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

In article ,
says...
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.

Make that "Moon low orbit" to "Earth low orbit" for that burn. The LEM
ascent stage already did the burn to go from "Moon" to "Moon low orbit".

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

This number will still be huge no matter what. You can add in "fly-
bys" to add in some "gravity assist" burns to lower that number a bit,
but not by enough to make the idea workable.

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.

William Mook wrote:

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.


Nobody but you is talking about shipping water back, Mook. Water is
too valuable where it is to ship it back to Earth, which has stupid
amounts of fresh water. Then you add in the shipping costs and it's a
REALLY dumb idea.



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


What the **** does an EARTH transportation system have to do with
putting things on the Moon?

snip MookSpew


--
"The reasonable man adapts himself to the world; the unreasonable
man persists in trying to adapt the world to himself. Therefore,
all progress depends on the unreasonable man."
--George Bernard Shaw

JF Mezei wrote:

On 2016-12-15 06:05, Jeff Findley wrote:

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.


Not asking to wave the laws of physics. Asking to understand in simple
words how returning from the moon is the same as de-orbiting from ISS
(aka: decelerate orbital speed around earth to drop altitude).


True. You'd have to *understand* the laws of physics before you could
ask to have them waived. The difference is that ISS doesn't have a
significant gravitational field and the Moon does. So from ISS you
just need to decelerate tangential to the orbital velocity to lower
your orbit, while from the Moon you first need to accelerate with
regard to your orbit around the Moon to raise your lunar orbit
velocity until Earth's gravity is stronger on you than the Moon's.


Here is a different question: When Apollo 13 ended its slingshot around
the moon, what direction was it going ? towards earth ? or against
orbital speed of moon going around earth to lower its orbital speed
around earth ?

Or did it aim "diagonal" to essentially follow a curve ball trajectory
such that it initially aims "behind" the Earth, knowing that its orbital
speed will accelerate as it drops altitude so would catch up with Earth ?


What you want is essentially what you want when you leave from lunar
orbit. You want to get into an orbit such that your aposelene is in a
region where Earth's gravity has more influence on you than the Moon's
gravity does. With regard to Earth, you want your velocity tangential
to Earth at apogee to be slow enough that you 'fall' out of orbit.
This is why the lunar return path looks sort of 's-shaped'.


When Apollo 13 did a burn midway in its return, what direction was it
burning towards ? Earth ? or again, against its orbital velocity around
earth ?


The latter (or actually probably a combination of the two, but the
main aim is to lower orbital velocity so that you 'fall').


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

On 16-12-15 13:15 , Jeff Findley wrote:
In article ,
says...

In article . com,
says...
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"!

That map seems not directly relevant to garbage disposal in the Sun. It
shows something like 600 km/s to reach the Sun, but I'm pretty sure that
if we get the garbage off Earth (11 km/s) and stop its circum-solar
orbital velocity (30 km/s more), it will fall into the Sun.

The 600-odd km/s in the map is probably for *landing* softly on the Sun
-- perhaps "landing" isn't the right word, so let's say "reaching a
point at the Sun's visible photosphere, at rest with respect to the
photosphere".

Of course 41 km/s is still a rather large delta-v.

--
Niklas Holsti
Tidorum Ltd
niklas holsti tidorum fi
. @ .

In sci.space.policy message -
september.org, Wed, 14 Dec 2016 06:25:54, Jeff Findley
posted:


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" suitable for calculating the
delta-V needed to go from earth's orbit to an elliptical orbit very
close to the sun. To that value, don't forget to add in the delta-V to
get from the earth's surface to earth escape velocity.



Not needed. The direct answer is obviously fairly close to the sum of
Earth escape speed plus the speed of the earth in its orbit, which is
(E&OE)
150e6 * 2 * 22/7 / (365.25 * 86400) km/s
about 30 km/s
which is quite a lot.

On the other hand, consider Ulysses, which went moderately close to the
Sun (from Jupiter's point of view) after a Jupiter gravity assist,
closest approach 6.3 RJ; I guess that a moderately closer approach could
have aimed Ulysses directly at Sol.

