100 meters in diameter, 24% Magnesium...  

I bet you would collapse the entire world magnesium market by introducing that much all at once.  I would not recommend using lasers to cut this rock either, that 6% water would most likely not do well around that much magnesium at high temperatures.

How do you know the makeup so well, by the way?

 

How do you know the makeup so well, by the way?

 This is a fictional construct based on spectrographic measurements of NEA’s cataloged by LINEAR

 

  I bet you would collapse the entire world magnesium market by introducing that much all at once. 

 Most of the materials harvested would be used in space, only those that would bring a profit would be sent to Earths surface & sold.

 

I would not recommend using lasers to cut this rock either, that 6% water would most likely not do well around that much magnesium at high temperatures.

 A C2 Clay matrix type asteroid has a hard crust that holds the volitiles in check. The first assignment when reaching the asteroid is to gain entrance without breaching that containment. And that will most be likely by drilling, not blasting or using a laser.

O.K., I did some researching and found some very useful information at wikipedia.org at:

http://en.wikipedia.org/wiki/Orbital_mechanics
As well as at:

http://www.braeunig.us/space/orbmech.htm

Orbital mechanics is something that I have personally tried to master for over a decade, and I've come to the conclusion that you don't need a PhD in mathematics to understand the nitty gritty of it--but it helps! The accurate calculation of orbits (called Astrogation--which is a contraction of AstroNavigation) is a complex, multivariat problem that requires intimate understanding of calculus, numerical methods (as applied to computers,) and knowledge of physics (Keplar's Laws.) And this just scratches the surface!

Anyways, to answer the question I must make some assumptions:

1) That the initial condition of 165,000 miles at 15 kps is the point of closest approach. (Actually I chose the number 150,000 miles as the initial point--which is about 240,000 km for those who are familiar with the metric system!)

2) Everything is in MKS units (meters, kilograms, seconds)

3) All delta-V's are instantly applied (sorry, I can't do the long integrations for low thrust burns...sorry, I've tried!)

4)The mass of the asteroid is neglible compared to the mass of the Earth, so we can assume it to be zero for the purposes of orbit computations...)

First of all, the initial point of closest approach is assumed because this drastically simplifies the math--dealing with the additional complexities of off tangent interception, long integrated burn times, and orbital plane changes is enough to give me and my poor computer a migraine headache!


Let's begin by looking at two basic equations for calculating the orbital velocity of an object in an elliptical orbit:

The velocity at the highest point in the orbit (the point is called Apoapsis--think apogee!) is given by an equation found in the second source:

Va=sqrt(2*G*M*Rp/(Ra*(Ra+Rp))) (1)
where G=6.672*10^-11 N*m^2*kg^-2 (Newton's constant of gravitation;) M=5.9742*10^24 kg (Mass of the Earth;)
Rp=80*10^6 m (radius of closest approach of orbit is 80,000 km;) Ra=240*10^6 m (radius of highest point in orbit which is our start point 240,000 km)

Doing the substitutions I get a Va=2733.8 m/s. So in order to put this asteroid into an elliptical orbit, we must capture it. To do this we must shed: 15,000 m/s-2733.8 m/s=12,266 m/s! In other words, the asteroid is moving faster than the escape velocity of Earth, so we must bleed off about 81.8% of its initial veloctiy!

Once the EOI (Earth Orbit Insertion) is complete, the asteroid will follow that elliptical orbit to its lowest point which occurs at periapsis (at 80,000 km.) The velocity it will be moving at with respect to the Earth's center of gravity will be:

Vp=sqrt(2*G*M*Ra/(Rp*(Ra+Rp))) (2)

Carefully substituting the same values before into equation (2) gives a Vp=2743.6 m/s.

Now if we want to circularize this orbit at 80,000 km, then we must further reduce the asteroids speed. To do this, we must calculate the velocity of a circular orbit of this height.

Vcirc=sqrt(G*M/R) substituting Rp=80*10^6 m for R, and all other values the same, gives a Vcirc=2232.1 m/s. But the asteroid will be moving at 2743.6 m/s, so we must reduce its velocity by: 2743.6 m/s - 2232.1 m/s= 511.5 m/s.

