Near Earth Asteroid Relocation

Page 3 - Seeking answers about space? Join the Space community: the premier source of space exploration, innovation, and astronomy news, chronicling (and celebrating) humanity's ongoing expansion across the final frontier.
Status
Not open for further replies.
B

Boris_Badenov

Guest
<br />I finally got this from another message board;<br /><br /> To relocate a 3 million ton asteroid into an Earth orbit depends entirely upon the initial trajectory and the final desired trajectory. The feasibility of this is dependent upon the actual change of velocity required and the propulsion system to be used. Now, generally, to even consider this several constraints must be made. First, choose a small rock. A 100m rock seems about right for a first attempt.<br /><br /><br />Second, choose an easily achievable destination trajectory. You mentioned the L1 Lagrange Point, which is the point at which the gravity of Earth is balanced by that of the Moon. This point exists about 61,500 km Earthward of the Moon. Another point, the L2 exists 61,500 km outward of the moon on a line between the moon and the Earth. It's interesting, but L1-L3 points are technically unstable. It is perhaps better to situate a mining operation at the L4 or L5 points which are located 60 degrees ahead and behind the moon respectively in its orbit about the Earth. Objects placed there tend to circulate about the center which is equidistant from the Earth and the Moon. This is a pretty good place for a mining operation, because no one wants a wayward rock crashing into anything valuable!<br /><br />A very good description of the gravitational "La Grange" libration points is availble at Wikipedia.org at: <br /><br />http://en.wikipedia.org/wiki/Lagrange_points#L1 <br /><br />Capturing an object into one of these points will still require a change of velocity of atleast 200 m/s, and probably more likely near 1km/s minimum, which is a bunch if you're moving a 3 Megaton rock around. Still, let's say we get lucky and its only going to take 200 m/s to park this thing.<br /><br />Third, choose a propulsion system. Let's say we got a really good deal on a Pratt and Whitney "Triton" Nuclear Thermal Rocket Engine with LOX afterburning. T <div class="Discussion_UserSignature"> <font color="#993300"><span class="body"><font size="2" color="#3366ff"><div align="center">. </div><div align="center">Never roll in the mud with a pig. You'll both get dirty & the pig likes it.</div></font></span></font> </div>
 
S

scottb50

Guest
To move it you have to get there, it would be a lot cheaper to get there and mine it in place, returning with a cargo of whatever you are mining than moving the mine to where it is more convenient.<br /><br />For the DV required to move it you could send multiple mining missions, if you move it the cost would be front loaded before you saw a dime in profitability. <div class="Discussion_UserSignature"> </div>
 
B

Boris_Badenov

Guest
Very true with this one. I think I bit off a little too much with this one. I am currently trying to find one in the 10-meter diameter, 300,000-ton range. I will then see how that one fares. <div class="Discussion_UserSignature"> <font color="#993300"><span class="body"><font size="2" color="#3366ff"><div align="center">. </div><div align="center">Never roll in the mud with a pig. You'll both get dirty & the pig likes it.</div></font></span></font> </div>
 
S

scottb50

Guest
I would think anything that small would be a rock. <div class="Discussion_UserSignature"> </div>
 
B

Boris_Badenov

Guest
That is a really big rock, about 1000 Shuttle payloads. <div class="Discussion_UserSignature"> <font color="#993300"><span class="body"><font size="2" color="#3366ff"><div align="center">. </div><div align="center">Never roll in the mud with a pig. You'll both get dirty & the pig likes it.</div></font></span></font> </div>
 
S

scottb50

Guest
I mean literally a rock. <div class="Discussion_UserSignature"> </div>
 
W

webtaz99

Guest
Even small ones can be 80% and better pure metal. <div class="Discussion_UserSignature"> </div>
 
