Once We Have A Self Sustaining Mars Colony - Then What?

On Tuesday, December 13, 2016 at 12:17:14 AM UTC+13, Jeff Findley wrote:
We've got a Mook on Mook on Mook reply here (minus the first two
Mooks)...

In article ,
says...
The sands of Mars are red. That's because they're made out of
hematite. Why wouldn't you mine iron there and send it back to
Earth with a rail gun?

Because steel made on earth is already quite cheap, so it would be
economic suicide to do what you propose.

Steel has been cheap historically, but is rising inexorably as raw materials are depleted here.

https://www.bloomberg.com/news/artic...-s-rapid-shift


Besides, one would think that a Mars colony would use such raw materials
to either build things they need on Mars or build things that are
actually worth exporting.

That's a false choice. In order to use or build things on Mars local steel is needed. So Martians would need to supply themselves with steel made from hematite on the surface and carbon in the carbon dioxide in the air - which means any surplus to their needs could be exported.

Raw materials aren't going to cut it as an
export unless there is a return on that investment.

Correct. Prices are rising on Earth and Earth's ability to produce low cost steel will be non-existant in 64 years according to the experts. Some believe shortages may be arriving in as little at 12 years.

Sorry Mook,

I feel your love.

but this entire idea is b.s.

No it isn't.

I don't know how you got to
visions of Mars colonies sending quite common raw materials like iron,
silicon, aluminum, and etc. to earth by railgun, but it's just not going
to be viable economically.

That's your problem that you don't know something. Perhaps if you listened to those who know more than you - that might help.

More below.

Jeff
--
All opinions posted by me on Usenet News are mine, and mine alone.
These posts do not reflect the opinions of my family, friends,
employer, or any organization that I am a member of.

The iron ore reserves of Earth at present seem quite vast, but continual exponential increase in consumption make this resource quite finite.

Lester Brown of the Worldwatch Institute has suggested iron ore will run out within 64 years based on an extremely conservative extrapolation of 2% growth per year. A 7% growth rate sees iron ore running out on Earth in 32 years. The Earth has 484.9 billion tons of economically recoverable iron ore.

So, to make steel on Earth you start with 2 tons of ore, 1 ton of carbon, half a ton of limestone and one ton of oxygen from 5 tons of air. There are 980 billion tons of economically recoverable coal in the world.

It takes 1.8 tons of metallurgical coal - or coking coal - to make 1.0 ton of coke used in steel production. 8% of all coal reserves are metallurgical variety. So, 24.3 billion tons steel possible to make with the proven reserves of coking coal we have today.

The Earth currently produces 350 billion tons per year of limestone and 7.5% of the Earth's crust is limestone, and there is plenty of economically recoverable limestone around. Only a small percentage of limestone is used in steel production - 0.8 billion tons per year.

So, the Earth's limit in steel production is about 200 billion tons and that's it, if limited by economically recoverable iron ore, ignoring coking coal limits. Only 24 billion more tons if limited by coking coal. This is 12 years at 7% growth and 14 years at 2% growth.

In contrast the Martian surface is littered with quadrillions of tons of hematite - with iron making up 2% of the Martian dust according to rovers sent there. There is more calcium in Martian dust than Iron! At 8% CaO matches Earth limestone in abundance. So, the limiting factor is carbon on Mars..

The Martian atmosphere has 26.4 trillion short tons of carbon dioxide. This can be reduced to CH4 by;

CO2 + 4 H2 -- CH4 + 2 H2O

and the CH4 may be reduced further by pyrolysis to;

CH4 + energy --- C + 2 H2

And the H2O made back into hydrogen and oxygen

H2O + energy --- H2 + 1/2 O2

Which makes 4 H2 in toto and takes us back to step one.

The energy needed is about 40 GJ/ton to power this process. Gotten from nuclear sources or solar sources, this works. The price of energy and manufacturing on Mars is the cost driver here.

So the Martian atmosphere yields 19.2 trillion tons of oxygen and 7.2 trillion tons of elemental carbon when extracted from the carbon dioxide. With recycling of carbon dioxide produced from burning the carbon and oxygen, sufficient carbon to make 1.8 quadrillion tons of low-carbon steel may be extracted from the planet using 72 quadrillion gigajoules.

As a side benefit, reducing CO2 from the Martian atmosphere yields 0.5 trillion tons of Nitrogen and 0.5 trillion tons of argon. Add 0.2 trillion tons of oxygen to the nitrogen, the oxygen extracted from the carbon and from the hematite and you have 0.7 trillion tons of Earth like atmosphere.