What was the total velocity increment (change) for the asteroid:

12,266 m/s + 511.5 m/s = 12,778 m/s. Whew!

Asteroid wrangling isn't going to be easy!

To compute propellant expended I will leave to another post...

Nice orbital viewer--I'm looking at using JAT (JAVA Astrodynamic Toolkit) to do some orbital stuff, but I've only just begun to play with it! I would really like to go for Analytical Graphics' STK 8.0 Professional with the Astrogator add on module--this is the professional version of the free software--but it's what NASA and JPL uses to plan actual mission trajectories.

Going back to my previous post, let's now look at propellant expended!

We found that the total velocity increment in two impulses would be around 12,778 m/s.
Now bringing back the orginal mass of the asteroid as 750,000 tons (I will assume this to be metric tons, so asteroid mass is 7.5*10^8 kg.)

Let's start with Konstantine Tsiokolvsky's rocket equation:

dV=Ce*ln(Mi/Mf) (1) where dV=the velocity increment,
Ce=the exhaust velocity,
Mi is the initial mass of the rocket,
Mf is the final mass of the rocket.

Just to get a first order of magnitude solution, just to see how big of a rocket we'll need, we can do something which seems completely illogical: neglect the rocket!

Taking equation one, and rearranging algebraicly to the useful form:

Mi/Mf=e^(dV/Ce) (1a) where e is the base of natural logorythms and is about
2.7183

But we can identify another useful quantity called the Mass Ratio usually denoted by

MR = Mi/Mf (2) or similarly:
MR=e^(dV/Ce) (2a)

The Mass Ratio is really important because it relates the empty mass of a vehicle to its fully fueled mass. Now we can do some useful computations. First, I want to use a NERVA style engine with a nominal specific impulse of 1000 seconds. The specific impulse is an interesting way to measure a rocket's performance: what it means is that with 1000 seconds of specific impulse, a rocket engine will generate 1000 lb of thrust (force) for every pound (mass) of propellant consumed each second. To sustain a thrust of 100,000 pounds requires consuming 100 pounds per second of propellant! Anyways, it is related to the exhaust velocity in meters per second (sorry about mixing units, I usually work in MKS) by the relation:

Ce=Isp*g (3) where Ce=exhaust velocity m/s,
Isp=specific impulse in seconds,
g=9.80665 m/s^2, the acceleration due to gravity at
Earth's surface
For our example Isp=1000*sec, so Ce=9807 m/s.

So, what is the required mass ratio needed to perform the impulses desired?
Taking equation (2a) and substituting dV=12,778 m/s as the required velocity increment, Ce=9,807 m/s for the exhaust velocity, we get an MR=3.68

Ouch this is not good! In order to use a NERVA-style engine with 1000 seconds of specific impulse requires us to expend a total of 2.68 (asteroid+empty vehicle) masses as propellant. Even assuming a completely neglible mass for the vehicle, the initial 'launch mass' at the start of intercept must be of order 3.68 times the initial asteroid mass: 2,760,000 metric tons for the 750,000 metric ton rock.

The only way we can reduce the required velocity increment is by 'cheating,' and believe it or not there is a way to cheat enough that it may actually be doable. I'll explain in the next post...

In the last posts, I looked at what it might take to capture and manipulate a small 750,000 metric ton rock into an earth orbit given an inbound relative speed of 15 km/s. This led to a very large propellant consumption (2.68 times the combined asteroid and dry vehicle mass as propellant.) However, this situation is a bit over simplified. By stepping back and taking a look at the bigger picture we see that in order for this situation to happen, the asteroid must be in a very highly elliptical orbit about the sun because the relative closure speed is so high. Earth orbits about the sun with an average orbital speed of 29.2 km/s. An asteroid in an approximately similar speed, or in a slightly elliptical earth orbit crossing configuration will have just about the same speed relative to the sun, say anywhere from 28-30 km/s, but relative to the Earth the actual speed may infact be only about 0.5-2 km/s. The actual speed of course will depend on the actual asteroid, the precise orbital elements, the precise moment when interception or Earth Orbit Insertion Occurs, etc. And this is where it gets dicey, mathematically speaking...

