-- By the Physicist blog Engineering Physics

Q: Could we get rid of CO2 if we pumped it through a pipe into space?

Physicist: Like several plumbing fix: it is dependent upon the pipe.

Evangelista Torricelli, the primary to measure the load of the air above us, once famously wrote that “Noi viviamo sommersi nel fondo d’un pelago d’aria.” (“We live submerged at the bottom of an ocean of air.”).  That’s a actually great way to consider the environment.  It’s held in place around the Earth in exactly the same method that the oceans are held where they’re; air has weight and gravity holds it down.  Gravity doesn’t instantly go away above the environment.  In low Earth orbit (where most satellites could be discovered) gravity is nearly as robust as it is on the floor.  So if you pump air or CO2 or any type of matter above the Kármán Line (the widely agreed upon, but arbitrary, boundary of area), it will still be topic to gravity and can fall.  You’ll have your self a CO2 fountain.

Pumping stuff to only above the environment isn’t an effective method to get rid of it, though it could be fairly if it’s visible from the ground.

Should you construct your pipe straight up from one of the poles, then it’ll be fountain-like even when it’s hundreds of thousands of miles tall.  Extra importantly, there are not any supplies that may help such a pipe (it’s essential, when constructing a CO2 pump to the celebs, to be real looking).  Nevertheless, a pipe constructed up from the equator would swing because the Earth turns, and that modifications things a lot.  Shorter pipes (a mere several thousand km tall) will nonetheless act like fountains, however ultimately that swinging motion becomes necessary.

Area elevators function on exactly this concept; combating gravity utilizing the Earth’s rotation.  So determining what happens to the CO2 flying out of the top of a very tall pipe boils right down to figuring out what would occur if you stepped out of a area elevator at the similar peak.  And since it is solely attainable (but troublesome) to build area elevators, this plan might work.  In truth, since that elevator would must be comprised of carbon nanotubes (no different materials is understood to be robust enough), we’d be utilizing carbon to get rid of carbon.  That’s virtually like being environment friendly!

Douglas Adams once sagely noticed that “There is an art to flying, or rather a knack. The knack lies in learning how to throw yourself at the ground and miss. … Clearly, it is this second part, the missing, that presents the difficulties.”  Ordinarily that is troublesome, but if the tower you throw your self off is a minimum of around 30,000 km from the middle of the Earth, then by the time you get to the ground it gained’t be there.  Sweeping out a larger circle every single day than a point on the bottom means the platform you bounce from might be shifting sideways quicker.  As soon as it’s fast sufficient, whenever you throw your self at the ground you’ll literally miss and find yourself in an orbit that skims just above the surface of the Earth at perigee (on the lowest) and returns up to the height of the platform at apogee (on the highest).  Missing an official identify, I do hereby declare this to be the “Adams Line”.

What happens to the CO2 whenever you pump it into area is determined by how excessive it is when it’s released.

Under the Adams Line, all of the CO2 pumped out of the pipe will fall back to Earth.  Above the Adams line the launched CO2 will find itself in orbit, and we’ll have given Earth a ring system.  This wouldn’t be a everlasting answer.  All of the fabric in a ring system needs to “stay in its own lane” and orbit in a near-perfect circle (therefore: “ring”) or things will run into each other, lose power, dropping into lower orbits and ultimately returning to Earth.  An entire collection of overlapping elliptical orbits swerving from the peak of the pipe to as low as Earth’s surface can be a very messy, very unstable ring system.

Gases launched above the Adams Line will orbit the Earth as an alternative of falling back to the bottom. Nevertheless, gases launched at totally different occasions can be on crossing (colliding) orbits, that might make Earth’s new rings very unstable.

However the larger the pipe, the more circular the orbit.  Once the highest of the pipe reaches geosync, it’s in orbit.  In the event you have been to step off of a platform you wouldn’t fall, you’d simply drift.  Fuel launched at this peak wouldn’t swerve between altitudes as it orbited Earth, it would kind of keep where it is and as an alternative would slowly refill the thin ring that types geosync.  The costliest and sought-after orbits around Earth would get dirtied up and be far much less useful for satellites, however it’s a small worth to pay.

