News on China's scientific and technological development.

Quickie · Sep 27, 2018

Anlsvrthng said:
I have to highlight this :

1 99.99 2018 AV2 3.758 km/sec

Ok, so same training :
Low Earth Orbit means 300 km orbit above the earth surface.
The orbital speed is 7.8 km/sec.
The above numbers showing how much you have to accelerate to reach the target from here.

So, is you want to reach the 2018 AV2 then you need to accelerate to 7.8 + 3.758=11.558 km/sec.
The escape velocity from earth is 11.186 km/sec.
.........

..........

But apart from hydrogen nothing else worth any effort, and actually there is NO data about how much ice is there, and in what form.

I think there is a planed Chinese probe as well to see the poles.

Maybe it is ex tractable.
------

It is very easy to calculate the trust of the ion engine.
You know for what speed it accelerate the mass ( 30 km/sec) , you have the kg/sec mass flow ( 0.00003 kg/sec) from the time needed to exhaust 1kg of xenon.
If you multiple the two you get the trust.....
F=m*a
It is 1 N.

Not this. You were saying "The rocket equitation doesn't need the trust, but it can be calculated with the help of it."

My point was that the rocket engine thrust can't be known from "the rocket equitation (equation)" .

Low Earth Orbit means 300 km orbit above the earth surface.
The orbital speed is 7.8 km/sec.
The above numbers showing how much you have to accelerate to reach the target from here.

So, is you want to reach the 2018 AV2 then you need to accelerate to 7.8 + 3.758=11.558 km/sec.
The escape velocity from earth is 11.186 km/sec

The 11.558 km/sec orbital speed would be way past the escape velocity, in the very beginning, at lower orbit!

I know what you are getting at, but it seems you are trying hard to recall what you've learned in high school/college physics.

This is kind of out of topics and detracting, so better stop.

Anlsvrthng · Sep 27, 2018

Quickie said:
Not this. You were saying "The rocket equitation doesn't need the trust, but it can be calculated with the help of it."

My point was that the rocket engine thrust can't be known from "the rocket equitation (equation)" .

The 11.558 km/sec orbital speed would be way past the escape velocity, in the very beginning, at lower orbit!

I know what you are getting at, but it seems you are trying hard to recall what you've learned in high school/college physics.

This is kind of out of topics and detracting, so better stop.

One second mass into the rocket equitation - calculate the acceleration for that.
It gives the m/sec^2 acceleration , from here it will be again F=m*a, ship acceleration multiplied with the ship mass.
It will gives an error as well, proportional to the choose delta V.
It is the other side of the calculation.

Quickie said:
The 11.558 km/sec orbital speed would be way past the escape velocity, in the very beginning, at lower orbit!
.

The escape velocity from earth is 11.186 km/sec.
It means that if you came back FROM the NEA asteroid to the L4/5 then it takes 11.558-11.186=0.372 km/sec .
You made mistake (or one mistake ) when you tried to use the LEO-NEA dV to calculate the NEA-L4/5 dV.
That was the reason of your 5 km/sec dV calculation. Just a reminder.

Anyway, the real speed to came back to L4/5 from NEA less than that , with proper timing and orbit.

It is way less than the moon landing /orbit /transfer orbit dV requirements.
IF you take off from the moon it takes 2.5 km/sec. It is way more than 0.372 km/sec for earth capture ( again, the L4/5 is less than that)
It is not possible to use ion engine for that ,

So, if we go back to the basic : the moon mining does't make sense. There can be same hydrogen , but there can be hydrogen on the NEAs as well. Actual the mankind has practically 0 knowledge about the NEAs volatile contents, because those are not survive the atmospheric entry in the meteorites.
If any country wants to do serious space mining then they have to start to think about how to send probes to NEAs ,and how to bring back few of them to earth (preferably to L4/5, considering those are stable only )

Quickie · Sep 28, 2018

Anlsvrthng said:
One second mass into the rocket equitation - calculate the acceleration for that.
It gives the m/sec^2 acceleration , from here it will be again F=m*a, ship acceleration multiplied with the ship mass.
It will gives an error as well, proportional to the choose delta V.
It is the other side of the calculation ....................................................

...................................................(preferably to L4/5, considering those are stable only )

It gives the m/sec^2 acceleration , from here it will be again F=m*a, ship acceleration multiplied with the ship mass.
It will gives an error as well, proportional to the choose delta V.
It is the other side of the calculation.

No. That is Newton's 2nd Law F=MA. Come on! You should know no one refers to the F=ma as the rocket equation!

From the start, it was always about the Rocket Equation. i.e. Tsiolkovsky's Rocket Equation.

You yourself has put out this equation earlier at the very start. Here,

" delta V [final] =V[exhaust speed] *ln(launch_mass/empty_mass)"

The escape velocity from earth is 11.186 km/sec.
It means that if you came back FROM the NEA asteroid to the L4/5 then it takes 11.558-11.186=0.372 km/sec .

Where did you get this kind of ridiculous simplistic calculation? The NEAs has a heliocentric orbit.

You can't do the calculation as if the NEA is orbiting Earth only. Even so, I don't see what this calculation is about!

This math of yours is just baloney. Give me break.

