Would terraforming Mars be self-defeating? Many articles discuss how to terraform Mars, but they seem to assume that a human-friendly atmosphere would just "stay there". Mars is smaller than Earth, therefore its gravity is weaker, so would a new atmosphere just evaporate into space because the planet could not retain it gravitationally?
The short answer is: Mars can retain a human-friendly atmosphere.
The long answer is detailed below. I originally wrote this article in January 2006 while examining Earth. Since that website will disappear in coming months, I wanted to republish it here in my blog so it stays online somewhere. This version, obviously, will focus on Mars.
Retention of Atmosphere by a Planetary Body
Whether we consider small atoms or huge rocket ships, there is just one property that determines if something can escape the pull of the planet's gravity: velocity. If something travels with enough speed directly away from the planet, then it will be able to overcome the persistent force that would otherwise send it back to the surface. It doesn't matter if we measure baseballs or rockets or molecules of air. They all obey this same basic principle. If it moves fast enough, then it will escape.
Reaching escape velocity is helpful to us when we launch probes to other planets. We might worry, though, about unintended loss of material from a planet. After all, air itself can escape out into space! If a molecule of air is travelling fast enough, it will leave the atmosphere forever. Two questions become very important in understanding this effect:
1) how fast does the molecule have to move in order to reach escape velocity, and
2) what can cause it to travel that fast?
The effect of gravity is well studied, and determining escape velocity is an easy calculation. The only information that we need about a planet is its mass and its radius. We insert those two values into a formula and then we find the speed needed to escape the surface of the planet while traveling straight upward.
|escape velocity formula||velocity = √2 * G * mass / radius||m / s|
|gravitational constant||G = 6.67 x 10-14||Nm2/g2 = m3/(g * sec2)|
|mars mass||mass = 6.42 x 1026||g|
|mars radius||radius = 3.40 x 106||m|
|now just plug into the formula...|
|mars escape velocity =||√2 * G * mass / radius|| |
|√2 * (6.67 x 10-14) * (6.42 x 1026>) / (3.40 x 106)||√(m3/(g * sec2)) * (g) / (m)|
|√2 * (1.26 x 107)||√(m2/sec2)|
|5.02 x 103||m / s|
So anything on Mars that moves straight upward at 5.02 kilometers per second will manage to escape the planet and travel on to other destinations. That speed is enormous! That's why space launch vehicles are so huge; they require large amounts of fuel to accelerate their payload to escape velocity. Could anything around us reach such speeds unassisted by human intervention? Could air itself travel that fast?
It is easy to measure the speed of a large object. For example, every car comes equipped with a speedometer that informs the driver of the car's rate of movement. It seems much more difficult to measure the speed of a very small object, though. How do you watch the movement of an atom, an object so small that you can't even see it? It turns out that we do it all the time. Rather than measure the speed of a single atom, though, we measure the average speed of lots of atoms together. It's simple, really. We just take its temperature.
A speedometer measures the movement rate of a large vehicle, and a thermometer measures the relative movement rate of lots of atoms together. The higher the temperature, the faster those atoms are moving. Likewise, the colder that something gets, the slower its atoms are moving around. The relationship between speed and temperature is more obvious in a gas, but even in solids it is still true that the individual atoms are jittering quickly in their place. The Kelvin temperature scale is anchored at zero degrees at the low end of the scale because that is the point at which the atomic movement has slowed as much as it can. It's not possible for the atoms to move any slower, so they are as cold as anything can get. Zero degrees Kelvin is called "absolute zero". The relationship between speed and temperature is easy to demonstrate.
Fill a balloon with air and tie off the end of the balloon so that no air escapes. Notice the size of the baloon. It keeps that size because of the air pressure inside it. That pressure is generated by all of the small molecules of air zooming off in random directions until they hit the wall of the balloon surface. They push outward and then bounce back to travel in another direction. They ricochet inside the balloon, exerting a constant pressure trying to expand the balloon outward. Similarly the air outside is bouncing against the balloon trying to crush it inward. They eventually reach equilibrium, but you can change that balance by changing the temperature in the balloon. Hold the balloon under hot tap water (or hold over a pot of boiling water). The faster molecules hit the wall of the balloon with greater force, expanding the balloon. Hold the balloon under cold tap water (or place in the refrigerator). The slower molecules hit the wall with less force, allowing the balloon to shrink. Temperature affects air pressure (but that's just a bonus lesson to be learned) because it affects the speed of molecules.
average molecular speed
The law of temperature and pressure is also well studied, so to calculate the speed of an air molecule we can once again use a simple formula. All we need to know is the air molecule's mass and temperature.