We even have examples of “pipes” releasing fuel into their very own orbits, so we know kind of what it would appear to be.  Several of the moons in our solar system are geologically lively and, having much decrease gravity, eruptions from their surfaces don’t necessarily fall again to the bottom.

Enceladus’ ice geysers release water into (roughly) the same orbit round Saturn, forming one of Saturn’s many rings.

So between the Adams Line and geosynchronous orbit we’d have a momentary ring system, maybe giving us time to assume up and even crazier plan, and above geosync we’d have a pretty secure ring system (no ring methods are solely secure).  But if we need to get rid of Earth’s CO2 eternally, and not simply orbitally sequester it, we need it to flee from Earth’s gravity solely.

At geosync objects orbit Earth once per day.  Under that they orbit quicker and above that they orbit slower (for instance, the Moon, which is approach past geosync, takes a month to orbit).  So if the pipe is taller than about 42,000 km, the CO2 it releases will truly fly upward earlier than falling into orbit, and the taller the pipe larger that orbit can be.  Beyond about 53,000 km, the top of the pipe shall be shifting at escape velocity; the CO2 that comes out will fly away from Earth to seek out its approach into orbit across the Solar.

Fascinating fun-fact: as measured from the center of any planet, the peak of the “Fling Line” is all the time the dice root of 2 (≈1.26) occasions the height of geosynchronous orbits.  The Fling Line is a actually good aim for anybody planning to construct a area elevator, as a result of it’s a free ticket to in all places.

Swing exhausting sufficient and ultimately centripetal pressure wins.  Above the “Fling Line” (not an official identify) stuff that’s released by no means comes back down.

You may worry about some small amount returning to Earth, however you need to be equally frightened about someone in Antarctica having dangerous breath; there’s a lot of room for that drawback to diffuse.

So that there is the reply.  If you wish to clear up international warming, you’ll be able to just gather up and pump CO2 straight up about 53,000 km.  Straightforward peasy.

The weak spot of sensible plans like this isn’t feasibility (this might technically be achieved), it’s the naysayers who point out that the remedy could be worse than the illness.  In any case, carbon is a vital part of the Earth’s biological and chemical techniques.  We’ve added about half-again as a lot CO2 into the environment in the previous few centuries, taking us from under 300 ppm to about 400 ppm at this time.  There’s been 400 ppm of CO2 in the Earth’s environment before now, most just lately within the Micene 10 million years ago.  Approach back then life on Earth was doing alright (we obtained grasslands and kelp forests out of it, which isn’t terrible).  Having a bunch of CO2 around isn’t essentially dangerous; like highway driving, the danger doesn’t come from any specific state of affairs a lot as sudden modifications in state of affairs.  Provided that, all of the sudden and irreversibly chucking a third of Earth’s CO2 into area has a truthful probability of being a actually dangerous concept.


Reply Gravy: In the event you’re questioning the place the varied heights came from, you’ll be thrilled to know which you could work out the height of both the Adams Line and Fling Line (not the official names), with nothing fancier than algebra.

A Keplerian orbit (a regular elliptical or hyperbolic orbit) may be described as:

r(theta) = fracp1+ecos(theta)

where r is the space from the middle of the Earth (or whatever you happen to be orbiting), e is the eccentricity (e=0 for a circle, 0<e<1 for an ellipse), theta is the angle between the bottom point in the orbit and the present position, and p is (principally) a convenient place holder that determines the general measurement of the orbit.

If V_t and V_r are the initial tangential (sideways) and radial (up/down) velocities, h is the initial radius, and phi is the initial angular position, then you’ll be able to remedy for the orbit using:

beginarrayllr(theta) = frac(hV_t)^2Gm[1+ecos(theta)] [2mm]ecos(phi) = frachV_t^2Gm-1[2mm]esin(phi) = frachV_rV_tGmendarray

Normally you don’t know what the preliminary angular position is, you just understand how high you’re and how you’re shifting.