A ton of air at sea level occupies 24,783 cubic feet. So, 0.7 trillion tons of air is 17.4 quadrillion cubic feet. That's 117,855 cubic miles. A layer of air 100 feet deep at this pressure covers 174 trillion square feet - or 6.24 million square miles. 11% of the Martian surface pressurised at a depth of 100 feet. The entire surface may be pressurised at 9 foot depth.

http://digitalcommons.calpoly.edu/cg...&context=cadrc

https://i.kinja-img.com/gawker-media...ysy9052jpg.jpg

The background radiation on Mars is quite high relative to Earth. That means the locals will find ways to shield themselves from this hazard. It also means that the production and use of fissile materials will not have the same adverse consequences that they do when used within the sheilded garden of Earth's biosphere. Further the presence of nuclear weapons is nonexistent on Mars and will remain so under current UN treaties. For this reason, experts have looked seriously at broad use of small nuclear power plants using high concentration fissile fuels that would be dangerous for these reasons to use on Earth.

https://www.nasa.gov/pdf/203084main_...11-07%20V3.pdf

http://www.world-nuclear.org/informa...-reactors.aspx

https://ntrs.nasa.gov/archive/nasa/c...9920005899.pdf

The Phoebus 2A produced over 4 GW of power from a reactor that weighed 800 pounds and was the size of a filing cabinet. A suitcase sized nuclear reactor that produces 750 MW of power has been made for use aboard US Navy Submarines using similar techniques.

These sorts of nuclear power plants will be produced and used in abundance on Mars. This will fulfill the promise of power 'too cheap to meter'

http://spectrum.ieee.org/energy/nucl...ower-revisited

This combined with a high degree of automation -

https://www.youtube.com/watch?v=tf7IEVTDjng

that will make commodities on Mars very cheap as well!

This occurs at a time when the same materials on Earth are running out and rising dramatically in price!

So, at 40 GJ per ton a 750 MW suitcase sized power plant will produce 67.5 tons of steel an hour from 135.0 tons of hematite along with 371.3 tons of oxygen of which 2 tons is surplus and not recombined with the carbon that remains in the steel. Extracting replacement carbon also produces 10 tons of nitrogen -11 tons of argon - and with the nitrogen produces 12 tons of air per hour.

That's 297,400 cubic feet of air at 1 atmosphere. An acre flooded at this pressure to a 7 foot depth every hour. Oxygen can be made in surplus and combined with the argon as well, producing a Mars flavoured atmosphere that is only half nitrogen. This may have an adverse impact on certain bacteria that fix nitrogen. The advantage is use of the argon capture will permit doubling the rate of atmosphere production.

Four cubic feet of steel represents one ton of steel. A rod 1 foot in diameter and 5 feet long contains about a ton of steel. Projecting this off Mars at 14,000 mph at the right time and in the right direction, using a maglev track, takes it to Earth in 26 weeks.

A shallow cone 12 feet across and less than 1/2 inch thick, projected off Mars edge first in the same way - arriving at Earth pointy end first, provides a more controlled entry when arriving at Earth, and naturally uses aerobraking.

https://en.wikipedia.org/wiki/Aerosh...yingSaucer.jpg

HAARP cannon tests in the 1960s routinely blasted instrumented shells out of cannons at 5,000 gees. We can do as well today. A maglev based cannon that operates at 5,000 gees takes only 1/8th second to accelerate a disk from rest to 14,000 mph. It takes 7.36 GJ of energy to blast a one ton disk like this off Mars. So, this takes 59.16 GW of power. 79 of the 750 MW suitcase sized modules. The rate of departure is 8 tons per second. That's 480 tons per minute. To have production match that ability to depart Mars, requires 430 suitcase sized power plants operating sweepers. 430 Acres of floor space may be pressurised with the air byproduct made every hour by this operation.

8 tons of steel per second is 0.25 billion tons per year. The Earth consumes 1.55 billion tons of steel per year. So, 6 setups like this produce sufficient steel for Earth today and pressurise 35,000 square miles of floor space with air.

Now steel requirements are between 8 pounds and 16 pounds per square foot depending on the nature of the construction - on Earth. On mars it is likely to be less than that due to lower gravity and partial support from pressurised structures.

http://www.steelconstruction.info/Co...ural_steelwork

Taking 4 pounds as an average per square foot, that means that 35,000 square miles per year of floor space construction on Mars will require 2 billion tons of steel consumed locally. So, we add 540 local sweepers per launcher, to supply local construction needs while sending material for export.

Now, this assuming continuous production and consumption along with continuous projection off world. The problem with this assumption is that the speeds increase dramatically when the planets are not in precisely the right positions. So, in practice we are limited to about 4 months out of every 26 months. So, the launchers send materials to Earth only 15.4% of the time. So, 40 launchers are required instead of 6 and only 66 miners per launcher are required to supply each launcher - and only 81 local miners per launcher are required as well. The surplus of launchers also mean that materials can be sent to other planets if desired as well.

The lead time for steel arrival will be 36 weeks, so time sensitive designs, like autobodies, will likely be made on Earth. For items that are not time sensitive, like rolled steel, this can be send. However, this means that we cannot merely form and ceramic coat an aeroshell and fit it with a reusable ACU.

