Thursday, July 10, 2014

Space topics from Dr. Plata

Doug Plata recently suggested some possible space exploration topics. All of them are very interesting.

Dr. Plata has good ideas and invests a great deal of time and effort looking at them. He's involved in two excellent websites: and .

July and August is a slow period for the Ajo Copper News, the weekly newspaper my sister and I publish. Most people with money and good sense leave Ajo, Arizona for the summer months. Hopefully I will have time to examine some of Plata's topics in my blog over the next few months.

Here they are:

Partial vs full reusability
Falcon 9 has nine engines on the first stage and one engine on the second stage.  So, if only the first stage is reused, it would seem to me that 9/10 engines would be recovered.  That's got to be a huge reduction in launch cost right there, yes?  Just how much?  Certainly achieving even partial reusability would make SpaceX even more competitive that it already is.  If the Falcon Heavy were to be partially reusable, reusing the lateral boosters would mean only 18 out of 28 engines would be recovered unless the central core could be reused as well.

Propulsion service options
For a cis-lunar transportation system we most often think of fuel depots in LEO.  One problem with this is the need for fuel depots in multiple LEO planes with those depots being used only occasionally.  However, if propellant were coming from ice harvesting operations at the lunar poles, then conceivably an OTV could bring propellant into any LEO inclination just prior to a launch into that orbit from Earth.  However, in this scenario, do we even need LEO depots?  Why couldn't the OTV dock with the launched satellite and then use its own engines to boost the satellite to GTO or even GEO?  Do we need fuel depots or could propulsion service be enough?

Power options for lunar mining
Say you are wanting to do ice harvesting operations in a lunar polar permanently shadowed crater with the rim of such having a peak of eternal light (PEL).  Great, but there's still potentially kilometers of distance between the source of power and the ice harvesting operations site.  How to deal with that gap?  RTGs?  Laser beaming of power?  Drive a rover laying a cable down the side of the crater?  Hop the lander from the rim to the floor while draping a (superconducting) wire?  Or forget a solar panel farm at the PEL and crack the water at a fuel depot in orbit?  An interesting trade analysis.

OTVs tend to be painted as broad, turtle-shaped craft.  But how do you launch and assemble such a thing?  Can aerobraking be done about as easily with a cylindrical-shaped OTV?  How about heat flaps popping out giving more surface area and control?  Necessary?  If one skims high enough in the atmosphere does one even need a heatshield?  What about using a lifting body form?

Travel times further out into the solar system
So if we develop the ability to safely send humans to Deimos, how much longer would it take to send them to Vesta, Ceres, and a Moon of Jupiter?
Heavy Lift vs Single Stage vs Reusable vs Gun
Air launch
Partial vs Full reusability
Chemical versus liquid rockets

Propulsion service options
Power options for lunar mining

The orbital dynamics of a Phobos vs Deimos vs surface mission

Mass calculations of open-loop, vs closed chemical, vs ECLSS
How could an RP5 be provided?
How do the space radiation numbers compare between locations (i.e. LEO, free-space, lunar surface, Phobos, Mars?)
Animal studies
Partial gravity options

O'Neillian vs lunar colony - Where first?
Travel times further out into the solar system

Wednesday, July 2, 2014

Kerbal Space Program

Lately this blog has been getting some hits from the Kerbal Space Program forum.

This looks like a good game. It seems based on the patched conics approach to orbital mechanics. It's good to see a popular game teaching users concepts like Hohmann or bi-elliptic transfers, sphere of influence, etc.

The art is appealing. Descriptions are entertaining. I purchased a copy for $27.00. It might be a way to become acquainted with some folks who share my orbital mechanics hobby. Hope it's a good investment!

Using the Kerbal Wiki I whomped up a HohmannKSP Spreadsheet. A few people like my spreadsheets for our solar system. Hopefully I'll be making some useful spreadsheets for this game.

Usual disclaimers apply:

My spreadsheets assume circular, coplanar orbits. Some of the game orbits are inclined and eccentric.

I occasionally make mistakes -- data entry as well as arithmetic errors. I'd be grateful if users check my efforts.

Wednesday, June 25, 2014

Travel on Airless Worlds Part II

This is a continuation of Travel on Airless Worlds where I looked at suborbital hops.

The surface of airless worlds will be exposed to radiation so it's likely the inhabitants would live underground.

Moreover, it is not as hard to burrow. The deepest gold mine on earth goes down about 4 kilometers. The heat and immense pressure make it hard to dig deeper. In contrast, the entire volume of a small body can be reached.

Courtney Seligman shows how to compute the pressure of a body with uniform density. The bodies we look at don't have uniform density but we'll use his method as a first order approximation.

Central pressure of a spherical body with uniform density is 3 g2/(8 π G)

Where G is universal gravitation constant, g is body's surface gravity and R is body's radius.

