Lunar Ice Vs NEO ice

Lunar Poles Aren't So Hard to Reach

"There might be ice in the lunar cold traps," moon haters like to say, "... but if there is, it's hard to reach. It takes way, WAY more delta V to reach the poles than the lower lunar latitudes."

This meme is wrong. Unfortunately it's widespread. The first part of this post will debunk this false notion.

I will describe a route from low earth orbit to the lunar north pole that takes 6.4 km/s and 6 days. It's by no means the only route or even the best route. But I'm using it because it's simple and easy to illustrate.

Start in an equatorial low earth orbit at 300 kilometer altitude. Do a TLI (Trans Lunar Injection) burn to reach an apogee of 1 lunar distance (384,400 kilometers from earth's center). Time the apogee so it is on the moon's line of nodes shortly before the moon crosses the equatorial plane. The ship will enter the moon's sphere of influence south of the moon. I am calling the radius of the moon's sphere of influence 60,000 km.

It pains me to make a cartoon illustration that is nowhere near correct scale. To atone for my sins the above is a scale drawing of what I'm trying to describe.

TLI

The first burn from LEO is TLI or Trans Lunar Injection.

Since the transfer orbit lies on the equatorial plane, the perigee velocity vector is parallel to the low earth orbit velocity vector. Therefore simply subtracting the LEO velocity from the transfer orbit's perigee velocity vector gives the TLI (Trans Lunar Injection) delta V.

But what is the transfer orbit's perigee velocity? And what is LEO velocity at 300 km altitude?

Both these can be found with the vis-viva equation: v = sqrt(GM(2/r - 1/a)

G is the gravitational constant which is about  6.67384e-20 km³ kg^-1 s^-2
M is the mass of the earth, about 5.972e24 kg.
r is distance from earth's center at TLI. At 300 km altitude this is about 6678 km.
a is the semi-major axis of the orbit. For the low earth orbit a is 6678 km, the same as r. For the transfer orbit, a is (6678 + 384,400)/2 or 195539 km.

Plugging these in we get 10.8 km/s for the transfer orbit's perigee velocity and 7.7 km/s for LEO.
10.8-7.7 = 3.1

The TLI burn is 3.1 km/s.

Entering the Moon's Sphere of Influence

When the ship arrives at apogee, r = 384,400 km. For the moon's orbit a = 384,400. Plugging these quantities into the vis-viva equation we get the moon's velocity as 1.018 km/s. The ship's velocity at perigee is .19 km/s.

But these two vectors aren't parallel. The moon's orbit is inclined to the equatorial plane anywhere from 18 degrees to 29 degrees. We'll pick the worst case scenario: 29 degrees.

Given two sides a and b of a triangle, separated by angle alpha, the third side can be found by the law of cosines:

a² + b² - 2ab cos (alpha) = c². For alpha = 90 degrees, cos(alpha) is zero. So the law of cosines is a more general version of the Pythagorean theorem.

For the triangle above, the third side comes to .86 km/s. The ship enters the moon's sphere of influence traveling .86 km/s with regard to the moon.

Inside the Moon's Sphere of Influence, LOI

Our ship enters the moon's sphere of influence at lunar longitude 90º. The lunar 90º longitude line lies in a horizontal plane with regard to the earth. At apogee the ship's vector is also horizontal with regard to the earth so the ship's velocity vector lies in same plane. Inside the moon's sphere of influence, the ship's path can be modeled as a hyperbola with the focus at the moon's center. The moon's center also lies in the 90º longitude plane. Since the hyperbolic orbit is coplanar with a plane that cuts through the moon's poles, this hyperbolic orbit is polar.

Using the Law of Sines we can find the angle between the velocity vector with regard to the moon and the local vertical with regard to the moon. It is 5.5 degrees.

Impact parameter of our hyperbolic orbit is sin(5.5º) * 60,000 km. Or about 6090 kilometers.