I see that Ulysses will again meet Jupiter in 2098 and be slung far
outwards.

--
(c) John Stockton, Surrey, UK. Turnpike v6.05 MIME.
Merlyn Web Site - FAQish topics, acronyms, & links.

On Dec/15/2016 à 6:41 PM, Dr J R Stockton wrote :
In sci.space.policy message -
september.org, Wed, 14 Dec 2016 06:25:54, Jeff Findley
posted:


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" suitable for calculating the
delta-V needed to go from earth's orbit to an elliptical orbit very
close to the sun. To that value, don't forget to add in the delta-V to
get from the earth's surface to earth escape velocity.



Not needed. The direct answer is obviously fairly close to the sum of
Earth escape speed plus the speed of the earth in its orbit, which is
(E&OE)
150e6 * 2 * 22/7 / (365.25 * 86400) km/s
about 30 km/s
which is quite a lot.

No, not the sum of Earth escape speed plus speed of Earth in its orbit.
If you do all the acceleration near Earth, it is the square root of the
sum of the squares of those two speeds. Which is about
(30^2 + 11^2)^(1/2) km/s or about 32 km/s. That is significantly less
than the sum, about 41 km/s.

On the other hand, consider Ulysses, which went moderately close to the
Sun (from Jupiter's point of view) after a Jupiter gravity assist,
closest approach 6.3 RJ; I guess that a moderately closer approach could
have aimed Ulysses directly at Sol.

I see that Ulysses will again meet Jupiter in 2098 and be slung far
outwards.

Yes, that would be the way to do it. In fact, Mr Mezei isn't all that
far away from the truth if you are willing to do multiple gravity
assist. Once you have just barely escaped Earth orbit, you are in a
solar orbit close to Earth orbit. You will eventually come close to
Earth again. If you do so just right, you get a gravity assist from
Earth (and the Moon). That gravity assist can set you up for another
gravity assist from Earth a few years later etc. If you are willing
to go through multiple gravity assist like that, and you have infinite
precision in your trajectory, you can reach the Sun with as little as
Earth Escape velocity plus epsilon, for any positive value of epsilon.
You must also accept travel times of a century or two (if you are lucky
and the planets are in favourable positions, it could be only decades).
So if you just barely escape Earth gravitational field, just exactly at
the right speed and direction, you can end up into the Sun. That is not
practical, but theoretically possible.


Alain Fournier

"Jeff Findley" wrote in message
...
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


These are very handy when answering questions like this for others.

I'll be honest though, even though I knew "launch into the Sun" made no
sense, I hadn't actually bothered to add that all up. It's even higher than
I had imagined!

Pretty impressive when you think about it.

JF Mezei, let me also suggest looking at the effort to get Messenger into
orbit around Mercury.


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

--
Greg D. Moore http://greenmountainsoftware.wordpress.com/
CEO QuiCR: Quick, Crowdsourced Responses. http://www.quicr.net

"JF Mezei" wrote in message
web.com...

On 2016-12-15 06:15, Jeff Findley wrote:

http://i.imgur.com/SqdzxzF.png
See the huge difference between the numbers? Well, there's your problem
with "sun disposal"!

These numbers are for circular orbit around sun. Notice no "red arrow"
to indicate one could use the sun to slow down to orbital speed of 0.


The line you want to look at is the last one: 440. Got that.. 440.
To get from Earth to the Moon takes:
9.4+2.44+0.68+0.14+0.68+1.73.

That's a grant total of 15.

The LAST step of intercepting the Sun is almost 30x that.
And that's for the last step.

So let's say you can do some form of "Coronasphere" braking (which is what I
think you're really suggesting, sorta like aerobraking into the Earth).

Let's say you can somehow get it down to 400 instead of 440. You're 27x
what it takes to get to the surface of the Moon.

And that's not including the 9.4+2.44+0.68+0.09+0.28+2.06+15.74+178, or
208.69 to get to the point where it's only going to cost you 440 (or 400).

In other words, it is 43x harder to get to the surface of the Sun than it
does to get to the surface of the Moon.


Would it be correct to state that orbital energy of a circular orbit of
X altitude could be equal to that of an elliptical orbit where perigee
is much lower than x, and apogee is much higher than x ?