At distance r from center, pressure is (1 - (r/R)^2) * central pressure.

What is the pressure 4 kilometers below earth's surface?
Earths's radius R is 6378000 meters. r is that number minus 4000 meters. g is about 9.8 meters/sec^2.
Plugging those numbers into
(1 - (r/R)^2) * 3 g2/(8 π G)
gives 2120 atmospheres.

Besides pressure, heat also discourages us from burrowing deeper. So it might be possible to dig deeper on cooler worlds but for now we'll use 2120 atmospheres as the limit beyond which we can't dig.

3/(8 π G) * g * R2 gives different central pressures for various worlds:

Ceres center is 1430 atmospheres, well below our 2120 atmosphere limit. And Ceres is a cooler world than earth. We would be able to tunnel clear through the largest asteroid in the main belt. Since smaller asteroids would have smaller central pressure, we would be able to tunnel through the centers of every asteroid in the main belt.

Imagine a mohole going from a body's north pole to south pole:

The diagram above breaks the acceleration vector into vertical and horizontal components. The mohole payload has the same vertical acceleration components as an object in a circular orbit with orbital radius R, R being body radius.

Somone jumping into this mohole could travel to the opposite pole for zero energy. Trip time would be the orbital period: 2 π sqrt(R3/μ).

Other chords besides a diameter could be burrowed. I like to imagine 12 subway stations corresponding to the vertices of an icosahedron:

The red subway lines to nearest neighbors would correspond to the 30 edges of an icosahedron .

Green subway lines to the next nearest neighbors would correspond to the 30 edges of a small stellated dodecahedron.

And there could be 6 diameter subway lines linking a station to stations to their antipodes.

The energy free travel time of all these lines would be the same as the diameter trip time:
2 π sqrt(R3/μ).

It would be possible have a faster trip time than 2 π sqrt(R3/μ). A train could be accelerated during the first half of the trip and decelerated the second half. During the second half, energy could be recovered using regenerative braking.

Most of small bodies in our solar system have internal pressures that don't prohibit access. But in some cases central pressure exceeds 2120 atmospheres. We'd be able to burrow only so deep. Here's my guesstimate of the maximum depth for various bodies:

Luna 40 kilometers
Mars 15 kilometers
Ganymede 77 kilometers
Callisto 97 kilometers
Europa 55 kilometers
Titania 318 kilometers
Oberon 61 kilometers
Pluto 200 kilometers
Haumea 82 kilometers
Eris 108 kilometers

Here is the spreadsheet I used to look at internal pressures.

The top four kilometers of earth's surface is only a tiny fraction of the accessible mass in our solar system.

Standup Maths did a nice look at moholes going through a body.

Tuesday, June 17, 2014

Travel on airless worlds

When we get to other worlds how will we get from point A to B? There are no roads on Ceres. No rivers or oceans on the moon. No airports on Enceladus or even air for an airplane to glide on.

Until transportation infrastructure is built, suborbital hops seem the way to go. A suborbital hop is an ellipse with a focus at the body center. But with much of the ellipse below surface.

For a suborbital hop from A to B, how much velocity is needed? At what angle should we leave the body's surface? At what angle and velocity do we return?

Points A and B are two points in a Lambert Space Triangle. The triangle's third point C is the body's center.

Two of the sides are radii of a spherical body so the triangle is isosceles.

A point's distance from one focus plus distance to second focus is a constant, 2a. 2a is the ellipse's major axis. Since A and B are equidistant to the first focus, they're also equidistant to the second focus. Points A & B, the planet and ellipse are all symmetric about the ellipse's axis.

The payload returns to the body at the same angle and velocity it left.

There are a multitude of ellipses whose focus lies at the center and pass through points A and B. How do we find the ellipse that requires the least delta V?

Specific orbital energy is denoted ε.

ε = 1/2 v2μ/r. 

1/2 v2 is the specific kinetic energy from the body's motion. μ/r is the  specific potential energy that comes from  the object's distance from C, the body's center. For an elliptic orbit, | μ/r | is greater than 1/2 v2, that is the potential energy overwhelms the kinetic energy. If potential and kinetic energy exactly cancel, the orbit is parabolic and the payload's moving escape velocity.

So to minimize 1/2 v2 we'd want to maximize | ε |.

It so happens that

ε = - μ/(2 a).

To grow |μ/(2 a)| we shrink 2a. So we're going for the ellipse with the shortest possible major axis.

The foci of all these ellipses lie on the same line. Moving the focus up and down this line, it can be seen the ellipse having the shortest major axis has a focus lying at the center of the chord connecting A and B. That would be the red ellipse pictured above. 