Given a .86 km/s Vinfinity vector and a 6090 km impact parameter, the hyperbola's perilune is 633 kilometers above the moon's surface or 2371 kilometers from the moon's center. A hyperbola's velocity is sqrt(V_escape² + V_infinity²). Our V_infinity is .86 km/s. Escape velocity is sqrt(2GM/r). Since we're in the moon's sphere of influence, we will use the moon's mass for M, 7.35e22 kg. At 2371 kilometers from the moon's center, V_escape is about 2 km/s. Sqrt(.86² + 2²) is about 2.2.

Our hyperbola's perilune velocity is 2.2 km/s.

We want to get from a 633 km altitude hyperbola perilune to an elliptical orbit having 633 altitude km apolune and 20 km perilune. This ellipse would have semi axis a = 2064 km. Then we want to move to a low lunar orbit at 20 km altitude. Using the vis-viva equation we can get these numbers:

.9 km/s for LOI (Lunar Orbit Insertion) from a hyperbolic orbit

and

.2 km/s for injection to a low lunar orbit.

We wait until our low lunar orbit takes us to the north or south pole. We need to kill the orbit's 1.6 km/s. As the orbit slows, the ship will lose altitude gaining vertical velocity from gravity. To make a soft landing we need to counteract gravity. I estimate gravity loss will cost .5 km/s.

.5 + 1.6 km/s = 2.1 km/s

2.1 km/s for descent and landing.

The total delta V from LEO to landing is the sum of 4 burns: TLI, LOI, insertion to LLO, and landing.

3.1 + .9 + .2 + 2.1 = 6.4.

6.4 km/s is the total delta V from LEO to landing at the moon's north pole.

Period of an elliptical orbit is 2 pi sqrt(a³ / (GM)). The transfer orbit from LEO to a 384,400 km apogee is about 10 days. We travel half that orbit so
5 days from LEO to 384,000 apogee.

We can find the angular momentum of the hyperbolic orbit about the moon by doing a cross product of it's position vector and velocity vector at perilune. Magnitude of the hyperbola's angular momentum vector is 2371 km * 2.2 km/s which comes to about 5200 square kilometers per second. The magnitude of an orbit's angular momentum is twice the area the orbit sweeps out in a given time.

To measure twice the area the hyperbola swept out from entering SOI to perilune, I did a scale drawing and measured in Photoshop:

310 million square kilometers divided by 5200 square kilometers per second comes to about 60,000 seconds or about .7 days.
.7 days from SOI to hyperbola perilune.

We can use 2 pi sqrt(a³ / (GM)) to get the period of the orbit from a 633 km altitude to a 20 km altitude. This orbit has period of about .1 days. We traverse half this orbit. .5 * .1 = .05
.05 days from 633 km altitude to 20 km altitude.

We can also use 2 pi sqrt(a³ / (GM)) to get the period of a low lunar orbit at a 20 kilometer altitude. The LLO period is .08 days (about 1.8 hours).
Less than .08 days to get to either the north or south pole.

5 + .7 + .05 + .08 = 5.83. I will round up to 6.

6 days is the total time from LEO to landing at the north pole.

Asteroid Ice Isn't So Easy to Reach

Main Belt Asteroids
I will look at water ice in the Main Belt asteroids, short period comets, and what I call accessible NEOs. By accessible NEOs I mean those objects retrievable by a vehicle as described in the Keck Report

There is evidence that some of the Main Belt asteroids have water ice. Three possibilities are Ceres, 24 Themis, and 65 Cybele.

Here is a table showing delta v, launch window frequency (aka synodic period), trip time, and surface gravity:

Asteroid Name	LEO to Transfer (km/s)	Transfer to Rendezvous (km/s)	Synodic Period (Years)	Trip Time (Years)	Surface Gravity (m/s^2)
Ceres	4.9	4.9	1.28	1.29	.28
24 Themis	5.2	5.1	1.22	1.48	.075
65 Cybele	5.4	5.3	1.19	1.65	.07

The above numbers come from a model that assumes circular, coplanar orbits. 24 Themis has less than 1 degree inclination, so coplanar orbits are a good approximation. But 65 Cybele is inclined 3.5º to the ecliptic and Ceres 10.6º. So the above table underestimates the delta V for these two asteroids.