1.66 x 10-27 * atomic mass =
|mass (in kilograms)|
|H, hydrogen atom||1.66 x 10-27 * 1.00794 =||1.67 x 10-27|
|H2, hydrogen gas||1.66 x 10-27 * 2.01588 =||3.35 x 10-27|
|He, helium atom||1.66 x 10-27 * 4.002602 =||6.65 x 10-27|
|C, carbon atom||1.66 x 10-27 * 12.0107 =||1.99 x 10-26|
|N, nitrogen atom||1.66 x 10-27 * 14.0067 =||2.33 x 10-26|
|O, oxygen atom||1.66 x 10-27 * 15.9994 =||2.66 x 10-26|
|H2O, water||1.66 x 10-27 * 18.01528 =||2.99 x 10-26|
|CO, carbon monoxide||1.66 x 10-27 * 28.0101 =||4.65 x 10-26|
|N2, nitrogen gas||1.66 x 10-27 * 28.0134 =||4.65 x 10-26|
|NH3, ammonia||1.66 x 10-27 * 31.03722 =||5.15 x 10-26|
|O2, oxygen gas||1.66 x 10-27 * 31.9988 =||5.31 x 10-26|
|CO2, carbon dioxide||1.66 x 10-27 * 44.0095 =||7.31 x 10-26|
|note: ultraviolet light can split water, carbon dioxide, and nitrogen molecules|
into their constituent atoms
The concept that we're examining is the "kinetic energy" (energy of motion) of molecules. Physics already has two simple formulas for measuring kinetic energy. If we compare them to each other, we can solve to find the velocity that we're needing.
(average for a gas)
|1/2 * (m * v2)||=||3/2 * (k * T)|
|m * v2||=||3 * k * T|
|v2||=||(3 * k * T) / m|
|v||=||√(3 * k * T) / m|
The value of "k" is the Boltzmann constant, 1.38 x 10-23
Joule / Kelvin. The average temperature "T" on Mars is 220 Kelvin. (For the rest of the world, that's -53 Celsius. And for backwater Americans, that's -63 Farenheit.) And those are the last two values that we need. Let's fill in the formula and see what we find out.
√(3 * k * T) / m =
|H, hydrogen atom||1.67 x 10-27||√(3 * (1.38 x 10-23) * (220)) /|
(1.67 x 10-27) =
|2.34 x 103|
|H2, hydrogen gas||3.35 x 10-27||√(3 * (1.38 x 10-23) * (220)) /|
(3.35 x 10-27) =
|1.65 x 103|
|He, helium atom||6.65 x 10-27||√(3 * (1.38 x 10-23) * (220)) /|
(6.65 x 10-27) =
|1.17 x 103|
|C, carbon atom||1.99 x 10-26||√(3 * (1.38 x 10-23) * (220)) /|
(1.99 x 10-26) =
|6.77 x 102|
|N, nitrogen atom||2.33 x 10-26||√(3 * (1.38 x 10-23) * (220)) /|
(2.33 x 10-26) =
|6.25 x 102|
|O, oxygen atom||2.66 x 10-26||√(3 * (1.38 x 10-23) * (220)) /|
(2.66 x 10-26) =
|5.85 x 102|
|H2O, water||2.99 x 10-26||√(3 * (1.38 x 10-23) * (220)) /|
(2.99 x 10-26) =
|5.52 x 102|
|CO, carbon monoxide||4.65 x 10-26||√(3 * (1.38 x 10-23) * (220)) /|
(4.65 x 10-26) =
|4.43 x 102|
|N2, nitrogen gas||4.65 x 10-26||√(3 * (1.38 x 10-23) * (220)) /|
(4.65 x 10-26) =
|4.43 x 102|
|NH3, ammonia||5.15 x 10-26||√(3 * (1.38 x 10-23) * (220)) /|
(5.15 x 10-26) =
|4.21 x 102|
|O2, oxygen gas||5.31 x 10-26||√(3 * (1.38 x 10-23) * (220)) /|
(5.31 x 10-26) =
|4.14 x 102|
|CO2, carbon dioxide||7.31 x 10-26||√(3 * (1.38 x 10-23) * (220)) /|
(7.31 x 10-26) =
|3.53 x 102|
According to these results, none of the atoms or molecules listed here could reach escape velocity on mars because they are all below the necessary 5.02 x 103
m/s. That's good news, right? Well, not really. The problem arises from the fact that we measure the temperature of a large sample of air all at once, not each individual molecule separately. The velocity that we compute is just the average velocity of all molecules in the group of air that we measured; it is not the velocity of any particular molecule. Some of them are moving faster than this calculated amount, and some are moving slower.