B

Boris_Badenov

Guest
I found this terrific website that has tons of useful info on asteroid orbits. <br /><br />http://hamilton.dm.unipi.it/cgi-bin/astdys/astibo <div class="Discussion_UserSignature"> <font color="#993300"><span class="body"><font size="2" color="#3366ff"><div align="center">. </div><div align="center">Never roll in the mud with a pig. You'll both get dirty & the pig likes it.</div></font></span></font> </div>
 
5

5stone10

Guest
<font color="yellow">33% silica, 24% magnesium / sulfur, 16% iron in the form of chunks, pebbles & granules, 6% H2O 3% carbon, 2% aluminum, & 16% trace element & minerals</font><br /><br /><br />I'm trying to determine which of these elements have any substantial value. The answer - they don't. They are commodities found in abundance right here at 1g.
 
B

Boris_Badenov

Guest
<font color="yellow"> I'm trying to determine which of these elements have any substantial value. The answer - they don't. They are commodities found in abundance right here at 1g. </font><br /><br /> If launched from the surface they are automatically increased in value by the cost per pound of launch. If they are harvested from an asteroid & utilized in space, you save the launch cost. <div class="Discussion_UserSignature"> <font color="#993300"><span class="body"><font size="2" color="#3366ff"><div align="center">. </div><div align="center">Never roll in the mud with a pig. You'll both get dirty & the pig likes it.</div></font></span></font> </div>
 
B

barrykirk

Guest
Other than He3, there aren't too many things that are worth more than the cost of launching them.<br /><br />By that reason alone, Near Earth Asteroids that can be tapped for resources are worth more than their weight in gold.
 
B

Boris_Badenov

Guest
I absolutely agree<img src="/images/icons/cool.gif" /><img src="/images/icons/smile.gif" /> <div class="Discussion_UserSignature"> <font color="#993300"><span class="body"><font size="2" color="#3366ff"><div align="center">. </div><div align="center">Never roll in the mud with a pig. You'll both get dirty & the pig likes it.</div></font></span></font> </div>
 
M

mdodson

Guest
I've amused myself today with a way to show equations in a web page while addressing something similar to your question. http://www.newoceans.net/space_truckin/rocket_eq/ shows that using propellants sucked from the asteroid, you can only bring back 21% of it. So you end up merely filthy rich instead of fantastically rich.
 
M

mdodson

Guest
Well, I got that math wrong. Using water as your propellant, you can bring back ~20% of the asteroid.<br />What you need is an engine that uses the more abundant solids as reaction mass. Remember the late 70's sketches of giant arms flinging pellets to provide propulsion? I don't know how you do that without leaving a stream of projectiles that could damage any craft that wandered into the vicinity.
 
B

Boris_Badenov

Guest
What you are thinking of is a Mass Driver http://en.wikipedia.org/wiki/Mass_driver <br /> We discussed that earlier in the thread, but decided this asteroid is too small & we would lose too much mass. But, a mass driver would be a good way to move a big rock. In this instance we are after as much raw materiel as possible for construction;33% silica, 24% magnesium / sulfur, 16% iron. Fabrication of products; 3% carbon, 2% aluminum. & export to Earth for profit; 16% trace element & minerals. This baby weighs in at 3,000,000 <i> <b> <font color="blue"> TONS </font></b> </i>. There is much we could do with it once we get it where we can utilize it at reasonable cost! <div class="Discussion_UserSignature"> <font color="#993300"><span class="body"><font size="2" color="#3366ff"><div align="center">. </div><div align="center">Never roll in the mud with a pig. You'll both get dirty & the pig likes it.</div></font></span></font> </div>
 
M

mdodson

Guest
But it seems to me that a mass driver leaves a very long rope of small, high-speed pebbles that would pulverize any craft that crosses its orbit. It won't disperse like the cloud of gas from a chemical rocket. Time your return missions carefully!
 