All three take considerably more delta V to reach than the lunar poles.

Lunar launch windows occur every two weeks from a given LEO orbit. Trip time is less than a week. By these metrics, the moon has a huge advantage over the main belt asteroids.

A common myth is that, due to their shallow gravity wells, return to earth from an asteroid takes virtually no delta V. Unless the asteroid has a near earth perihelion or aphelion, the heliocentric transfer orbit have a different velocity at asteroid rendezvous. For example, 24 Themis moves about the sun at about 16.8 km/s. The transfer orbit is moving about 11.7 km/s at rendezvous. Regardless if the transfer orbit is 24 Themis to Earth, or Earth to 24 Themis, the needed delta V is 5.1 km/s.

Some asteroids may be amenable to rendezvous via low thrust but high ISP ion rockets. But not these. The vehicle described in the Keck Report has a thrust of 2 newtons and a dry mass of 5.5 tonnes. That comes to about .0004 newtons per kilogram with no propellant. So to have a thrust/weight ratio greater than 1, surface gravity would need to be less than .0004 meters/second^2. It's hard to imagine an ion rocket getting off the ground on any of these three asteroids.

Short Period Comets
Next I will look at short period comets. A body that has recently outgassed is likely to have volatile ices.

Among the short period comets, Comet Encke has the shortest known semi-major axis, 2.22 A.U. It has a high eccentricity, e = ~.85. This means it has a perihelion quite close to the sun at .34 AU inside the orbit of Mercury. It has an aphelion at about 4.1 AU, out past the main belt.

It gets quite hot at perihelion, a black body temp of 460K (assuming an albedo of .15). And fairly cold at aphelion, 132K. But it spends more time in the aphelion neighborhood, so it's average black body temp is 167 K. If it had a circular orbit of radius 2.22 AU, the average black body temp would be 180K. So more eccentric orbits have the effect of lowering average black body temp throughout the orbit.

While the big eccentricity makes the comet a little cooler, it also makes for healthy delta V. Here are two possible routes from earth to Encke:

 Where rendezvous is at Encke's aphelion....

 ....or perihelion, there is big delta V. This isn't considering Encke's healthy inclination. Most comets have a good inclination that will boost delta V.

But if an asteroid/comet has a 1 A.U. perihelion, that changes the picture quite a lot. For the sake of argument I will imagine there is a comet JohnD with 0º inclination, 1.9 AU semi-major axis and a 1 AU perihelion. The fictitious comet JohnD is named after John DeLaughter, a planetologist I've been arguing with. Much of this blog post comes from that argument.

You may have noticed I've been drawing the transfer orbits sort of purplish. Like Hohmann orbits, the transfer orbits have been tangent to the departure and destination orbits. Thus delta V is only needed for speed change, not direction change. Well, Comet JohnD's orbit is already tangent to earth's orbit. Thus when the comet comes around our neighborhood, the only delta V needed is to leave earth orbit for TJI (Trans JohnD Injection). Rendezvous with the asteroid is virtually no delta delta V. In this case Vinfinity would be about 6.4 km/s. Therefore TJI from LEO is around 5 km/s. It would take less delta V to reach Comet JohnD than the moon.

Likewise for the return trip, wait until asteroid Johnd passes near earth at perihelion. A slight nudge can send a cargo to an earth atmosphere grazing orbit. Let aerobraking take care of most the delta V. Pretty sweet, huh?