--
Greg D. Moore http://greenmountainsoftware.wordpress.com/
CEO QuiCR: Quick, Crowdsourced Responses. http://www.quicr.net

On Friday, December 16, 2016 at 9:42:54 AM UTC+13, Fred J. McCall wrote:
William Mook wrote:

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.


Nobody but you is talking about shipping water back, Mook. Water is
too valuable where it is to ship it back to Earth, which has stupid
amounts of fresh water. Then you add in the shipping costs and it's a
REALLY dumb idea.

Someone asked the question, I analysed it.

What someone thinks is dumb or not doesn't determine the profitability of a business.

Would people buy water bottled on Mars at some price? Yes. How do we know that? People ship water from Iceland to California and pay $2.22 per litre for the privilege. 900,000 litres per day is being shipped from an Icelandic glacier to all points around the world at this price - and a rather large business exists.

As mentioned, I am responding to another party's observation that water could be shipped from Mars. I agree. It might even be a subsidiary to the Icelandic water company. It would pay dividends and support a small base of operators near this glacier.

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

Iceland Glacier water is cheap in comparison to these brands;

http://www.therichest.com/expensive-...-in-the-world/


At $40 per 750 ml bottle, and the bottle is manufactured on Mars as well! It would be quite a deal. A ton of water would generate $48,485 - not $2,200.

At those prices we could dispense with the big maglev launcher and use the ability to make steel on Mars from Martian hematite, to build a rocket drone with a 3D printer - fuel it with LOX/LNG made from Mars' atmosphere and local water...

4 H2O + energy --- 4 H2 + 2 O2
CO2 + 4 H2 --- CH4 + 2 H2O

So 2 O2 is 64 amu whilst CH4 is 16 amu - a mass ratio of 4 to 1. An idealised LOX/LNG engine operating in near vacuum has an Isp of 368.9 a mix ratio of 3.45 a propellant density of 830 kg/m3.

So, for each metric ton of propellant for launch we process 1,011.2 kg of water and 618.0 kg of carbon dioxide into 224.7 kg of methane, 775.3 kg of LOX for use on board the rocket, 123.6 kg of LOX for use on the colony.

With a 6% structure fraction and a 6.1 km/sec final velocity, and a 368.9 sec Isp we can calculate that to send a 1000 litres to Earth requires a drone take of weight of 7,985.6 kg. With 319.4 kg structure, 1,462.1 kg methane made on site, 5,044.4 kg of LOX made on site, from 6,579.6 litres of water - consuming a grand total of 7,579.6 litres of water when you include the bottles of water anyway.

A tank holding this much propellant consists of a sphere 2.465 meters in diameter of combined LOX/Methane mix - MOX monopropellant rocket.

An automated system that landed on Mars, and did one system a week - knowing that it takes 40 MJ to process Mars' soil into 1 kg steel - and it takes 15.9 MJ to process one kg of propellant. 12.8 GJ for the rocket, 127.2 GJ for the propellant and water. 140 GJ per launch. One drone per week (all launched during Synodic alignment for a Hohmann transfer orbit) That's a 231.48 kW average power unit. 452.95 kW peak output under Mars lighting conditions, it covers 1,055 square meters. A 26 meter diameter inflatable concentrator that focuses on to a 300 mm diameter wafer that is 68% efficient converting sunlight to power.

http://www.dlr.de/Portaldata/55/Reso...5-0212prop.pdf

https://www.youtube.com/watch?v=a3j4y6uJQmg
https://www.youtube.com/watch?v=gs2OVwT_R4Q

The same systems that Electrolux uses to clean your home is modified to find particles of hematite and water - bring it back to the 3D printer that produces a rocket and fuels it up. The rocket has its own landing gear and can flex that gear to move away from the construction stage to a storage point where it waits until the right time to return to Earth.

It takes 779.94 days for Mars to align properly with Earth so that the rockets can launch. That's 111 weeks - so the 111 rockets built during the 111 weeks between synodic periods - launched over the 2.94 day period sees a launch every 38 minutes 8.4 seconds - before the cycle starts again.