The angle separating A and B is labeled θ. Recall the black length and red length sum to 2a

2a = r + r sin(θ/2).

a = r(1 + sin(θ/2))/2.

Now that we know a, we can find velocity with the vis-viva equation:

V = sqrt(μ(2/r - 1/a)).

At what angle should we fire the payload?

It's known a light beam sent from one focus would be reflected to the second focus with the angle of incidence equal to the angle of reflection:

Angle between position vector and velocity vector is (3π - θ)/4. A horizontal velocity vector has angle π/2 from the position vector. 
 (3π - θ)/4 -  π/2 = (3π - θ)/4 -  2π/4 =
 (π - θ)/4

 (π - θ)/4 is the flight path angle departing from point A. If A and B are separated by 180º (i.e. travel from north pole to south pole), flight path angle is 0º and the payload is fired horizontally. If A and B are separated by 90º (i. e. travel from the north pole to a location on the equator), flight path angle is (180º - 90º)/4 which is 22.5º

When A and B are very close, θ is close to 0. (180-0)/4 is 45º. When A and B are close the the payload would be fire at nearly 45º. With one focus near to the surface and the other at the body center, the ellipse would have an eccentricity of almost 1. The trajectory would look parabolic.

Here's a few possible suborbital hops on our moon:

Minimum energy ellipse from pole to pole is a circle. Launch velocity is about 1.7 km/s. It'd take the same amount of delta V for a soft landing at the destination. Trip time would be about 54 minutes

From the north pole to the equator would take a 1.53 km/s launch. Trip time would be about 27 minutes.

Period of a circular orbit (T) is 2 π * sqrt(r3/(Gm)) where m is mass of planet.

Mass can be expressed as ρ * volume where ρ is density.

π * sqrt(r3/Gm) = 2 π * sqrt(r3/(G * 4/3 *π * r3 * ρ).

The r3 cancels out and we're left with T = sqrt(π/(G * 4/3 * ρ).

Thus trip times for a given separation rely solely on density. Period scales with inverse square root of density.

I did a spreadsheet where the user can enter angular separation on various airless bodies in our solar system. Delta Vs vary widely depending on size of the bodies. But trip times between comparable angular separations are roughly the same. This is because most the bodies have roughly the same density. Denser bodies like Mercury will have shorter trip times while icey, low density bodies will have longer trip times.

I got some help on this from a Nasa Spaceflight thread. My thanks to AlanSE and Proponent.

After time we would establish infrastructure on bodies and burrow into their volume. With tunnels we could reach various destinations with very little energy. I look at this in Travel On Airless World Part II.

Friday, May 30, 2014

Reusable Earth Departure Stage

My notion of a reusable Earth Departure Stage (EDS) assumes a staging platform at Earth Moon Lagrange 2 (EML2). Propellent, water and air at EML2 might come from an carbonaceous asteroid parked in lunar orbit and/or volatiles in the moon's polar cold traps.

Pictured above is Robert Farquhar's route between EML2 and LEO. It's time reversible so it could be to or from EML2.

A .15 km/s burn at EML2 will drop a spacecraft to a perilune 111 km from the moon's surface. At this perilune the spacecraft is traveling nearly lunar escape velocity with regard to the moon and so enjoys an Oberth benefit. A .19 km/s perilune burn suffices to send the spacecraft earthward to a perigee deep in earth's gravity.

At perigee the spacecraft is traveling about 10.8 km/s, just a hair under earth escape. A burn at this perigee enjoys a huge Oberth benefit. A .6 km/s burn would suffice for Trans Mars Injection (TMI). So delta V from EML2 to TMI is (.15 + .19 + .6) km/s. I will round .94 km/s up to 1 km/s to give a little margin and also 1 is an easier number to type.

After TMI the EDS as well as it's payload is moving 11.5 km/s. To reuse the EDS we would need to return it to EML2.

Farquhar notes the trip from perilune to perigee takes about 140 hours. In that time the moon will advance 76º and the space craft 180º. So in my shotgun orbit simulator I set the perigee 104º ahead of the moon. My first try had pellets ranging from 10.7 to 10.9 km/s and then I'd narrow the blast to the pellets coming closest to the moon. After a few iterations I arrived at a perigee velocity of about 10.85 km/s. This gives an apogee of about 396,000 km and a period close to 2/5 that of the moon. After 50 days, the pellets return to a near moon fly by:

Thus braking about .6 km/s drop the EDS hyperbolic path to a trajectory where it will do a near moon fly-by after 50 days. At the near moon fly by it can do a .14 km/s burn for lunar capture. Then when it reaches an apolune near EML2, a .19 km/s burn to park at EML2.

Thus the EDS' delta V for returning to EML2 will be about 1 km/s.