Not so fast. The period of Comet JohnD is 2.619 years. When Comet JohnD returns to perihelion, earth will have advanced (360 + 360 + 222.8) degrees. The earth is nowhere near the comet, it's location at the 2nd perihelion is denoted by the number 2 in right part of the graphic above. Each circuit of the comet sees earth advance 222.8 degrees. Earth and the comet won't be in the same neighborhood for 22 periods! 22 * 2.619 = 57.  57.6 years.

If Chris Lewicki of Planetary Resources sets up a water mine on Comet JohnD, he'd have to wait 58 years to send the first shipment back to earth. Or else pay a delta V penalty.

Also microgravity mining in a vacuum is something the human race has zero experience doing. Acquiring the needed experience will be a trial and error process. Thus multiple trips would be needed to establish infra-structure. In the case of Comet JohnD, 3 trips to the comet to establish infra-structure would take 173 years.

The rarity of launch windows more than nullifies the slight delta V advantage this comet has over the moon.

Moreover, it is less likely Comet JohnD would have water ice in it's interior. Recall the average black body temperature of Encke was 167 K. And this is the shortest period comet known. Assuming an albedo of .15, Comet John D would have an average temp of 190 K:

This is 23 K warmer than Encke. A 23 K difference is the same as a 41.4º F difference. The difference between freezing and comfortable room temperature. Water ice in a vacuum starts sublimating at healthy rate at around 150 K. Being surrounded by clay and dust might mitigate sublimation loss for a time. But less so when the average temperature is boosted by 41º F. Encke is a rare comet having the shortest known semi-major axis at 2.22 AU. In my opinion a dead comet with a 1.9 AU axis is less likely to keep volatile ices at its core.

Accessible Asteroids: small asteroids with an earth-like orbit.
While there are many near earth objects quite close in terms of delta V, launch windows to these close objects are rare. Which makes establishing infra-structure more difficult. Also very rare are the opportunities to deliver the asteroid's resources to earth's neighborhood. Due to these considerations, NEOs fell off my radar screen.

Then in April 2012 the Keck Report was published.

Keck Report Authors include Chris Lewicki -- chief engineer of Planetary Resources, John S. Lewis - author of Mining the Sky and Rain of Iron and Ice, Don Yeomans - Manager of NASA's Near-Earth Object Program Office, Rusty Schweickart - chair emeritus of the B612 Foundation. There are many respected engineers and scientists among the authors.

The authors did the numbers demonstrating it's possible to park a small asteroid in high lunar orbit. I've examined the numbers and they are only mildly optimistic. In my opinion the vehicle described is doable.

Parking the rock in lunar orbit completely changes the picture. Now the rock has launch windows each two weeks. Trip time is less than a week.

Moreover, light lag latency from earth's surface is only 3 seconds. Since signal strength scales with inverse square of distance, the rock's proximity makes for good bandwidth. The rock is amenable to being worked by telerobots.

The Keck Report reversed my opinion. I now believe mining a retrieved asteroid is doable.

Retrievable asteroids would be rocks similar to 2008 HU4 -- small and having a semi major axis close to 1 AU. Also small eccentricity and inclination.

Here's a look at 2008 HU4's average temperature (assuming .15 albedo):

This is 117º F hotter than Comet JohnD. What's the life span of a small ice ball at this temperature?

John DeLaughter cites The Stability of Volatiles in the Solar System which says, in part, "a 1 km sphere at 1 a.u. is stable for 3,000 years,"

It's possible that near passages with the earth or other planets could lower a comet's aphelion. I would guess there are some dead comets with orbital elements similar to 2008 HU4. But how many of these were perturbed into their earth like orbits within the past 3,000 years?

And retrievable rocks are much smaller than 1 kilometer. More like 5 to 7 meters.

Equation (6) from Stability of Volatiles:
t_max = r₀/(dr/dt) = r₀ρ/É.

The number of interest here is r₀, initial radius. The ice ball's life scales with initial radius. 7 meters/1000 meters = .007. .007 *3000 = 21 years. A 7 meter ice ball at 1 a.u. would last 21 years.