That is a slightly streamlined version of the answer that may be found here.  Leaping off of a platform that’s h above the middle of the Earth and hooked up to stated Earth signifies that V_t=frac2pi ht, where t is the length of one sidereal day (the time it takes Earth to completely rotate as soon as).  This is just the space (the circumference of a giant circle, 2πh) over time (in the future, t).  Because you’re stepping off of a platform, your initial radial velocity is zero, V_r=0.  Lastly, because the query is concerning the minimal peak you possibly can step off such that you would miss the Earth, you’re stepping off on the highest level in the orbit and then dropping, so phi=180^o.  Usually you need both the second and third equations, but figuring out you’re staring on the highest level means you already know what phi is.  One much less equation to fret about!

Use the second equation to unravel for e:

beginarrayrclecos(phi) &=& fracrV_t^2Gm - 1 [2mm]-ecos(180^o) &=& frach(frac2pi ht)^2Gm - 1 [2mm]-e &=& frach(frac2pi ht)^2Gm - 1 [2mm]e &=& 1 - frach(frac2pi ht)^2Gm [2mm]e &=& 1 - frac4pi^2 h^3Gmt^2 [2mm]e &=& fracGmt^2 - 4pi^2 h^3Gmt^2endarray

and then plugging that into the primary equation provides us the orbit:

r(theta) = frac(rV_t)^2Gm[1+ecos(theta)] = frac4pi^2 h^4Gmt^2[1+ecos(theta)]=frac4pi^2 h^4Gmt^2[1+fracGmt^2 - 4pi^2 h^3Gmt^2cos(theta)]=frac4pi^2 h^4Gmt^2+(Gmt^2 - 4pi^2 h^3)cos(theta)

In case you plug in theta = 180^o, you discover that the height of the platform is in reality the height of the platform:

r(180^o) = frac4pi^2 h^4Gmt^2+(Gmt^2 - 4pi^2 h^3)cos(180^o) = frac4pi^2 h^4Gmt^2-(Gmt^2 - 4pi^2 h^3) = frac4pi^2 h^44pi^2 h^3 = h

It’s good to double examine.  However to truly work out what h is (this was the whole level) you need the lowest point within the orbit to be the Earth’s radius, R=6371 km (any lower and also you hit the ground).

beginarrayrclR &=& frac4pi^2 h^4Gmt^2+(Gmt^2 - 4pi^2 h^3)cos(0^o) [2mm]R &=& frac4pi^2 h^42Gmt^2 - 4pi^2 h^3 [2mm]R(2Gmt^2 - 4pi^2 h^3) &=& 4pi^2 h^4 [2mm]0 &=& 4pi^2 h^4 + 4pi^2 Rh^3 - 2Gmt^2Rendarray

This is a fourth degree polynomial in h, and while it is technically attainable to unravel this by hand, you never need to clear up this by hand.  So plugging in the gravitational fixed, G=6.67times10^-11fracm^3s^2 kg, and all of the small print about Earth, R=6.37 times 10^6 m, m=5.97times 10^24kg, t=86,164s, after which giving this to a pc to unravel for you, you discover that one of the 4 options is sensible: h=29,825 km.  So if you must leap into an empty swimming pool, that’s a good peak to do it.

The peak of the Fling Line (not an official identify), f, is a bit simpler.  You don’t need to mess around with orbits, you simply need to set the velocity your platform is shifting at a given peak, frac2pi ft, equal to the escape velocity at that peak, sqrtfrac2Gmf:

beginarrayrclfrac2pi ft&=&sqrtfrac2Gmf[2mm]frac4pi^2f^2t^2&=&frac2Gmf[2mm]h^3&=&frac2Gmt^24pi^2[2mm]f&=&sqrt[3]frac2Gmt^24pi^2endarray

You possibly can calculate the height of geosync, s, by setting the centrifugal pressure equal to the gravitational pressure for an object orbiting exactly once per (sidereal) day, t:

beginarrayrcl fracleft(frac2pi stright)^2s&=&fracGms^2[2mm] frac4pi^2s^2ht^2&=&fracGms^2[2mm] s^3&=&fracGmt^24pi^2[2mm] s&=&sqrt[3]fracGmt^24pi^2 endarray

In order that’s a cute, in all probability ineffective reality; for any planet, the Fling Line is all the time at sqrt[3]2 occasions the radius of geosyncronous orbits.