B

Boris_Badenov

Guest
This asteroid is between 25 & 50 meters in diameter, 2006 UJ185. Depending on composition, it could weigh in at between 100,000 to 1,500,000 tons. I approaces to within 165,000 miles of Earth at a speed of around 15kps. Can anybody do the math on what kind of Dv it would take to put it into Earth orbit at between 150,000 & 50,000 miles. Lets put the weight at 750,000 tons for the exersise. <div class="Discussion_UserSignature"> <font color="#993300"><span class="body"><font size="2" color="#3366ff"><div align="center">. </div><div align="center">Never roll in the mud with a pig. You'll both get dirty & the pig likes it.</div></font></span></font> </div>
 
S

scottb50

Guest
I,m sure Spacester is hard at work on it. Just as a guess I would say a whole lot. <div class="Discussion_UserSignature"> </div>
 
M

mrmorris

Guest
<font color="yellow">" I approaces to within 165,000 miles of Earth at a speed of around 15kps."</font><br /><br />A scalar isn't good enough for this. To do a calculation like this, you'd need a vector -- speed <b>and</b> direction. In fact -- you really need to have the full orbital path of the asteroid so that it can be overlapped onto Earth's orbit to determine where and how to add/subtract velocity such that it approaches Earth with a vector that would take it deep enough in Earth's gravity well to be captured, but avoid going so deep as to impact. And no -- I have no intention of trying to work it myself -- too many flipping calculations. You pretty much need a program dedicated to the task.
 
M

mdodson

Guest
To have captured it, we need to slow it to something less that Earth's escape velocity, which is 11.2 km/sec. Let's call it a delta V of 5 km/sec. Plugging engines that perform as well as SSMEs into the rocket equation, it takes 1, 750, 000 tons of propellant to lasso that one. At shuttle prices, that's $10 trillion to tote up the propellant. That's enough to make me not want to spend any time refining my calculations!
 
M

mikeemmert

Guest
Agreed, that will not work with archaological artifacts.<br /><br />Let's try some plutonium, here. What, 100 kg? No, wait, rocket fuel is...uh...more powerful than TNT...<br /><br />Couple or three tons, maybe?<br /><br />What I like about plutonium is that it's very concentrated. So the gear needed to use the stuff is manageable. Sunlight is available, but it's just too spread out. Too much machinery needs to be ferried up.<br /><br />As far as plutonium's restricted availability, I've got an idea on that. Let's steal it from Saddam Hussein!
 
M

mikeemmert

Guest
Thanks for the simulation, Boris1961. I was hoping it was from GravitySimulator so that I could plug it into my machine and start working on it. I can tell at a glance, though, that there are more suitable asteroids. But I don't know which ones. I'm mostly working with the Kuiper belt, and none of those objects are suitable.<br /><br />I think asteroid recovery is a better project than trying to work with the Moon, we just got a report from radar that they couldn't find ice at the poles. We know where the platinum is, though. In the core.<br /><br />Jokes aside, I am very much in favor of such projects. Good luck.
 
B

Boris_Badenov

Guest
Here are the basic equations, I got them from another website I post on;<br /><br />Let's begin by looking at two basic equations for calculating the orbital velocity of an object in an elliptical orbit:<br /><br />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:<br /><br />Va=sqrt(2*G*M*Rp/(Ra*(Ra+Rp))) (1)<br />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;)<br />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)<br /><br />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!<br /><br />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:<br /><br />Vp=sqrt(2*G*M*Ra/(Rp*(Ra+Rp))) (2)<br /><br />Carefully substituting the same values before into equation (2) gives a Vp=2743.6 m/s.<br /><br />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. <br /><br />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. <br /><br />What was the total velocity increment (change) for the asteroid:<br /><br />12,266 m/s + 511.5 m/s = 12,778 m/s. Whew!<br /><br />Asteroid wrangling isn't going to be ea <div class="Discussion_UserSignature"> <font color="#993300"><span class="body"><font size="2" color="#3366ff"><div align="center">. </div><div align="center">Never roll in the mud with a pig. You'll both get dirty & the pig likes it.</div></font></span></font> </div>
 
Status
Not open for further replies.

ASK THE COMMUNITY

Latest posts