Now the steel, when produced by automated systems, powered by nuclear power that is too cheap to meter, will make the steel itself be too cheap to meter - ON MARS! It won't be the case OFF MARS!

For much the same reason that petrol is $0.16 per litre in Saudi Arabia, and $1.83 per litre in New Zealand steel will be cheap on Mars!

At today's prices ON EARTH ($300 per ton) the value of the steel exported to Earth in this way is worth $465 billion per year presently. Divided among 40 launcher complexes this is $11.6 billion per complex per year! With a town of 35,000 people per complex this is $332,143 per person - allocating half this to income and half this capital utilisation (4.5% discount over 40 years) over $107 billion may be spent per complex to produce 8 tons per second for export to Earth and 10 tons per second to be made for use on Mars (along with air to flood the constructed space).

Mars would have 1.4 million inhabitants under this scenario, each earning $166,000 per year on average, have Earthside investors putting $4 trillion in bets on the planet, and building pressurised floor space at 16 acres per year per person! The US stock market engages in $2.8 quadrillion in trades each year. The shortages and difficulties are well known in the industry. The presence of iron on Mars is well known!

Only a small fraction would be employed in the mining operation. Most would be free to develop a wide range of ideas in the 16 acres of pressurised space each year.

A similar analysis can be done for aluminum, and other materials. The low hanging fruit is the rare earth and other precious metals abundant on Mars.

Power is an interesting possibility. One approach would be to build nuclear power stations and send them to Earth. This is an interesting approach. A 750 MW station processes 120 tons per hour of water into 13.3 tons per hour of hydrogen. This hydrogen combines with 73.15 tons of carbon dioxide in the air per hour to produce 28 tons per hour (4700 barrels per day) of synthetic petrol made from atmospheric CO2

http://newatlas.com/audi-creates-e-d...rom-co2/37130/

Enough to support 83,190 Earthlings at US rates of petrol consumption. At $30 per barrel of crude and adding $20 per barrel for refined product (which this unit produces) each unit generates $235,000 per day in revenue. A fully automated system with 4% operating cost, financed at 4.5% discount rate over a 40 year useful life - with a $10 million cost to return the device to Mars for rebuild and reuse - we can see that up to $597.4 million may be charged for each unit! Recalling that the cost of building anything on Mars is too cheap to meter - we can count this as profit to Mars. The 4.5% discount is counted as profit for Earth based capitalists and financiers.

The export of 778 of these per year matches the income earned from the export of raw steel and doubles the income of the planet! 64.72 million people are provided with petrol (and fertiliser and plastics) at the rate used in the USA per person at stable fixed long term costs. With 7.5 billion persons on Earth a total of 90,155 of these units may be deployed to provide 423.7 million barrels per day equivalent production of oil.

If operated on the oceans, near the shore, sea water may be desalinated. The GTHTR300 nuclear plant produces 300 MWe and uses its waste heat to desalinate water with multi-stage fractionation;

http://www.world-nuclear.org/informa...alination.aspx

A 750MWe plant produced on Mars - and operating in the ocean near the shore - produces more than double the amount of fresh water using the same process - and recovers and uses salts. Brine electrolysis produces sodium hydroxide and hydrochloric process - the chloralkali process. The source of all bleaches. The production of soaps and scents - basically everything produced by Procter & Gamble - is achieved by this. So, not only does a compact power plant produce abundant petrol, but located in the ocean, produces fresh water and many other items as well.

Setting up a production plant to produce one 750 MW plant of this type per hour, and shipping it to Earth, one launch out of every 28,800 launches on one maglev track - is sufficient to send one ton to Earth - which is far less than the size of the plant! (recall the weight of the Phoebus 2a weighed 800 lbs and produced 4100 MW!) This stems from using 100% pure fissile materials instead of 'reactor grade' fissile materials that are far less efficient.

One per hour deploys sufficient reactors to transform the Earth in 11 years.. Over the same period steel consumption would increase to US per capita rates so global steel demand would rise from 1.55 billion tons per year to 3..43 billion tons per year - a 7% rise per year compounded over the period.

Now, 83,190 persons consume 125.7 MWe - about 25.7 MWe for home use - the balance - for industrial, commercial, and other uses. The 750 MWe unit can easily be modified to supply this as well. At $0.11 per kWhe this adds $121.2 million per year to the revenue stream. Adding $2.23 billion to the value of the product when that revenue is discounted at 4.5% over 40 years.

A few million people on Mars doing a few things right, will transform life on Earth for billions. They will reap rewards amounting to millions of dollars per person on Mars. A Mars colony of this type will rescue Earth's moribound markets and provide real stable long term growth in the quadrillions of dollars as billions of people begin to have the fundamental resources needed to live a good life.