Interestingly, this leads us to the conclusion that inbound impactors must have predominantly been moving rather slowly with respect to earth, with a few moving much faster (in highly elliptical orbits) and with a tiny fraction of impacts really smokin' (impacts of 60-65km/s) because of the retrograde solar orbits. Let's hope we don't get any 5km diameter retrograde visitors!

O.K., if we tailor our capture scenario so that we have a Near Earth Orbit crossing asteroid that, say, crosses the earth orbit with a heliocentric speed of say 30.5 km/s, if at that same point Earth's heliocentric speed is say 29.3 km/s, then the velocity we need to 'get rid of' is only about 30.5 km/s - 29.3 km/s = 1.2 km/s. The actual deficit requires intimate knowledge of the actual interception angle (the angle the asteroid's trajectory makes with resepct to the earth's orbit at the point of interception) but for our purposes we can approximate this as zero so this is a 'minimum' but close to reality. 1.2 km/s delta-v is still a hefty chunk of velocity change so can we 'cheat again' and reduce this further? Sure--by selecting a destination orbit where the orbital veloctiy is close to this 1.2 km/s deficit at closes approach. Think of it like 'merging' with traffic. So previosuly we saw that one desirable orbit with an apoapsis of 150,000 km had Va=2733.8 m/s. So if our asteroid is approaching Earth's vicinity at 1.2 km/s, we will need to increase it's speed by 2733.8 m/s-1200 m/s = 1533.8 m/s to capture it into prograde orbit. If we can arrange to use the Moon's gravity we might be able to cheat further and subtract or add the Moon's orbital speed as well, but in this business timing is everything, and the more gravity assissts you want to perform requires more and more carfeul planning and computations--beyond the scope of this post...

So an 'actual' velocity change of 1533.8 m/s sounds more realistic than 12,778 m/s originally explored. What mass ratio do you need for that one?

Referencing equation (2a) of the previous post:

MR=e^(dV/Ce) where dV is the velocity increment, and Ce=9807 m/s (1000 sec Isp)

MR=1.17

For our 750,000 metric ton asteroid this implies an initial mass of 877,000 tons---of which 177,000 tons is propellant. This is still a really hefty vehicle!

Could we increase the size of the initial capture orbit so that the deficit is lower? Sure...let's look at what it would take to capture the rock into an initial circular orbit about the earth...let's begin by asking is there an orbit about the earth where the speed is about 1533.8 m/s?

Referencing my previous posts we see that for a circular orbit:

Vcirc=sqrt(G*M/R) but since we know Vcirc and don't know R we'll have to solve for it:

Doing some algebra we find that R=G*M/Vcirc^2. Substituting known values as before G=6.672*10^-11 N*m^2/kg^2, M=5.9742*10^24 kg, Vcirc=1533 m/s we get an R=1.696*10^8 m or about 169610 km (or about 106,000 miles.) This would be the speed of the desired orbit that most closely matches the relative speed of closure of the asteroid. So this would be about where we would want to initially capture the rock--about halfway to the moon--and this seems to be about where some of the closest asteroid flyby's have occured in recent memory, interesting...! Of course to make this happen, this means changing the asteroid's actual flight path just a tiny bit so that it passes into this precise 'Entry Coridoor' so that the Earth Orbit Insertion Burn can be effective. This is where the complex mathematics, computers, Arecibo RADAR ranging pulses anf JPL Ephemerides comes in...!

So if we can cheat and pare down the total delta-v required to something manageable like 1533-1533.8 m/s = 0.8 m/s, (OK lets generously round up to 1 m/s!) then asteroid capture is a little more realistic. A capture delta-V of 1 m/s would require a MR=1.000102 which means that the intial mass must be 750076.5 tons, so that this implies about 76.5 tons of propellant must be expended. As an order of magnitude estimate this is much, much more realistic.

Once that the asteroid is safely stashed in a high earth orbit, it becomes possible to use it as a base, a mining post, or even a motel. To bring it down lower requires expending a lot more propellant, or tailoring the Entry Coridoor to a suitable orbit that more closely matches the relative inbound velocity. What you can capture is entirely dependent on what your tug is capable of and what orbit you can get it into. Everything else is just accounting...