In Capturing Near-Earth Asteroids around Earth, Hasnain, Lamb and Ross look at two delta Vs:
1) Delta V to nudge an asteroid's heliocentric orbit so the rock passes through the earth's sphere of influence.
2) Once in Earth's sphere influence, the delta V to make the hyperbolic orbit an elliptical capture orbit about the earth.
They don't try to find minimum delta V to reach an asteroid. Rather they try to determine whether low thrust ion engines can impart the needed delta V within plausible time frames.
I want to find asteroids that take the least delta V.
A good resource is JPL's NEO Close Approach page. To find orbits that already pass close to earth's sphere of influence, choose Nominal Distance less than or equal to 5 Lunar Distances. To find asteroids that don't need to shed too much velocity once in earth's sphere of influence, Sort by V-infinity:
Near the top of the resulting page is 2008 HU4's close encounter in 2015. Here is a picture of 2008 HU4's 2015 fly by:
To draw this picture I used position vectors generated by Horizon's Ephemeris page. I asked for position vectors in 3 day increments ranging from a month before to a month after the fly by:
Both the very helpful pages cited are JPL pages. The folks at JPL are worth their weight in gold, IMHO.
Nudging 2008 HU4's Heliocentric orbit
Let's take a closer look at the close encounter:
The asteroid is a little ahead of the earth and moon. If it's orbit could be slowed by about a day, it'd be neck and neck with us. Increasing the perihelion speed by .02 kilometers/second would boost aphelion by 500,000 kilometers. This would increase the asteroid's orbital period by about a day -- from 420 days to 421 days.
Now that we have the asteroid running neck and neck with the earth and moon, we need to rotate the asteroid's line of nodes. This is where the asteroid's orbital plane intersects earth's orbital plane. The asteroid is above or below earth's orbit everywhere but at the line of nodes.
I want the asteroid to graze the moon's orbit. At about 6 degrees from the line of nodes, the moon is furthest from the sun, a full moon. Doing a plane change burn 90 degrees ahead of this can rotate the line of nodes where we want the close encounter. 2008 HU4's inclination is about 1.3º. The plane change needed is (1 - cos(6º)) * 1.3º. A .008º plane change would be more than enough to rotate the line of nodes 6º. At the region where we want to do the plane change, the asteroid's moving a little less than 30 kilometers/second. Delta V for plane change is 2 * sin(.008º/2) * 30 kilometers/second = .005 kilometers/second.
The asteroid's a little further from the sun than the earth and moon. Decreasing the aphelion speed by .015 kilometers/second would shrink the perihelion by 300,000 kilometers. More than enough to get the asteroid to a moon grazing orbit.
So the delta V to nudge 2008 HU4's heliocentric orbit is (.02 + .005 + .015) kilometers/sec. About .04 kilometers/sec.
Parking The Rock After It Enters Earth's Sphere Of Influence
Now for the second phase of capture. Now that we have the asteroid passing through our sphere of influence, we want it to stay. We need to get the hyperbolic velocity down below earth escape velocity.
2008 HU4's Vinfinity with regard to earth is about 1.4 km/sec. But by the time it reaches the moon's sphere of influence, it will have picked up some speed from earth's gravity. Speed of a hyperbola is sqrt (Vescape2 + Vinfinity2). In the moon's neighborhood, earth escape velocity is about 1.4 km/s. So as the asteroid nears the moon's sphere of influence it's moving sqrt(1.42 + 1.42 ) kilometers/second. This is about 2 kilometers/second.
But the moon is moving about 1.02 km/s with regard to the earth. If the asteroid enters the moon's sphere of influence at 5 degrees from horizontal, it's speed will be about .97 km/s with regard to the moon.
When the asteroid enters the moon's sphere of influence, it's path can be modeled as a hyperbola about the moon with a Vinfinity of about .97 km/s. Let's aim the asteroid so the hyperbola's perilune is about 1800 kilometers from the moon's center, about 70 kilometers above the moon's surface.
When this turned velocity vector is added to the moon's velocity vector, we have a speed under earth escape, about 1.19 km/s.
Doing a .08 km/s burn a little after apogee would put the rock into a 450,000 km by 1,800,000 km. This ellipse's perigee matches the altitude and speed of EML2.
With the gravity assist from the moon, the second phase takes a total of about .21 km/sec.
Total Delta V
With .04 km/s to nudge the heliocentric orbit and .21 km/s to park the asteroid at EML2, total delta V is around .25 km/sec.
Can Ion Engines Do The Burns?
I've called for burns at the asteroid's perihelion, aphelion, and a plane change 84º from the node. Also high apogee burns after the rock's been captured to earth's orbit. Do these need high thrust impulsive burns or can ion engines do the job?
Heliocentric and high earth orbits have a much more leisurely angular velocity than low earth orbit.
LEO angular velocity is about 4 degrees per minute. In earth's neighborhood, low eccentricity heliocentric orbits are about 1 degree per day. LEO orbits have around 6000 times the angular velocity. A 5 minute burn in LEO scaled up to heliocentic proportions would take around 20 days.
Doing a velocity change of 20 meters/sec over 20 days is an acceleration of about .000012 meters/sec2 . To accelerate a 500 tonne rock this much, we'd need a thrust of about 5 newtons.
The Keck study for retrieving an asteroid calls for 5 10 kW Hall thrusters, of which 4 would operate at a given time. According to this Nasa page, a 10 kW Hall Thruster gives about half a newton. Four engines would give two newtons.
It's plausible higher thrust ion engines will be developed in the future. It's also possible the nudges to the heliocentric orbit could be done over several orbital periods.
How about the burns needed once in earth's sphere of influence?
At a 1,500,000 km apogee, the rock is moving about .6 degrees a day.
A burn over 20 degrees would take about 34 days. A .13 km/s burn would be an acceleration of .00001 meters a second. Accelerating a 500 tonne rock this much would take about 5 newtons.
If the last ellipse is allowed to return to apogee, it is likely the sun's influence would tear the asteroid loose from earth's capture orbit. 1,500,000 km is the outer edge of the earth's sphere of influence.
If I understand the Keck pdf correctly, they give .17 km/s as the delta V to return a 500 tonne rock, and they say the 5 10 kW Hall thrusters are up to the task. There are some very clever people contributing to that paper and it is likely they've come up with more effective lunar assists then mine. But until I have a better understanding, I'll say 2008 HU4 is .25 km/s from EML2 and retrieving a 500 tonne rock takes 5 newtons of thrust.
How many rocks can be caught?
2008 HU4 is near the very top of the list of low delta V asteroids. How many other rocks could be grabbed as easily?
In general smaller bodies are more common than larger bodies. If we successfully expanded our asteroid inventory to include most the asteroids 5 meters or greater, we would probably know of 100s or even 1000s of asteroids just as catchable as 2008 HU4.
It is the aim of Planetary Resources to take just such an inventory. They plan to launch a fleet of orbital Arkyd telescopes to accomplish this task.
It's likely rocks fall into temporary earth capture orbits without human intervention. New Scientist says any given time we could have several captured asteroids as tiny moons.
I applaud the efforts of Planetary Resources and Deep Space Industries. Their goals are doable. I'm also pleased NASA has expressed an interest in the Keck Study. Hopefully NASA will award PR and DSI contracts. In the near term I'm hoping NASA will buy time on the Arkyd Telescopes to get a more complete inventory of Chelyabinsk sized rocks.