This page still a work in progress, I'm getting good comments and info from a NASA spaceflight thread. Cryogenic boil off was an issue raised in that thread. A sixty day round trip goes well beyond what present hydrogen/oxygen upper stages do.

Revisiting Kirk Sorensen's EML2 thread I noticed he had posted a Farquhar illustration suggesting reusable booster stages could be returned to EML2!

The United Launch Alliance has done work on hydrogen/oxygen upper stages that could do longer missions. See Advanced Cryogenic Evolved Stage (ACES). Cyrogenic boil off might be mitigated by Multi Layer Insulation (MLI). Another cooling device is a Thermodynamic Vent System (TVS). Those who live in the southwest are familiar with "swamp coolers" where water soaked pads cool by evaporation. In a similar fashion hydrogen boil-off can be used to cool the cryogens. The hydrogen boil off can be vented in a specific directions and used for station keeping or attitude control.

Besides these passive thermal control systems the ACES might also utilize a two stage turboBrayton cryocooler.
"This design was based on the Creare NICMOS cooler that has been flying on the Hubble Space Telescope for the last ~4 years. The turboBrayton cycle uses GHe as the working fluid and this cooled gas can be easily distributed to the loads (i.e. the 22K and 95K shields). The ACES cryocooler configuration, shown in Figure 3-1, has 3 compressors in series and 2 expansion turbines in parallel, one for the 22K load, and one for the 95K load".

In this ULA pdf an ACES 41 propellant tanker has 5 tonnes dry mass and 41 tonnes propellent. I will be much more conservative in my hypothetical reusable EDS. A Centaur has 2.25 tonnes dry mass, 21 tonnes propellent and 99.2 kilo newtons. I will use the same but boost the dry mass to 5 tonnes for MLI, cryocooling, solar arrays, etc.

Specs for Hop's EDS
5 tonnes dry mass
21 tonnes hydrogen/oxygen
99.2 kilo newtons thrust

After the EDS sends the payload on its way, it will need 1 km/s of propellent of delta V to return to EML2. Exhaust velocity of hydrogen and oxygen is about 4.4 km/s. Exp(1/4.4) - 1 is about .255. To get back the EDS' 5 tonnes of dry mass we'd need 1.3 tonnes of propellent.

So for the first leg of the trip we have (21 - 1.3) tonnes of propellent or 19.7 tonnes. The first leg is also a 1 km/s delta V budget. With a 1 km/s delta V budget, 19.7 tonnes of propellent can do 19.7tonnes/.255. That's about 77 tonnes. But recall 6.3 tonnes is EDS dry mass plus propellent for the return trip. That's (77 - 6.3) tonnes of propellent available for payload. Let's call that 70 tonnes.

This little EDS could impart Trans Mars Insertion (TMI) to 70 tonne payload. Two of these EDS stages could send a 140 tonne payload on its way to Mars. Wilson and Clarke imagine a Mars Transfer Vehicle (MTV) of 130 tonnes.

Of the MTVs 130 tonnes, about 60 tonnes is propellent and consumables. If propellent, water and air are available from an asteroid or lunar volatiles, it would only be necessary to send the MTV's 70 tonne dry mass to EML2.

Wilson and Clarke also call for two EDS stages (they call them TMS -- Trans Mars Stages). Their stages are 110 tonnes and not reusable.

130 + 2*110 = 350. 350 tonnes to LEO for each (non reusable) conventional MTV. Vs 70 tonnes to LEO for an MTV that relies on extra terrestrial propellent and consumables. And an MTV departing from and returning to EML2 would have a much lower delta V budget. Making the MTV reusable would be much more doable.

The EDS would zoom through the perigee neighborhood very quickly. Would it have enough time to do the burns and enjoy an Oberth benefit?

The EDS and payload would spend about 54 minutes in the shaded region above.

A 70 tonne payload plus a 26 tonne EDS total 96 tonnes. The thrust of the engine is 99.2 kilonewtons. Acceleration is newtons/kilograms. 96/99.2 is ~.96. .96 meters/second^2 is about a tenth of a g.

Delta V imparted is acceleration * time of burn. Recall the perigee burn is about .6 km/s or 600 meters/second. We solve for t.

a * t = v
.96 m/s^2 * t = 600 m/s
t = 600/.96 seconds = ~620 seconds, a little over 10 minutes. The ten minute neighborhood just preceding perigee is all close to 10.8 km/s.

After separating from payload, the EDS and it's return propellent mass 6.3 tonnes. 99.2 kilonewtons divided by 6.3 tonnes is 15.75 meters/second^2 or nearly two g's. The deceleration burn to brake the hyperbolic orbit to an elliptical capture orbit would take about 40 seconds.

Near Earth Asteroid Retrieval

The Near Earth Asteroid retrieval described in the Keck Report uses xenon as a propellent. The exhaust velocity would be 30 km/s. What possible use could an EDS with a measly 4.4 km/s exhaust velocity be for such a vehicle?