John DeLaughter believes a blanket of loose soil could reduce water loss by a factor of 10 to 20. He argues the soil's permeability and tortuosity would slow sublimation. But he neglects to mention that it would also lower albedo. The paper he cites assumes a .6 albedo for ice balls. A comet's exterior mantle more typically has albedo of .1. Dark objects absorb more light and get hotter than pale objects. But for the sake of argument, I will give him his factor of 20. That's 21 * 20 years. How many comets have been perturbed into an earth like orbit within the past 420 years?

5 to 7 meter diameter rocks with an ~1 AU semi-major axis are unlikely to have water ice. This is not to say such rocks have no water! I believe there are many accessible rocks with water in the form of hydrated clays. But such rocks are water rich in the same way concrete is water rich. Ice deposits are more easily exploited than hydrated clays. If the lunar cold traps do indeed have large, thick ice deposits, I believe the moon would be a better source of extra-terrestrial propellant.

Study of Escher's Print "Gravity"

In his print Gravity, Escher placed 12 monsters on the star faces of the small stellated dodecahedron. He had thought of using turtles. I wish he had. With all their polygons, turtles are walking polyhedra.

So I did this polyhedra study using polyhedral turtles:

I've tried to follow Escher's patterns. Each turtle is looking at his neighbor's right leg, which forms rings of three. For example the yellow turtle is looking at the green turtle which is looking at the purple which is looking at the yellow. There are 4 sets of triple turtles and these four sets suggest a tetrahedron. There are other polyhedra suggested by other turtle groupings.

Here is a picture of a turtle along with the net used to make his shell:

To make this net I used rotation matrices, vector norms, vector cross products as well as dot products and an equation for finding the intersection of two coplanar lines. It was hard but fun.

Even MIT students can make misteaks.

Finding Delta V between various places in the solar system has been one of my hobbies. I like to learn as well as share what I know about this topic. So I occasionally Google search strings that include the term "Delta V".

Googling: Delta V Mars. The first hit is a Wikipedia article, the second hit is How Much Delta V do you need to get to Mars - Yahoo! Answers.

A fellow who calls himself ronwizfr confidently asserts that the delta V from earth orbit to Mars transfer is 6.6 km/s. Ronwizfr cites an MIT student project for a class called Solving Complex Problems.

The MIT page cited is a well done presentation until the students get to constants:

Then the students made an arithmetic error. The distance from the earth to the sun is 1.49X10¹¹ meters, not 1.49X10¹⁰ meters. Their radius for Mars' orbit is also off by a factor of 10.

The MIT students correctly plug in these wrong quantities to get speeds that are off by sqrt(10):

The velocity of the earth is not 94,384 m/s but closer to 29,846 m/s. Likewise velocity of Mars isn't 74,467 m/s but closer to 23,548 m/s. Their quantities would have been correct if the earth and Mars were .1 A.U. and .152 A.U. from the sun.

Even the very accomplished can make errors. So we should examine every assertion, no matter the source. My dad used to tell me we all put our pants on one leg at a time. His way of saying we're all human and capable of making mistakes.

Surgical Robots

At the Leprecon 39 Science Fiction Convention I attended a talk by Dr. Bruce Davis, a general/trauma surgeon in the Phoenix area.

Dr. Davis has done many laparoscopic surgeries using the da Vinci robot.  Instead of opening up a patient's belly, a small incision is made. A pair of robotic arms as well as a binocular pair of cameras are inserted through a small opening.

Davis reports that the two cameras give good depth perception. The telepresence is so immersive he often forgets he's not physically present inside the patient's body. He finds himself trying to turn his head to look about the cavity. Davis expects the robots will soon have motion capture for the head and neck so the camera motion will mimic the motion of the surgeon's actual eyes.

The arms are operated by motion capture. Robotic wrists mimic the motion of the surgeon's wrists. Instead of a thumb and four fingers, the robotic hand looks more like a crab's pinchers. The surgeon moves these pinchers with his thumb and index finger. Still, a lot can be accomplished with this simple hand.