above-average molecular speed
Most molecules will have a velocity near the average in their group, but a few will have twice that velocity. Fewer will have four times that velocity. Even fewer will have six times that velocity. The general rule (for which I have not yet found a mathematical explanation) is that a molecule will escape into space if 6X
its average velocity is sufficient to reach the planet's escape velocity. Once that threshold is reached, the planet cannot retain that type of molecule because random exchanges of energy will give too many individual molecules the boost of kinetic energy that they need to speed away from the planet.
It's time for the big question...
so how does Mars fare under this new scheme?
|6X average speed|
|> escape speed?|
(5.02 x 103 m/s)
|H, hydrogen atom||2.34 x 103||1.40 x 104||279%|
|H2, hydrogen gas||1.65 x 103||9.89 x 103||197%|
|He, helium atom||1.17 x 103||7.02 x 103||140%|
|C, carbon atom||6.77 x 102||4.06 x 103||81%|
|N, nitrogen atom||6.25 x 102||3.75 x 103||75%|
|O, oxygen atom||5.85 x 102||3.51 x 103||70%|
|H2O, water||5.52 x 102||3.31 x 103||66%|
|CO, carbon monoxide||4.43 x 102||2.66 x 103||53%|
|N2, nitrogen gas||4.43 x 102||2.66 x 103||53%|
|NH3, ammonia||4.21 x 102||2.52 x 103||50%|
|O2, oxygen gas||4.14 x 102||2.49 x 103||50%|
|CO2, carbon dioxide||3.53 x 102||2.12 x 103||42%|
By these calculations, Mars retains all but hydrogen (whether in its atomic or molecular form) and helium. So if we found a way to mimic Earth air composition on Mars, the planet could gravitationally retain the atoms and molecules. But there's still a catch...
The mean temperature on Mars is 220 Kelvin. What if we wanted to improve the temperature as well as the chemical composition of the atmosphere? Would the air at 282 Kelvin (Earth's current average) still remain bound to the planet? Plugging the new temperature into the formulas, I find very similar results. The next heaviest common atom is Carbon, and it reaches only 92% of the necessary velocity to escape. Since it doesn't usually exist in a gaseous state, we have no worries about its approach to 100%. Nitrogen reaches 85% which might be troublesome, depending on just how long a time period we want to examine, or how accurate is the "6X" rule of thumb. If "6X" is just guesswork, then Nitrogen retention might be problematic on Mars.
Atomic and molecular hydrogen reach either double or triple the needed velocity to escape Mars. They will leave "quickly", so it's important that hydrogen is always bound together in heavier molecules or is replenished regularly. Of course, there is the ongoing problem of natural geologic processes that will affect the air composition, but we'll assume that the terraforming plan takes them into account.
Future exercise: What is the rate of hydrogen loss on Mars? How much hydrogen and helium does our sun deposit on Mars via the solar wind? Would it offset the natural gravitational losses?
Disclaimer: I am not responsible for errors caused by variation in temperature due to global warming, solar nova, or universal collapse. The rest of it is definitely my fault though.
"Universe", 6th ed, by Roger A Freedman and William J Kaufmann III, 2002, ISBN 0-7167-4647-6
, for formulas and conceptshttp://physics.nist.gov/PhysRefData/Compositions/index.html
, for standard atomic weights