment isn't as easy as it sounds. For one, the water will sublimate from the body as soon as it is exposed to the vacuum of space and any of the bacteria or other organic hangers on will die in the process as well. Also, cows are not a complete eco-system in and of themselves, so assuming that you can get the cow to drop to the surface of Mars with its water intact, you still need to seed the general area with plants, bacteria, animals, and the other necessities for a functioning ecology. And, the temperature and atmospheric pressure have to be conducive to their survival before you seed them.

In short, if you have a functioning magnetic field already established on Mars, and you're NOT trying to actually terraform Mars via bovine orbital bombardment but merely provide a large supply of water and organic compounds, then it's possible that if you had 10¹⁸ cows on hand ready to drop onto Mars from a great height, AND you follow that up with some form of extremophile algae² capable of it's own Great Oxygenation Event, then you might have a start on the terraforming process. That said, it's still going to take thousands of years. Also, a qunitillion cows are hard to come by, even on Earth. There are currently only around 1.5 billion on Earth, so we're short by around 9 orders of magnitude. Just saying.

Point of note however, please if you are going to do this, euthanise your cows prior to dropping them on Mars. It's the humane approach, AND you won't have to feed and water them on the trip from Earth.

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^{1. Yes, it's true that the absence of a magnetic field isn't going to have an immediate impact, but over time the loss still be significant because you can't just dump 10^18 cows onto Mars in a single block of mass - that many cows will have about half the mass of Ceres. You have to introduce them over time, and even if you release a hundred cows a second, all over the planet, it's going to take somewhere in the order of a billion years to finish. In that time the lack of a magnetic field definitely has an effect, meaning you'd need even more cows to finish the job. Also, the cows that are NOT burnt to ash in that time still need working bacteria to break them down via putrefaction, and having the magnetic field in place reduces the risk of that life dying off before they can do their work. So, perhaps it's not necessary to have a magnetic field, but it is certainly recommended. I do concede though that the lack of gravity is going to have to be addressed for this plan to work as the gases released into a warming planet of that size will tend to drift away over time.}

^{2. From comments, it is noted in some scientific papers that there are some forms of bacterial life on Earth that could survive on Mars before the cows get there, and therefore it's possible that the seeding of putrefactive organisms could occur before the bovine bombardment begins - certainly it could occur during according to the article.}

Answer 2

Frame Challenge: Your first two points don't hold up.

1) Heat generated from crashing meteors into the surface is generated from the extreme speed at which the meteors travel through the air; Cows flying through the sky at that speed would not only burn up too quickly to be of any use, they would also be dead, and unable to contribute to points 2 and 3.

2) Methane is not a significant portion of global warming. The main greenhouse gas that traps heat is actually water vapor; it is the most abundant greenhouse gas in our atmosphere, both by weight, and by volume. This is why one reason why deserts are super cold at night, even though they're extremely hot in the daytime; there is no water vapor in the air to retain the heat during the night.

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Terraforming mars requires much more than just increasing the heat of the atmosphere with greenhouse gases.

To begin with, greenhouse gases make up less than 1% of the atmosphere; and of that 1%, water vapor is 95% of it. You'd need to add enough water to the planet such that the the evaporation and condensation cycles can maintain about 1% water vapor in the air.

You also need to pump in carbon dioxide, or you won't be able to grow any plants.

Of course, there are other factors, but if we only consider the ones I listed, you'd have much better luck crashing comets into the planet than asteroids or cows.

Answer 3

You are literally putting the cart before the cows! To reach a state where cows can be mass imported on a scale needed for this project, you would already have done 90% of the terraforming effort.

Answer 4

Rather than terraforming Mars through just the composition of dropped cows, I propose the ambitious project to knock Mars into Earth Orbit, thereby making it easier to start the terraforming process:

The Gravitational Steak Slingshot Project

In summary, this project will require the use of 20.4 quintillion cows over the next 4,084,481,927 years, and is fully sustainable, with the only caveat being the use of QPDs(quantum portal devices) and most of our planet food produce and space being dedicated to raising cows to jettison into space.

These are the basic premise:

• A cow weighs ~910kg (average of a bull and female cow).
• Earth has ~1.0 billion cows
• We modify Earth to henceforth only focus on cows, sustaining a peak of 10 billion cows
• Each cow gives birth to roughly 1 calf per year
• With this, we can produce roughly 5 billion cows every year, keeping roughly 5 billion to breed (this is possible because we only roughly need 1 bull per 50 cows for breeding
• We euthanize and jettison our 5 billion spare cows from our planet's space elevator after wrapping them in highly temperature resistant metals
• The cows are precisely shot in the direction of our sun, and we use it as a gravitational slingshot, similar to what is proposed with the Parker Solar Probe
• The cows will reach a peak speed of 692,000 km/h and sling around the sun, becoming wellx10⁹⁰⁰¹ done steaks inside the foil
• through a carefully calculated trajectory, it will shoot through a quantum-portal set up near the Sun, with the other end pointing to the right of the trailing side of Mars, right after reaching peak velocity in the gravitational slingshot
• the wellx10⁹⁰⁰¹ done steak will impact the surface of Mars from the side, propelling Mars towards Earth orbit and vaporizing into its base components
• The results from roughly 360 trillion cows will change the orbit of Mars to coincide with that of Earth's in 72,040 years