The cameras give up to 10X magnification. When higher magnification is invoked, the motion of the robotic hands become more minute -- in effect shrinking the surgeon's avatar. We saw a video where a grape seemed like a large watermelon. The surgeon easily cut and peeled back a section of the grape's skin.

The robotic hands lack a sense of touch. Davis says the robot's designers are trying to figure out how to do haptic feedback but it's a challenging problem.

The da Vinci robot Davis uses costs 1.2 million dollars. I noted design and development is a large expense for early technologies. As more robots are made, design cost is amortized over more units and unit price can fall. Davis replied that this is happening to some extent. But da Vinci is the sole vendor in his field of robotic surgery and so has less incentive to drop their price.

I asked Dr. Davis if he foresaw uses of this technology outside of surgery. For example snaking robotic arms and eyes to a problem inside a utility line. This might avoid busting up a busy intersection with jack hammers and backhoes. Davis replied that this sort of application is likely and will probably lead to cheaper robots. Liability is a big expense in surgery. For example, a titanium screw breaking in a hip replacement can result in a major law suit. The same titanium screws that cost air lines $4.00 apiece are sold to surgeons at $400 apiece. Not having a heavy insurance burden, robots for plumbers and electricians can be less expensive and more common place than surgical robots.

I asked Davis if he thought telerobots could be used to build infrastructure on the moon or on an asteroid parked in lunar orbit. He said due to the 3 second light lag, motions would have to be very deliberate. But he thought it was doable. He mentioned there is a.i. being developed that mitigates slow reaction time caused by latency. For example, Big Dog's balance or Google Cars collision avoidance.

In Puppets, Telerobots and James Cameron, I opined that telerobots will be the game changer that enables use of space resources. I wrote that there are many uses right here on earth that are advancing telerobotic technology. So Davis' presentation was encouraging to me. Dr. Davis is also a science fiction writer.

Cartoon Delta V Map

This is the second cartoon delta-v map I've drawn. Clicking on the above can give a larger version.

My first cartoon map gave a lot more space to EML5 and little to L1 or L2. I knew of L4 and L5 through fiction like Gundam which was probably inspired by Gerard O'Neill's The High Frontier. Since then I've become less interested in L4 and L5 and more interested in L1 and L2. This new map reflects that shift in focus.

I had heard of the Interplanetary Transport Network as well as Shane Ross, Martin Lo, and Edward Belbruno. But I knew almost nothing about the low delta V routes achieved with n-body mechanics. I had a vague notion that Lagrange points were involved but that was about it.

Then in 2009 I came across a thread in Nasa Space Flight entitled An Alternative Lunar Architecture. In that thread Kirk Sorensen wrote at length about EML2 and work done by Robert Farquhar. Farquhar's 3 body work was done in the late 1960's and early 70's, decades before Ross, Belbruno and other modern advocates of 3-body mechanics.

Here's one of the Farquhar graphics Sorensen posted to that thread:

This is a 9 day route from LEO to EML2 taking delta V of about 3.5 km/s. It's time reversible so .4 km/s can drop a payload from from EML2 to an atmosphere grazing perigee. There's a 4 day route to EML1 that takes 3.8 days. It was surprising to me that EML2 could be reached with less delta V even though it's on the far side of the moon.

Also on Nasa Spaceflight in 2009, Martijn Mmeijering started a thread looking at an online 3 body text book by Koon, Lo, Marsden and Ross. 

It was in 2009 that I became more interested in L1 and L2.

There are routes between LEO and EML1&2 taking even less delta V, but these are time consuming. These are described by Andreas Stock's Investigation on Low Cost Transfer Options to the Earth-Moon Libration Point Region. 3.1 km/s seems to be the minimum between LEO for EML1 as well as EML2. In the map above I've depicted these routes with darker brown branches.