Along with xenon's high exhaust velocity comes very low thrust. It would take nearly two years to spiral from Low Earth Orbit (LEO) to escape velocity. A good part of that long spiral would be spent in the Van Allen Belts. Low Earth Orbit also has a relatively high debris density.

Low thrust rockets don't enjoy any Oberth benefit. So the spiral from LEO to C3=0 would take about 7 km/s. Recall the exhaust velocity of the xenon rockets is around 30 km/s. Exp(7/30) - 1 is .26. Using an EDS would leave the asteroid fetcher with about 33% more xenon.

Many NEAs are much closer than Mars in terms of delta V. So perigee burn would be much less than .6 km/s for TMI, probably more often in the neighborhood of .2 or .3 km/s.

Saturday, March 8, 2014

Murphy's reply

In general, Murphy's Do The Math crowd don't do math. Rather they appeal to Murphy's authority and hurl insults. Mike Stasse's forum was no exception. Is Murphy an authority? Does his PhD means he's qualified?

I responded with The Most Common Delta V error. High school seniors typically mispatch conics the same way Murphy does. Murphy's level of expertise is somewhere below Orbital Mechanics 101 for liberal arts majors.

Stasse passed this on to Tom Murphy. And got a reply. Stasse quotes Murphy:
I don't dispute the more careful approach used on hopsblog. I put pieces together very simply, which may not represent more clever ways to manage interplanetary trajectories. That said, I stated clearly what I was doing, so that it's an easy job to pick it apart. I'm fine with that. I hope I never appealed to my authority as an orbital mechanics expert, because I am not
By his own admission, Murphy's no expert. Perhaps Murphy hasn't appealed to his authority. But Stasse and his friends certainly have. Murphy goes on:
I just try to put scales on things and sort out roughly how hard things are. At the pace of a post a week (during that time)--on top of a busy job--I could not spend time polishing. 15 km/s (allowing a bit of rounding) is still frikin' hard, so my main point is barely scratched.
Murphy's point isn't merely scratched, it's gored. 15 km/s is about what it takes to put a geosynchronous communication satellite in place. This is doable as demonstrated by the large number of geosynchronous sats. Murphy's 20 km/s is about what it takes to land on the moon's surface and come back. 15 km/s and 20 km/s are vastly different delta V budgets.

But 15 km/s vs 20 km/s is isn't the worst Murphy error. He's done much worse. From Grab That Asteroid! in Murphy's Stranded Resources post:

The asteroid belt is over 20 km/s away in terms of velocity impulse. If the goal is to use the raw materials for production on Earth or in Earth orbit, we have to supply about 10 km/s of impulse. We would probably try to get lucky and find a nickel-metal asteroid in an unusual orbit requiring substantially less energy to reel it in. So let's say we can find something requiring only 5 km/s of delta-v.  . . .

To get this asteroid moving at 5 km/s with conventional rocket fuel (or any "fuel" that involves spitting the mass elements/ions out at high speed) would require a mass of fuel approximately twice that of the asteroid. As an example, using methane and oxygen,  . . .

Does fetching an asteroid take twice the rock's mass in propellent?

Ratio of propellent to dry mass can be found with Tsiolkovsky's rocket equation:
(Mass propellent)/(dry mass) = e(delta V/exhaust velocity) - 1

Let's see -- in Murphy's example delta V is 5 km/s. Exhaust velocity of oxygen and methane is about 3.4 km/s.

e(5 km/s / 3.4 km/s) - 1 = 3.35. So for every ton of asteroid, we'd need more than 3 tons of propellent. At first glance it looks like Murphy is being kind and even under estimating propellent needed.

But methane and oxygen isn't the only propellent. Xenon from an ion engine has an exhaust velocity of around 30 km/s.

e(5 km/s / 30 km/s) - 1 = .18

So about .2 tonnes (or 200 kilograms) of propellent to park a tonne of asteroid. 2/10 is not a "rough approximation" of 2.

But Murphy's error gets worse.

Murphy thinks we'd be lucky to find an asteroid outside of the Main Belt that takes 5 km/s to retrieve. Evidently he hasn't heard of Near Earth Asteroids. There are many asteroids that take much less.

The Keck study for retrieving an asteroid notes some asteroids take as little as .17 km/s. Let's plug in .17 km/s delta V:

e(.17 km/s / 30 km/s) - 1 = .006

So 6 kilograms of propellent to park a tonne of Asteroid. Now Murphy's guesstimate of twice the asteroid's mass is off by a factor of about 350.

Six kilograms is about the mass of two chihuahuas. Two tons is about the mass of two large horses, big horses as in Budweiser clydesdales.