JF Mezei wrote:

On 2016-12-14 06:25, Jeff Findley wrote:

People that propose this somehow assume that once you reach "earth
escape" velocity that you'll magically fall right into the sun as long
as you "escape" in the right direction.


I realize that the garbage ship will be in the sun's orbit after
escaping earth.


And yet you don't seem to.


But consider Apollo returning from the moon. It doesn't need delta-V to
land on earth because it is aimed such that its elliptical orbit scapes
earth's atmosphere which then gives it the delta-V needed to end orbit.


What? I suspect you are not saying what you mean clearly, because the
preceding makes no sense.


So I am not advocating that the garbage ship lower its circular orbit, I
am advocating that it transform its circular orbit into highly
elliptical one. And in that case, would the delta-V requirement be
significantly lower ?


That is precisely the delta-V requirement everyone is referring to. It
is an order of magnitude larger than the change required to go
anywhere else.



Please stop hand waving and DO THE MATH! Lucky for you since this is
the 21st century, I'm willing to bet that if you did a bit of Google
searching, you'd find an "orbit calculator"


My confusion:

I understand de-orbiting from ISS. You fire de-orbit engines against
your orbital speed which not only reduces orbital energy, but puts you
into elliptical orbit with perigee sufficiently inside atmpsphere to
continue to slow you down.


Essentially correct. Most of the delta-V required to hit Earth (which
you are already orbiting very close to) is provided by atmospheric
braking. Remember, ISS is so close to Earth that it must periodically
reboost due to atmospheric drag.


But when Apollo came back from the moon, it did not fire engines against
Earth orbital velocity, as a ship does to drop out of orbit from ISS, it
accelerated toward the Earth, and once past laGrange, let Earth
accelerate it. Very little delta-V was involved, it merely had to
ensure it got to the right place at right moment to scrape the
atmosphere just right to slow it down enough to kill what would be an
elliptical orbit.


Of course it did. That's how it got out of the orbit the Moon is in.
This 'straight line' stuff you keep doing in your head is WRONG.


I assume that as Apollo got accelerated towards earth, its orbital
velocity around earth accelerated (since altitude dropped). Yet, it
still managed to aim itself to scrape the atmosphere on first pass
instead of endlessly spinning in an elliptical orbit that never touches
the upper reaches of atmosphere.


Not how it worked.


So (in simple words for my simple mind), why can't a garbage truck be
sent to the sun in the same way that Apollo returned to earth ?


It can. That way just isn't what you think it is and it has a delta-V
requirement on the order of 29 kps plus whatever it takes to get into
LEO first.

Several people have told you that it doesn't work the way you think it
does and yet you still want to insist that it does.


--
"Ignorance is preferable to error, and he is less remote from the
truth who believes nothing than he who believes what is wrong."
-- Thomas Jefferson

William Mook wrote:

On Tuesday, December 13, 2016 at 12:17:14 AM UTC+13, Jeff Findley wrote:
We've got a Mook on Mook on Mook reply here (minus the first two
Mooks)...

In article ,
says...
The sands of Mars are red. That's because they're made out of
hematite. Why wouldn't you mine iron there and send it back to
Earth with a rail gun?

Because steel made on earth is already quite cheap, so it would be
economic suicide to do what you propose.

Steel has been cheap historically, but is rising inexorably as raw materials are depleted here.

https://www.bloomberg.com/news/artic...-s-rapid-shift


Pretty sure it's never going to exceed a million dollars a tonne, so
it's always going to be cheaper to do it here than to bring it back
from space. The sensible thing to do with space resources is, well,
space stuff.



Besides, one would think that a Mars colony would use such raw materials
to either build things they need on Mars or build things that are
actually worth exporting.


That's a false choice. In order to use or build things on Mars local steel is needed. So Martians would need to supply themselves with steel made from hematite on the surface and carbon in the carbon dioxide in the air - which means any surplus to their needs could be exported.


Except no one would buy it at the prices they'd have to charge.



Raw materials aren't going to cut it as an
export unless there is a return on that investment.


Correct. Prices are rising on Earth and Earth's ability to produce low cost steel will be non-existant in 64 years according to the experts. Some believe shortages may be arriving in as little at 12 years.


It's always going to be cheaper to do it here once you factor in
transport costs. It's why there will be a local steel industry on
Mars; because it costs too bloody much to bring it from Earth.



Sorry Mook,


I feel your love.


Keep your hands to yourself!



but this entire idea is b.s.


No it isn't.


Yes it is. Think about the transportation costs. Get back to me when
the price of steel on Earth exceeds a million dollars a tonne and we
can start thinking about it.



I don't know how you got to
visions of Mars colonies sending quite common raw materials like iron,
silicon, aluminum, and etc. to earth by railgun, but it's just not going
to be viable economically.


That's your problem that you don't know something. Perhaps if you listened to those who know more than you - that might help.