EML1 moves slower than an ordinary earth orbit at that altitude. An EML1 object nudged a little earthward will fall into an approximately 100,000 by 300,000 km elliptical orbit about the earth. A  .3 km/nudge suffices to send to it to a 36,000 km perigee where a 1 km/s burn can circularize the payload at geosynch orbit. A .7 burn can drop an EML1 payload to a LEO grazing orbit. If the LEO grazing orbit passes through the upper atmosphere, aerobraking can provide the 3.1 km/s needed to circularize at LEO.

EML2 moves faster than an ordinary earth orbit at its altitude. Nudged a bit away from the moon, a payload from EML2 will sail to a 1.8 million km apogee. The Sun Earth Lagrange 1 and 2 are 1.5 million kilometers from earth, so transfer from EML2 to SEL1 or 2 can be done with little delta V. Or an EML2 payload can sail through SEL1 or 2 completely out of earth's sphere of influence.

Nudge either EML1 or 2 a little moonward and they will fall into an approximately 5,000 x 60,000 km lunar orbit. Since the moon's rotating about the earth, a 60,000 km apolune can pass by both EML2 and EML1 over time. Thus it's possible to move between EML1 and 2 with very little delta V.

An object falling to a 300 km earth altitude from either EML1 or 2 will be traveling just a hair under escape when it reaches low altitude. Both EML1 and 2 make a complete circuit each 27.3 days so by timing your drop it's possible to choose longitude of perigee during a launch window. Plane changes are much less expensive at high altitudes so the velocity vector can be pointed in the right direction at perigee. Starting at EML2, injection into Mars or Venus Hohmanns can be done with around .9 km/s delta V (.4 km/sec to drop and a .5 km/s burn at perigee). On arriving at Mars, .7 km/s suffices to exit Hohmann for a 300 x 570,000 km Mars capture orbit.

A Near Earth Asteroid with a Vinfinity of 2 km/s or less can be dropped into an earth capture orbit using a lunar swing by. From there repeated lunar swing bys and little delta V can park the rock in high lunar orbit. Planetary Resources hopes to search for smaller rocks with their Arkyd orbital telescopes. If successful, they will likely find a multitude of rocks within .2 km/s of EML2.

Many NEAs are water rich and the cold traps at the lunar poles may have minable water deposits. So there are a number of potential propellent sources close to the earth-moon L1 and 2. Propellent sources high on the slopes of earth's gravity could break the exponent in Tsiolkovsky's rocket equation. This would give us mass fractions much easier to deal with. The highest delta V budget we'd have to endure is the 9.5 km/s from earth to LEO. Round trips between most other orbits would be in the neighborhood of 4 or 5 km/s.

Some notes on Venus: An earlier version of this map indicated delta V from Venus' surface to a capture orbit was 11.6 km/s. But then I came across an excellent delta V map by a fellow who calls himself Curious Metaphor.  In a discussion of his map, I was convinced Venus' thick dense atmosphere would make for a slower, steeper ascent from the planet surface. I've added 20 km/s gravity loss between Venus' upper atmosphere and surface. In Venus' upper atmosphere I've added a location labeled Landis Land. Named for Geoffrey Landis who noted there is a layer in Venus' atmosphere with earth like temperature and pressure. Moreover, a nitrogen/oxygen mix such as we  breath would be buoyant in Venus' CO2 atmosphere. Landis is a scientist as well as a science fiction writer. Some of his fiction takes place on the cloud cities of Venus.

Catching an Asteroid

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 (V_escape² + V_infinity²). 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.

The turning angle is about 96º.

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.

This would toss the rock into a 113,000 kilometer by 1,110,000 kilometer elliptical orbit about the earth. A .13 km/s apogee burn will boost perigee by about 180,000 kilometers. This path will take about two months so when the asteroid falls back, the moon will there to provide another gravity assist. This new gravity assist will boost the perigee as well as apogee. The new ellipse is a 311,000 x 1,500,000 km ellipse.