Tom Murphy is a busy guy. So he uses furious handwaving to excuse info that's completely wrong.

Murphy's figures are often off by several orders of magnitude his PhD notwithstanding.

Neither Stasse's appeal to authority nor Murphy's "rough approximation" defense salvage Murphy's arguments.

Monday, February 24, 2014

The most common delta V error

Patching Conics
Time and time again I've watched people patch conics by using straight addition and ignoring the Oberth benefit. It is a very easy mistake to make. I'll give an example of this error for an earth to Mars delta V budget.

At perihelion an earth to Mars Hohmann orbit is moving about 33 km/s. 3 km/s is needed to leave earth's 30 km/s heliocentric orbit and enter this transfer orbit. In a similar fashion it takes 2.5 km/s to leave Hohmann transfer and match velocities with Mars.

But before we go from one heliocentric orbit to another, we need to escape the planet's gravity well. Earth's surface escape velocity is about 11 km/s. Mars' surface escape velocity is about 5 km/s.

The novice will look at these 4 quantities and simply add them. 11 + 3 + 2.5 + 5 is 20.5. They'll tell you it takes about 20.5 km/s to get from earth's surface to Mars surface.

But to accurately patch conics you need to use the hyperbolic orbit that takes you out of the planet's sphere of influence to a heliocentric orbit. Hyperbolic orbit speed is sqrt(Vescape2 + Vinfinity2). But what's Vinfinity? In this example it's the 3 km/s needed to go from earth's 30 km/s orbit to a Hohmann's 33 km/s. In Mars' neighborhood Vinfinity is the 2.5 km/s needed to exit Hohmann and match velocities with Mars.

If you remember high school math, sqrt( a2 +  b) should look familiar. It's the hypotenuse in the good old Pythagorean theorem! And that's what I use to visualize hyperbola speed:

The novice will tell once you've achieved the 11 km/s to escape earth's gravity well, you need another 3 km/s. The informed will tell you only another .4 km/s is needed.

For Mars to Earth the naive will tell you after you've reached 5 km/s to escape Mars then you need another 2.5 km/s to send the ship earthward. The savvy will tell you an additional .6 km/s is needed.

.4 vs 3 and .6 vs 2.5. In this case the novice method results in a 4.5 km/s overestimation of the delta V budget.

Erik Max Francis
I used to call this the Erik Max Francis Error. Delta V budgets from one planet to another would often come up in space usenet groups. Erik would use his Python BOTEC and give an answer to ten decimal places. People would ooooh and ahhhhh. Wow! Accurate to 10 significant figures! In reality Erik's answers were accurate to zero significant figures. I finally prodded him to correct his error. Now his BOTEC is accurate to 1 or 2 significant figures. So far as I know, he still gives answers to 10 decimal places.

Later I called it the Rune error. The total Vinf can be roughly estimated by subtracting Mars 24 km/s from earth's 30 km/s. And this is what Rune uses on the New Mars forum when Louis asks how much delta V is needed after you get out of earth's gravity well.
Brute force" trajectories would take about as much delta-v as is the difference between the orbital speeds of mars and earth, so about 29.8km/s (for earth) - 24km/s (for mars) = 5.8km/s
I have explained to the New Mars Forums many times that the speed of a hyperbola is sqrt(Vescape2 + Vinfinity2). Will Rune ever learn it? I doubt it.

But no matter, Rune's a member of Zubrin's cult. People expect Zubrinistas to be innumerate, nobody takes  them seriously. But sadly they are loud and high profile. John Q Public can be misled into thinking they speak for all space advocates.

It's more damaging when someone in authority commits this error.  Now I'm talking about Dr. Tom Murphy.

Professor Tom Murphy
This is from a graphic from Tom Murphy's Stranded Resources:

Murphy writes:
For instance, we travel around the Sun at a velocity of 30 km/s, while Mars sails at a more sedate 24 km/s. So to meet up with Mars, we have 6 km/s of extra velocity to burn, helping us up the hill. We speak of this as a Δv (delta-vee) adjustment to trajectory.
Same method as Rune for getting the total Vinf: Subtracting Mars' 24 km/s from earth's 30 km/s to get 6 km/s. An over estimation but not wildly inaccurate. And Murphy correctly shows earth's escape as 11 km/s and Mars escape as about 5 km/s.

But then Murphy straight up adds 11+6+5:

Crudely speaking, we must have the means to accomplish all vertical traverses in order to make a trip. For instance, landing on Mars from Earth requires about 17 km/s of climb, followed by a controlled 5 km/s of deceleration for the descent. Thus it takes something like 20 km/s of capability to land on Mars
Sounds like he's being generous to the poor deluded space cadets by rounding 22 down to 20. But the distance from earth's C3=0 to Mars C3=0 is not 6 km/s. It's about 1 km/s. Here's Murphy graph corrected for the Oberth benefit:

For comparison, Murphy's erroneous graph is left in but a shade lighter. From surface of Earth to Surface of Mars is about 16 km/s.