Great advice. One wishes YOU would take it once in a while.


More below.


Nope. Dumping the Magic Mookie Multiplication Math.

Massive MookSpew Munched


--
"Some people get lost in thought because it's such unfamiliar
territory."
--G. Behn

On Tuesday, December 13, 2016 at 7:29:48 AM UTC+13, Fred J. McCall wrote:
JF Mezei wrote:

Possible Mars exports:

- Cans filled with martian air. "Fresh CO2 from Mars".
- Bottled water

AKA: novelty items.

Heck, if Coke/Pepsi can market tap water as upscale water, surely
someone will market water from mars glaciers as something that is very
desirable.

It would make for an intreresting debate on whether humans should strip
materials from one planet to bring back to earth.


It just doesn't pay.


Can we steal a few rocks ? Extract a few kilos of unobtainium ? Ship
tonnes and tonnes of water back to earth ? Steal a planet's atmosphere
with giant "mega maid" to bring back to replenish earth's atmosphere ?


Uh, what needs 'replenished'?


For that matter, could be take over Mars and use it as a giant trash can
to dispose of all of earth's garbage and hazardous materials, spent
uranium ?


The Moon is better suited to that sort of thing, but it's still
hideously expensive trash.


--
"Some people get lost in thought because it's such unfamiliar
territory."
--G. Behn

http://icelandicglacial.com

https://www.amazon.com/icelandicglacial

$40 USD plus shipping for 12 bottles each 1.5 litres. plus shipping

That's $2.22 per litre (kg) plus shipping.

Olfus Spring - the source of Iceland Glacial Water produces 900,000,000 litres of fresh water per day. Icelandic Glacial Water collects only 0.1% of this 900,000 per day - that has a street value of $2 million per day. This supports a global industry worth $8.2 billion - earning $730 million per year.

http://icelandicglacial.com/pages/the-source

Similar Water supplies exists on Mars

http://astronomy.com/news/2015/04/ma...f-frozen-water

https://en.wikipedia.org/wiki/Glaciers_on_Mars

Setting down on the right spot it takes 334 MJ to melt and reprocess 1 ton of ice. That's 1000 litres - worth $2,220 if Icelandic water is any measure. Melting the ice purifying the liquid and forming it into a spherical ball of pure ice and putting it into a 2 steel container with 12 foot aeroshell that is coated with ceramic TPS and then the entire apparatus is levitated by a maglev track and accelerated at 5000 gees edgewise - it attains 14,000 mph at Mars' surface which means that if its directed in the right direction at the right time and the right speed - it will slam into Earth's atmosphere and descend to the surface landing softly there - to be drained and the steel container to be sold as well. So, the water is worth $2,220 - the 200 pounds of steel worth $30 - $2,250 per disk. Far more than the $300 ton of steel disk alone would be.

Now, it takes 1/8th of a second to project an object to 14000 mph when accelerating at 5000 gees over a quarter mile length - this is 8 tons per second - which requires 56 GW (I calculated this earlier in my steel plant calculation) 900 shipments a day is only one launch every 96 seconds. Reducing power consumption to 73 MWe. Shipping over a three month period every 26 months - the synodic period - means one launch every 14 seconds. This reduces power consumption to 500 MWe.

The Phoebus 2A produced 4100 MW thermal in a package that weighed only 800 pounds. A suitcase sized nuclear power plant using these techniques is possible that produces 750 MW and weighs only 146 pounds.

The important detail is - one of these disks may be substituted for a solid steel disk their value is 7.5x as much as a one ton disk at these prices.

A mission to Mars may be undertaken to take this glacier

and melt it - and send it back to Earth

https://upload.wikimedia.org/wikiped...mage_field.JPG

Offering a 50% discount from retail - using the Icelandic Glacier as a price point - means that $4 billion might be spent to establish a base that does this. A small group of 100 people would make use of carbon dioxide in Mars' air along with water.

If we wanted to bottle the water on Mar, and assure its authenticity - we can make PET plastic from Martian air.

4 H2O + electricity --- 4 H2 + 2 O2
CO2 + 4 H2 --- CH4 + 2 H2O
CH4 + energy -- CH2* + H2

Which can be polymerised into polyethylene terepthalate (PET) and made into a variety of things - including water bottles (so the bottle itself is made from Martian Air!)

https://en.wikipedia.org/wiki/Methylene_(compound)
https://en.wikipedia.org/wiki/Polyet..._terephthalate

The steel is made from hematite that occurs freely on the martian surface and from calcium oxide that also occurs freely there. The nuclear power source uses some water to make hydrogen and uses that hydrogen to reduce carbon dioxide to methane, but then uses pyrolysis to reduce the methane to elemental carbon recovering the hydrogen. 45 packages of 12 count 1.5 litre bottles can be projected per ton - with a 200 pound mass budget for the shipping drone.