At a value of $40 per 750 ml bottle this is $53.33 per litre. That's $5.92 million per synodic period.

Acqua di Cristallo $60,000 $8,878.50
Kona Nigari------ $ 402 $ 59.49
Fillico------------ $ 219 $ 32.41
Bling------------- $ 40 $ 5.92
Veen------------- $ 23 $ 3.40

According to Reuters the USA spends $40 billion on space.

USA $40.0 billion
China $11.0 billion
Versace - Mars Water - $8.8 billion ====per synodic period
Russia $8.6 bilion
India $4.3 billion

A Reusable Falcon Heavy costs 30% less than an expendable that's $63 million instead of $90 million. The rocket can put 54,400 kg into LEO. Deploying the large solar collector that is to be used on Mars, at 1 AU produces 1 MW continuously!

Okay, so using an electro-spray ion rocket array - MEMS propulsive surface that - that projects inert propellant at a speed of 15.6 km/sec produces 128.2 Newtons of thrust. It has 13,110 kg of propellant and masses 41,290 kg - arriving at Mars. It accelerates 8.48 m/sec per hour.

It takes 566.6 hours - 23.61 days to boost to Trans Mars Injection. When it arrives at Mars it folds its array away after the final course correction, and begins its descent from Mars orbit. It slows and then uses MEMS bipropellant rocket array to touch down at the Glacier edge. The inflatable solar concentrator re-deploys on the surface, and the micro robot array begins operation.

This would be a piloted system.

A totally drone system - could be 1/12th the size - and each one would have a solar array as large as the one described here. So it has 12x the power - so it boosts in 48 hours instead of 24 days. A dozen drones sent instead of one drone increases chances of success - and production rate. 111 kiloliters times 11 drones - with one support drone in transit - Is 1221 kilolitres per synodic period -at $60,000 per bottle this is $97.68 billion - At these prices, each bottle is equipped with a drone to deliver the bottle directly to a designated locations that is beamed to the system on Mars. People who buy can access the water factory and track their production. The system launches 1000 litres per drone rocket, but each 750 ml bottle has built around a drone. So, like a MIRV each drone enters the Earth's atmosphere, slows to subsonic speeds in the stratosphere, and the drone delivers the package directly to the designated location.

http://www.usatoday.com/story/tech/n...rone/95401366/

Communicating with the buyer via bluetooth and wifi directly to the cell phone - and through the internet.

That way the buyers can confirm the entry of the drone capsule into the Earth's upper atmosphere, and watch it as each bottle flies directly to its client.

* * *

http://pubs.acs.org/doi/abs/10.1021/...nalCode=iecred

We can synthesis other things on Mars - and send them to Earth at great price. Alcohol comes to mind. Tobacco as well.

Check it out;

http://www.therichest.com/luxury/mos...-in-the-world/

Johnny Walker Diamond Jubilee at $165,000 - is more expensive than the water. An ice cube made on Mars, mixed with some Martian Water and good Martian Scotch - would cost as much as a small used car! I can see this being quite popular.

$90.0 billion per year is spent on alcohol in the USA - that's more than double what is spent on space.




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


What the **** does an EARTH transportation system have to do with
putting things on the Moon?

snip MookSpew

Some people like spending money on the most expensive things. They feel its exclusivity confers value.



--
"The reasonable man adapts himself to the world; the unreasonable
man persists in trying to adapt the world to himself. Therefore,
all progress depends on the unreasonable man."
--George Bernard Shaw

In article om,
says...

On 2016-12-15 06:05, Jeff Findley wrote:

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.


Not asking to wave the laws of physics. Asking to understand in simple
words how returning from the moon is the same as de-orbiting from ISS
(aka: decelerate orbital speed around earth to drop altitude).

Because absent leaving the solar system completely, you're *always* in
an orbit around something. Leave lunar orbit with the smallest burn
possible puts you in an earth orbit similar to the moon's orbit. Leave
earth orbit with the smallest burn possible puts you in a solar orbit
similar to the earth's orbit.

The really tricky bits are when your orbit is being influenced heavily
by more than one object (like "halo orbits" around Lagrange points).

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.