"But wait!" a Murphy apologist might say. "Murphy didn't include the delta V needed to rise above earth's atmosphere. That's 1.5 to 2 km/s! That makes the budget more like 18 which can be rounded up to 20."

An atmosphere does indeed add to delta V for departing a planet. On the other hand, an atmosphere is a big help for planet arrival. Park in a capture orbit with periapsis velocity just a hair under escape. Position the periapsis in the planet's upper atmosphere. Each orbit at periapsis, atmospheric friction slows the ship. This is known as aerobraking. Almost all of Mars' 5 km/s descent can dealt with via aerobraking. Here is Murphy's graph corrected for atmospheric influence:

Now it takes 13 or 14 km's to reach escape. But with descent taken care of with aerobraking it only takes another 1 km/s to reach Mars' surface. A more realistic delta V budget from earth surface to Mars surface is about 14 or 15 km/s.

And in fact numerous Mars landers and orbiters have used this method. I am stunned that Murphy, a self proclaimed space insider, has never heard of aerobraking.

Is a 6 km/s error a big deal? Since the exponent of the rocket equation scales with delta V, it's a very big deal. Murphy himself would tell you exponential growth can be dramatic.

Above graph assumes hydrogen/oxygen bipropellent. Each 3 km/s added to the delta V budget about doubles propellent needed. Each 5 km/s added nearly triples the amount. Murphy's 20 km/s delta V budget would need a little more than triple the propellent of the actual 15 km/s delta V budget.

In my opinion the limits to growth is the most important issue facing us. Can space resources raise the ceiling on our logistic growth? If so, it is worthwhile to invest in building space infra-structure. If not, expensive space infrastructure is a waste of money. We should look at the question seriously. That is why I get bent out of shape when a so-called authority makes common mistakes that would embarrass a freshman aerospace student.

Tom Murphy's arguments against space are often cited in discussions of limits to growth. When I find such a discussion, I will chime in that a bright high school student could tear apart Murphy's arguments. The most recent visit was at Mike Stasse's Damn The Matrix. As usual, the Do The Math crowd response was insults, appeal to authority, but no math.

Judging by Murphy's fans, his blog should be renamed "Don't do the Math. Take my word for it because I'm a Ph. D."

Saturday, February 1, 2014

Terraforming Mars vs Orbital Habs


Those who advocate Mars settlement like to say Mars can be terraformed. First I will take a look at what it would take to terraform Mars.

How much air do we need to add to Mars?

From NASA's Mars Fact Sheet, surface density of the Martian atmosphere is about .02 kg/m3. That is about 1.5% of Earth's surface air pressure of 1.27 kg/m3. Mars' atmosphere is virtually a vacuum.

Mars surface gravity is about 38% earth gravity. That means given an atmosphere of comparable temperature and composition, Mars atmosphere scale height is 264% earth atmosphere scale height. But Mars surface area is about about 28% that of earth's. 2.64 * .28 is about .75. To get comparable air density, we would need Mars' atmospheric mass to be about three quarters that of earth's atmosphere.

The total mass of the Martian atmosphere is about 2.5 x 1016 kg. Earth's atmosphere is about 5 x 1018 kg. So to make Mars surface air density earth like, we'd need 3.6 x 1018 kg of air added to Mars.

But do we need sea level air density? No, there are people who survive at higher elevations. This list of the world's highest cities show several places at around 5000 meter elevation. Granted the dwellers of the highest city La Rinconada, Peru don't live comfortably. But they demonstrate humans can endure air density half that of sea level. If half is sufficient, Mars only needs 1.8 x 1018 additional kilograms of air.

Would be Mars terraformers like to point at the frozen CO2 at the Martian poles. If Mars temperature is raised just a little, they hope the vaporized carbon dioxide would create a greenhouse effect that would cause more carbon dioxide to be vaporized. Their hope is that a runaway greenhouse effect could substantially boost Mars' atmosphere from frozen volatiles already in place.

According to Wikipedia, there is thought to be a 1 meter thick layer of CO2 at Mars north pole, a cap about 1,000,000 meters in diameter. At the south pole there is an 8 meter thick layer of CO2 over a cap having a 350,000 meter diameter. That's about 1.6 x 1012 cubic meters of CO2. Dry ice has a density of 1.6 thousand kg/m3. If all of that CO2 is vaporized (an optimistic assumption) that totals about 2.5 x 1015 kg of atmosphere. Short by almost 3 orders of magnitude, a miniscule contribution toward the needed 1.8 x 1018 needed kilograms.