At this early stage, the Martian steel from the re-entry vehicle, wouldn't be sold as bulk steel, but broken down and sold as jewelry. Steel jewlery bearing important symbols sell from between $8 and $80 and mass 4 ounces - that's $32 to $320 per pound - $6,400 to $64,000 for 200 pounds made from the break up of the re-entry vehicle. It might be designed to efficiently break apart into forms that are immediately recognisable and promoted. Using the lower price, and assuming only half the vehicle may be made in this way, and assuming only half the money is retrived at the wholesale level, this adds another $1,600 per shipment for the Martian colony.

This makes a $10 billion investment a real possiblity of payback - raising it from the $4 billion value estimated previously.

Of course the colonists accumulate significant portion of this total, even after paying back their backers, and use the ability to make steel, PET and water, to create habitats and farms, factories and forests.







The elemental carbon is combined with the hematite and calcium oxide and oxygen to produce steel. The steel is used to make things locally, and to house the

the cost of steel has dropped a lot.

5 years ago i junked a old dodge caran at the scrap yard. got 450 bucks for it

these days a junked caravan is worth under 120 bucks.

so little the tow truck drivers have quit picking up junk vehicles. the costs to pick it up leave no profit

William Mook wrote:

On Tuesday, December 13, 2016 at 7:29:48 AM UTC+13, Fred J. McCall wrote:
JF Mezei wrote:

Possible Mars exports:

- Cans filled with martian air. "Fresh CO2 from Mars".
- Bottled water

AKA: novelty items.

Heck, if Coke/Pepsi can market tap water as upscale water, surely
someone will market water from mars glaciers as something that is very
desirable.

It would make for an intreresting debate on whether humans should strip
materials from one planet to bring back to earth.


It just doesn't pay.


Can we steal a few rocks ? Extract a few kilos of unobtainium ? Ship
tonnes and tonnes of water back to earth ? Steal a planet's atmosphere
with giant "mega maid" to bring back to replenish earth's atmosphere ?


Uh, what needs 'replenished'?


For that matter, could be take over Mars and use it as a giant trash can
to dispose of all of earth's garbage and hazardous materials, spent
uranium ?


The Moon is better suited to that sort of thing, but it's still
hideously expensive trash.


http://icelandicglacial.com

https://www.amazon.com/icelandicglacial

$40 USD plus shipping for 12 bottles each 1.5 litres. plus shipping

That's $2.22 per litre (kg) plus shipping.


Yes, but when you start talking Martian water that 'plus shipping'
around a million dollars per tonne or $1,000 per litre. Now who do
you think will pay that?

Magic Mookie Math and Machines Munched


--
"Ordinarily he is insane. But he has lucid moments when he is
only stupid."
-- Heinrich Heine

JF Mezei wrote:

On 2016-12-14 23:08, Fred J. McCall wrote:

Not how it worked.

So why have all the NASA documents showed a more or less straight line
(or in slight S shape) transit between moon and earth ?


Because they're STATIC ILLUSTRATIONS, you dip****.


In the case of Apollo 13, they did a slingshot around the moon which put
them into a free return trajectory, after which they only had to fire
engines once to adjust trajectory/speed to meet Earth at the right
moment/place.


Gravity slingshots are still ORBITS, you tiny ****.


Pardon my confusion if that didn't happen and they came back from the
moon the same way a ship de-orbits from ISS.


You're not confused. You are stupid and adamantly ignorant.

Go learn something about the subject and get back to me. Of course,
if you learn something about the subject, you won't have to.


--
"Ignorance is preferable to error, and he is less remote from the
truth who believes nothing than he who believes what is wrong."
-- Thomas Jefferson

In article . com,
says...

On 2016-12-14 06:25, Jeff Findley wrote:

People that propose this somehow assume that once you reach "earth
escape" velocity that you'll magically fall right into the sun as long
as you "escape" in the right direction.

I realize that the garbage ship will be in the sun's orbit after
escaping earth.

But consider Apollo returning from the moon. It doesn't need delta-V to
land on earth because it is aimed such that its elliptical orbit scapes
earth's atmosphere which then gives it the delta-V needed to end orbit.

So I am not advocating that the garbage ship lower its circular orbit, I
am advocating that it transform its circular orbit into highly
elliptical one. And in that case, would the delta-V requirement be
significantly lower ?

I know what you're proposing. I took the 500 level Orbital Mechanics
class in college. This whole "waste disposal in the sun" idea might
have even been a class exercise just to show how ludicrous it is since
it's not a difficult calculation.

You need to RUN THE NUMBERS. When you do, you'll see that the delta-V
is so high that this idea is just not economically viable.

Please stop hand waving and DO THE MATH! Lucky for you since this is
the 21st century, I'm willing to bet that if you did a bit of Google
searching, you'd find an "orbit calculator"

My confusion:

I understand de-orbiting from ISS. You fire de-orbit engines against
your orbital speed which not only reduces orbital energy, but puts you
into elliptical orbit with perigee sufficiently inside atmpsphere to
continue to slow you down.