Calculations:

Cows:
5,000,000,000 cows/year
910 kg/cow
= 4.55 x 10^12 kg/year

Speed:
692,000 km/h
6.062 x 10^9 km/year

Mars Weight:
6.39 × 10^23 kg

Distance from Mars Orbit to Earth Orbit:
54,600,000 km

Simplifying impact calculation to find resulting velocity, assuming the cow collision is perfectly elastic, with no loss of energy involved, for 5,000,000,000 cows a year (and luckily with no air friction, assuming our portal is placed flush against the surface of Mars):

M_cows*V_cows = M_Mars*V_Mars

V_Mars = M_cows*V_cows / M_Mars

M_cows = 4.55 x 10^12 kg

V_cows = 6.062 x 10^9 km/year

M_Mars = 6.39 × 10^23 kg

V_mars = 4.55 x 10^12 kg * 6.062 x 10^9 km/year / 6.39 × 10^23 kg
V_mars = 0.04316 km/year

We know that these numbers are for our yearly cow rate, so we know that V_{cow_speed/year} is equal to this V_mars / 1 year

Assuming we shoot out 5,000,000,000 cows a year, and add this speed to Mars each year, the distance traveled by Mars can be plotted out by a linear line, where the slope is V_{cow_speed/year}.

The area of this function is the distance traveled, which we want to equal half of the distance from Mars to Earth, 28 million km.

To find the years needed to achieve this distance (with years as y), the formula for this function is 28 million km = (V_{cow_speed/year})*y²/2.

28,000,000km =  (0.04316 km/year^2)*y^2 / 2
y^2 = 28,000,000km * 2 / 0.04316 km/year
y= sqrt(1297497683 year^2)
y= 36020 year

This is only for half of the distance traveled, once this is done, we must employ more cows from the opposing end, for the next 36,020 years to bring Mars to a stop.

Thus, to send Mars into a similar orbit around Earth, we will need 72,040 years and (72,040 * 5,000,000,000) ~= 360 trillion cows.

Edit: It seems that we need to revise our answer. To change the orbit of a planet does not depend on its distance from the sun, but its orbital speed. As referenced from here, we will need to change the orbital velocity to perform a Hohmann transfer of Mars:

The most efficient way to move from one orbit to another is via a Hohmann transfer. We'll apply a delta-V to Mars to slow it down and put the planet into an elliptical transfer orbit that just intersects Earth's orbit, then another delta-V once Mars reaches perihelion. Assuming Mars is orbiting circularly at 1.524 AU, a retrograde delta-V of 2.65 km/s will put Mars on that transfer ellipse. Half an orbit later, another retrograde delta-V, this time 2.94 km/s, will put Mars in a 1 AU circular orbit. No problem! All we have to do is change Mar's velocity by 2.65 km/s and then later by 2.94 km/s, or a total delta-V of 5.59 km/s, and voila! we have Mars orbiting at 1 AU.

To shift the orbital speed of Mars by 5.59km/s (equivalent to 176,286,240 km/year) we will need to divide this by V_{cow_speed/year} instead, to get the number of years needed. This comes out to 4,084,481,927 years and 20.4 quintillion steaks, although the gravitational pull of Mars and Earth should greatly reduce this number. This comes off a lot more than the previous number due to us needing to change the orbital velocity of the entire planet, rather than simply shifting its trajectory over time.

_{Please don't take the calculations too seriously, it's obvious we don't have any QPDs, space elevators, etc. With 360 trillion collisions equaling to roughly 2518880000000 nuclear bombs, we would be lucky if any of Mars remained by the time it arrived. We also don't consider the inelastic nature of a 910 kg steak hitting the surface of Mars, nor the needed distance from the sun for Mars to reach similar temperatures as Earth, considering a difference in atmospheric gases, surface area, etc. We also don't consider the potential consequences of Mars being in a similar orbital distance from the Sun as Earth.}

Answer 5

Assuming cows would survive (which is a false assumption), there are a number of possible answers:

2 would be enough potentially (one male, one female), if they had enough food, water and air, because they would soon become 4, 8, 16, 32, etc. until their population expanded to fill the whole planet.

There is a risk in that one might die before they bred offspring of the right sex, so at least 2 of each sex would be a sensible precaution.

With such a small breeding stock, as well as risk of death of individual animals, there is the risk of inbreeding causing unwanted genetic defects, leading to sickness and infertility. To counteract this, you would ideally want to start from a genetically diverse breeding stock of at least 64 individuals (approximately). This sort of population size is the absolute minimum to capture the wide diversity of all cows, if you ensure that no two cows are closely related. You would probably also want to at least double this number to accommodate the risk of individuals dying without reproducing.