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.

What the heck is Vinf?

The internet is a great resource for isolated students like me without access to university libraries or college classrooms. But often Google will land me on a page over my head. I often see strange and unfamiliar terms.

Years ago I found myself gnashing my teeth and crying "What the heck is Vinf?!"

Then, as now, Vinf was a common term appearing on web pages teaching orbital mechanics.

Here is a screen capture from Atomic Rockets, a popular resource for science fiction writers:

(red underlines added by me)

And here's a screen capture from a Wikipedia article on hyperbolic orbits:

(red underlines added by me)

In the Wikipedia article they use the symbol for infinity instead of inf. By now you may have guessed Vinf is short for a hyperbola's velocity at infinity.

But why would we be interested in something an infinite distance away?

A hyperbola gets closer and closer to a straight lines called asymptotes. And as an object moving along a hyperbolic orbit gets farther from the earth, it's speed gets closer and closer to Vinf. A million kilometers out, the actual speed is so close to Vinf that we might as well call it Vinf. From page 36 of my orbital mechanics coloring book:

Page 36 from my coloring book Conic Sections and Celestial Mechanics

The formula for a hyperbola's velocity can be easily remembered if you picture V_hyperbola as the hypotenuse of a right right triangle with V_escape and V_infinity as legs:

V_escape²  +  V_infinity²  =  V_hyperbola²

I like to talk about hyperbolas but some people don't see the point. A fellow who calls himself Rune has informed me that transfer orbits aren't hyperbolas.

For most transfer orbits, Rune's right. A Hohmann transfer orbit from Earth to Mars is an ellipse with the sun at a focus:

Planet and asteroid orbits are also ellipses about the sun. So what's the fuss about hyperbolas?

To see let's take a closer look at our Hohmann to Mars:

At one scale we see the path as an ellipse about the sun. But magnify the Hohmann path in earth's neighborhood and we see a hyperbola about the earth.

What is the Vinf and Vesc we use to figure the velocity of this hyperbolic orbit? Let's take another look at our Hohmann ellipse about the sun:

The speed at the Hohmann perihelion (closest point to the sun) is about 33 kilometers/second. The earth is moving about 30 kilometers/second. So at perihelion, the Hohmann orbit is moving 3 km/s faster than earth. This 3 km/s is the Vinf of the hyperbolic orbit about the earth.

At the hyperbola's perigee (closest point to the earth), earth escape velocity is about 11 km/s.
So sqrt (V_escape² + V_infinity²) = sqrt (11² + 3²) km/s = sqrt (121 + 9) km/s
Which is about 11.5 km/s.
Only .5 km/s greater than the 11 km/s earth escape velocity

And how about the Mars hyperbolic orbit?

Speed at the Hohmann ellipse aphelion (furthest point from the sun) is about 2.8 km/s slower than Mars orbit. This 2.8 km/s is the Vinf of the hyperbolic orbit about Mars.

At a low periaerion, Mars escape velocity is around 4.8 km/s.
So hyperbola speed at the Mars end of the Hohmann is
sqrt (V_escape² + V_infinity²) km/s = sqrt(4.8² + 2.8²) km/s 
Which is about 5.5 km/s.
Only .7 km/s greater than Mars escape velocity.

Rune as well as illustrious people like Dr. Tom Murphy like to say the delta V from earth C3 = 0 to Mars C3 = 0 is around 6 km/s. Those who know how to patch conics will tell you it's closer to 1.2 km/s.

And to properly patch conics going from planet centric to heliocentric orbits we need to know Vesc and Vinf to find the velocity of planet centered hyperbolas. Since it's important, I'll repeat this:

The formula for a hyperbola's velocity can be easily remembered if you picture V_hyperbola as the hypotenuse of a right right triangle with V_escape and V_infinity as legs:

V_escape²  +  V_infinity²  =  V_hyperbola²