But when Apollo came back from the moon, it did not fire engines against
Earth orbital velocity, as a ship does to drop out of orbit from ISS, it
accelerated toward the Earth, and once past laGrange, let Earth
accelerate it. Very little delta-V was involved, it merely had to
ensure it got to the right place at right moment to scrape the
atmosphere just right to slow it down enough to kill what would be an
elliptical orbit.

I assume that as Apollo got accelerated towards earth, its orbital
velocity around earth accelerated (since altitude dropped). Yet, it
still managed to aim itself to scrape the atmosphere on first pass
instead of endlessly spinning in an elliptical orbit that never touches
the upper reaches of atmosphere.

So (in simple words for my simple mind), why can't a garbage truck be
sent to the sun in the same way that Apollo returned to earth ?

Calculate the delta-V needed to change from earth's orbit to an orbit
which grazes the sun. Now, compare that to the delta-V needed by the
Apollo CSM to go from lunar orbit to its earth return trajectory.

No one is saying you can't do this. We're saying it's hideously
expensive to do this due to the huge delta-V needed. All the hand
waving in the world won't change the laws of physics, so stop waving
your damn hands.

Jeff
--
All opinions posted by me on Usenet News are mine, and mine alone.
These posts do not reflect the opinions of my family, friends,
employer, or any organization that I am a member of.

In article ,
says...

In article . com,
says...

On 2016-12-14 06:25, Jeff Findley wrote:

People that propose this somehow assume that once you reach "earth
escape" velocity that you'll magically fall right into the sun as long
as you "escape" in the right direction.

I realize that the garbage ship will be in the sun's orbit after
escaping earth.

But consider Apollo returning from the moon. It doesn't need delta-V to
land on earth because it is aimed such that its elliptical orbit scapes
earth's atmosphere which then gives it the delta-V needed to end orbit.

So I am not advocating that the garbage ship lower its circular orbit, I
am advocating that it transform its circular orbit into highly
elliptical one. And in that case, would the delta-V requirement be
significantly lower ?

I know what you're proposing. I took the 500 level Orbital Mechanics
class in college. This whole "waste disposal in the sun" idea might
have even been a class exercise just to show how ludicrous it is since
it's not a difficult calculation.

You need to RUN THE NUMBERS. When you do, you'll see that the delta-V
is so high that this idea is just not economically viable.

Since you're so lazy as to not even attempt a Google search, here is a
link to a "delta-V map of the solar system". Sure, there are lots of
simplifying assumptions behind the numbers, but it's a good starting
point for someone when they really have no clue how much it costs to go,
well anywhere, in the solar system.

http://i.imgur.com/SqdzxzF.png

The way this works is you add up the numbers along the path from "Moon"
to "Low (earth) Orbit" to get the approximate delta-V the Apollo CSM
used to get from lunar orbit back to earth.

Now, add up all the numbers along the path from "Earth" to the "Sun".

See the huge difference between the numbers? Well, there's your problem
with "sun disposal"!

Jeff
--
All opinions posted by me on Usenet News are mine, and mine alone.
These posts do not reflect the opinions of my family, friends,
employer, or any organization that I am a member of.

On Tuesday, December 13, 2016 at 9:37:17 AM UTC+13, JF Mezei wrote:
On 2016-12-12 13:29, Fred J. McCall wrote:

The Moon is better suited to that sort of thing, but it's still
hideously expensive trash.


Water is far more abundant on Mars than the Moon, but for 900,000 litres per day, we can likely find a place on the Moon to mine for water - and send part of it to Earth for consumption there. The problem with the moon is we don't have iron or carbon dioxide readily available to make return capsules or water bottles from. We do on Mars.



It is hideously expensive to launch spent radioactive garbage and have
it crash onto the moon (there is no need to land, is there ?) compared
to all the regulatory red tape and long term costs of maintaining
uranium dump site on earth ?

That depends on the details. Using a self-supporting hyperloop



Different slant: the sun is said to be a big fusion reactor. If one
were to send a tonne of uranium to the sun, would it remain as uranium
(either molten or vapour) or would the extreme conditions cause any type
of atom to break apart and form hydrogen ?

Your understanding of nuclear vs chemical reactions is lacking. This is not unusual in today's world. You can correct this ignorance by reviewing the following

http://www.differencebetween.net/sci...ical-reaction/

So, the uranium won't change as you imagined, also, uranium already exists in the Sun,

http://adsabs.harvard.edu/full/1969SoPh....6..381G

and our ability to add uranium from Earth to the sun in any quantity is limited. The sun wouldn't notice at all! Also, nuclear waste doesn't consist of uranium anyway. Typically, some 44 million kilowatt-hours of electricity are produced from one tonne of natural uranium. Natural Uranium is as common as tin. Only 0.7% of natural uranium is the type that reacts promptly U235. U238 is the heavy isotope and removed from the equation - though when exposed to fission of U235 it can convert into Plutonium.