Zubrin and McKay believe runaway greenhouse could boost Mars atmosphere to 300 to 600 millibars. Besides the polar dry ice, they also mention CO2 in Martian regolith. I believe most of Zubrin's optmistic estimates are influenced more by wishful thinking than hard data. But for the sake of argument I'll grant 300 millibars of CO2. 300 millibars of CO2 is not breathable. But let's say green plants combine Martian water and CO2 to make sugars and starches plus oxygen. Taking the carbon out of 300 millibars of CO2 leaves about 220 millibars of oxygen. Earth's 1000 millibar atmosphere is 1/5 oxygen, so perhaps a 220 millibar oxygen atmosphere would be breathable. But it would also be an extreme fire hazard. Apollo 1 taught us a pure oxygen atmosphere isn't a good idea.

Even with Zubrin's very optimistic scenario, it seems we'd still need to import 1.5 1x 1018 kilograms of nitrogen.

Can we add to Mars' air with comets?

Zubrin and McKay suggest  it'd take .3 km/s to nudge an ammonia asteroid in the outer solar system towards Saturn and then Saturn's gravity could throw the ammonia snowball Marsward.

"Consider an asteroid made of frozen ammonia with a mass of 10 billion tonnes orbiting the sun at a distance of 12 AU. Such an object, if spherical, would have a diameter of about 2.6 km, and changing its orbit to intersect Saturn's (where it could get a trans-Mars gravity assist) would require a DV of 0.3 km/s. If a quartet of 5000 MW nuclear thermal rocket engines powered by either fission or fusion were used to heat some of its ammonia up to 2200 K (5000 MW fission NTRs operating at 2500 K were tested in the 1960s), they would produce an exhaust velocity of 4 km/s, which would allow them to move the asteroid onto its required course using only 8% of its material as propellant. Ten years of steady thrusting would be required, followed by a about a 20 year coast to impact. When the object hit Mars, the energy released would be about 10 TW-years, enough to melt 1 trillion tonnes of water (a lake 140 km on a side and 50 meters deep). In addition, the ammonia released by a single such object would raise the planet's temperature by about 3 degrees centigrade and form a shield that would effectively mask the planet's surface from ultraviolet radiation. As further missions proceeded, the planet's temperature could be increased globally in accord with the data shown in Fig. 12. Forty such missions would double the nitrogen content of Mars' atmosphere by direct importation, and could produce much more if some of the asteroids were targeted to hit beds of nitrates, which they would volatilize into nitrogen and oxygen upon impact. If one such mission were launched per year, within half a century or so most of Mars would have a temperate climate, and enough water would have been melted to cover a quarter of the planet with a layer of water 1 m deep."

This scheme presupposes we could land a 20 gigawatt power source on a rock in the outer solar solar system. For comparison the Palo Verde Nuclear Power Plant, the largest nuclear power plant in the United States, produces about 3.3 gigawatts. So we're sending 6 Palo Verde Nuclear Power Plants out past Saturn. McCay's scheme stipulates using the comet's mass as reaction mass. So now we have a mining and transportation infra structure on the comet that digs up the ice and places this reaction mass in the nuclear rocket engine.

If we have the wherewithal to establish such infrastructure, we certainly have the ability to build habs on these rocks.

Asteroidal Real Estate

How much asteroidal real estate could 1.5 1x 1018 kilograms of air give us? An O'Neill cylinder 8 kilometers in diameter and 32 kilometers long would give us 804 square kilometers of real estate. Such a cylinder would have a volume of 1.6e12 cubic meters. On earth's surface, our air has a density of about 1.27 kg per cubic meter. So that volume at 1 bar density would be 2e12 kilograms of air.

1.5e18/2e12 = 750,000. Three quarters of a million O'Neill habitats. Recall each cylinder has 804 square kilometers of real estate. 750,000 * 804 km2 = 603 million km2. Mars' surface area is 145 million km2. So if we put the asteroidal resources to use where they're at, we get 4 times as much real estate.

Some would point out that O'Neill cylinders are very extravagant pieces of mega-engineering. I completely agree! It's my belief that humans don't need a full g to be healthy, I believe .4 g (a little more than Mars' gravity) would suffice. In which case the hab radius could be 1.6 km. Such a hab would have only  321 km2 of real estate but a volume only 2.6e11 cubic meters. 2.6e11 m3 * 1.27 kg/m3 = 3.3e11 kilograms. 1.5e18/3.3e11 = ~4.5 million. 4.5 million of the smaller O'Neill habitats. 4.5 million * 321 = 1460 million square kilometers. Or about as much real estate as 10 Mars planets.

If the goal is to provide more real estate and resources for humanity, terraforming Mars is an extravagant waste. We should ditch planetary chauvinism and go for the small bodies.

Robert Walker also takes a look at terraforming Mars.