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
= 4.55 x 10^12 kg/year
6.062 x 10^9 km/year
6.39 × 10^23 kg
Distance from Mars Orbit to Earth Orbit:
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):
Mcows*Vcows = MMars*VMars
VMars = Mcows*Vcows / MMars
Mcows = 4.55 x 10^12 kg
Vcows = 6.062 x 10^9 km/year
MMars = 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 Vcow_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 Vcow_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 = (Vcow_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 Vcow_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.