In a reactor fuel rod 98.8% of the Uranium is burned up. Other radioactive by products remain. Uranium 238 isotope, Thorium, Plutonium, are all by products, but they can be removed and reprocessed into fuels for different reactor types.

Instead of a once through cycle we can imagine recycling the fuel elements extracting the short lived elements (and even using them in new reactor designs that make use of their very high specific energies) - this was all imagined in the early days of nuclear but have been forgotten!

http://www.world-nuclear.org/informa...anagement.aspx

http://fissionenvironmentalists.files.wordpress.com

This fuel cycle might be completed off world. On the moon or on Mars.

Each fission event releases around 200 MeV. 181 MeV is prompt energy. A mole of Uranium 235 weighs 235 grams. It also contains 6.03*10^23 atoms. All the atoms in a mole of Uranium 235 fissioning produce primarily Barium and Krypton, and release 17.45 trillion joules of energy. That's 4.85 million kWh per mole. That's 20.64 million kWh per kg. 20.64 billion kWh per metric ton!

20,635,548 kWh/kg - Pure Uranium 235 metal
20,635,548,128 kWh/t - Pure Uranium 235 metal
148,575,947 kWh/t - 0.72% Concentration in ore
47,544,303 kWh/t - 32% conversion efficiency to electricity

To get reactor weight, of a reactor that uses 100% pure uranium metal we just multiply the second number above by 38% - which is the efficiency of high temperature reactors that are compact and don't require powering huge facilities to work appropriately

7,841,508,288 kWh/t - Pure Uranium 235 metal - 38% conversion efficiency

This is sufficient to power 750 MW power plant for 1.19 years. A gas core fission reactor using MHD for a portion of its energy conversion can produce 70.1% of its heat energy to electricity. So, a ton of uranium metal may be converted to 14.47 billion kWh of electrical energy - so a reactor such as this could operate for 2.2 years continuously - which is helpful because this is nearly the synodic period of Earth/Mars.

https://en.wikipedia.org/wiki/Gaseous_fission_reactor

14,465,519,237 - kWh/t - Pure Uranium 235 metal - 70.1% conversion efficiency

So, 91,000 of these reactors launch over a few days period after being constructed - and they arrive on Earth - operate for a period of time - and then fly back to Mars for refueling.

A one ton gas core reactor that has a cryogenic storage tank for hydrogen - is equipped with a propulsive skin which is powered by the gas core reactor. The exhaust speed is fixed at 15.3 km/sec - and at 750 MWe the propulsive skin produces 10 metric tons of thrust. To fly to Earth from the surface of Mars requires that the drone power plant chemical processing station attain a speed of 6.1 km/sec relative to the surface of Mars. This requires a propellant fraction of 0.329 - so a 1.5 metric ton (3,300 pound) system requires 735 kg of liquid hydrogen propellant. This requires 122.7 GJ of energy (34,083 kWh). A small fraction of the 14.47 billion kWh available. The spacecraft lands on Earth and operates continuously for a period of 2.15 years - and returns to Mars - being replaced by a fresh replacement. The delta vee required to leave Earth's surface for Mars is 12.9 km/sec. With a 15.3 km/sec exhaust speed a propellant fraction of 0.57 of the take off weight is required. This requires 1,986 kg of liqiud hydrogen. A sphere 3.8 meters (12.5 ft) in diameter. Enough to store 6.3 minutes of full scale production of hydrogen. This requires 232.5 GJ (64,570 kWh) of energy.

An automated 'fog' of micro-scale robots maintain these devices and provide support for their operation as well as maintenance and even construction.

Here's an array of tiny robots that clean your home -
https://www.youtube.com/watch?v=gs2OVwT_R4Q

It takes a lot of energy to dispatch an object to the Sun, relative to the energy it takes to send an item to the Moon or Mars. Using a hyperloop - that extends above the atmosphere - that uses magnetic or electrical acceleration to project objects off world - we require 10.9 km/sec (24,300 mph) to reach the moon, 12 km/sec (26,600 mph) to reach mars, and 31.8 km/sec (71,115 mph) to reach the Sun, from the surface of the Earth. That's because the moon bound craft doesn't have to reach escape velocity to reach the moon, the Mars bound craft only has to achieve 4 km/sec (8,945 mph) excess speed beyond escape velocity (energies add, not velocities) but the Sun bound payload has to cancel ALL the Earth's 66,557 mph orbital speed to fall reliably into the Sun.

Now there are ways to nuance that. One is to launch to Jupiter, and use Jupiter's sling shot to drop the payload into the Sun. This requires far less energy than the direct approach.