You made mistake (or one mistake ) when you tried to use the LEO-NEA dV to calculate the NEA-L4/5 dV.
That was the reason of your 5 km/sec dV calculation. Just a reminder.

Anyway, the real speed to came back to L4/5 from NEA less than that , with proper timing and orbit.

It is way less than the moon landing /orbit /transfer orbit dV requirements.
IF you take off from the moon it takes 2.5 km/sec. It is way more than 0.372 km/sec for earth capture ( again, the L4/5 is less than that)
It is not possible to use ion engine for that ,

No, you are the one who made the following "mistakes".

1. Your first mistake is to claim that NEA has the ridiculous LEO-to-NEA Delta-V value of 382 m/s when the bulk of them has around a 7 km/s value.

2. Your second mistake is your narrow view that a trip from the NEO --> Lagrange Point is
all there is to it, when it always has to be round trip. (You have to bring back the raw material, remember?)

1. LEO --> NEA --> LEO

or,

2. (Going through the NEA-to-LPt way) LEO --> LPt --> NEA --> LPt --> LEO.

But you only focused on the "NEA --> LPt" part!

i.e. The lower energy capture part! (Low Delta-V gravity capture, which is not always low energy, by the way, depending on the transfer orbit and the NEA heliocentric orbit)
That's why it seems so magical to you!

Remember, the Delta-V for LPt --> NEA can be even higher than LEO --> NEA , depending on the heliocentric orbit of the NEA.

When you consider the complete round trip the saving in energy is not that much, or even higher, with the cost of a more complicated maneuver i.e. the addition of LPt --> NEA --> LPt.

So, if we go back to the basic : the moon mining does't make sense. There can be same hydrogen , but there can be hydrogen on the NEAs as well. Actual the mankind has practically 0 knowledge about the NEAs volatile contents, because those are not survive the atmospheric entry in the meteorites.
If any country wants to do serious space mining then they have to start to think about how to send probes to NEAs ,and how to bring back few of them to earth (preferably to L4/5, considering those are stable only )

Wow, you never give up, don't you, despite the few articles I posted showing you're dead wrong on this opinion.

Anlsvrthng · Sep 29, 2018

Quickie said:
No. That is Newton's 2nd Law F=MA. Come on! You should know no one refers to the F=ma as the rocket equation!

From the start, it was always about the Rocket Equation. i.e. Tsiolkovsky's Rocket Equation.

You yourself has put out this equation earlier at the very start. Here,

" delta V [final] =V[exhaust speed] *ln(launch_mass/empty_mass)"

I have no clue what you want to say here.

The rockets working on a simple principle: they accelerate M.exhaust mass to V. exhaust, and it accelerate the rocket weighting M.rocket by a.rocket.

means there is two balancing force, the engine "pushing"force against the rocket inertia.
So, you can either calculate the engine force by using the acceleration of reactive mass and the mass flow, OR you can use the speed change of the rocket and the mass of rocket ( this is actually beyond the high school level if anyone wants it precisely )

Quickie said:
Where did you get this kind of ridiculous simplistic calculation? The NEAs has a heliocentric orbit.

You can't do the calculation as if the NEA is orbiting Earth only. Even so, I don't see what this calculation is about!

This math of yours is just baloney. Give me break.

The earth escape velocity is 11 186 meters / seconds.

It calculated from the surface of earth ( sea level) , and means that if you accelerate on object horizontally on a rail ( with 0 air friction) by this mach it will enter a heliocentric orbit .

Now, to enter low level orbit takes 7800 meter/sec from sea level. By accelerating this much you enter a few hundred km height orbit around earth.

So, if you came back from the solar system then 3386 km/sec to go back to LEO.

However this is nonsense from the startin point of the topics, considering if you bring back mass from moon, then you have to d othe next to came back leo:
(moon surface-orbit)+(orbit-earth transfer)+(earth transfer-LEO)
The (moon surface-orbit) on its own 2.4 km/sec.

The NEA-LEO looks like
(solar orbit-earth capture) + (earth capture-LEO) .
The first leg in this case is 0.382 km/sec.

So, is it make sense to mine on moon?

Quickie said:
No, you are the one who made the following "mistakes".

1. Your first mistake is to claim that NEA has the ridiculous LEO-to-NEA Delta-V value of 382 m/s when the bulk of them has around a 7 km/s value.

2. Your second mistake is your narrow view that a trip from the NEO --> Lagrange Point is
all there is to it, when it always has to be round trip. (You have to bring back the raw material, remember?)

1. LEO --> NEA --> LEO

or,

2. (Going through the NEA-to-LPt way) LEO --> LPt --> NEA --> LPt --> LEO.

But you only focused on the "NEA --> LPt" part!

i.e. The lower energy capture part! (Low Delta-V gravity capture, which is not always low energy, by the way, depending on the transfer orbit and the NEA heliocentric orbit)
That's why it seems so magical to you!

Remember, the Delta-V for LPt --> NEA can be even higher than LEO --> NEA , depending on the heliocentric orbit of the NEA.

When you consider the complete round trip the saving in energy is not that much, or even higher, with the cost of a more complicated maneuver i.e. the addition of LPt --> NEA --> LPt.

Wow, you never give up, don't you, despite the few articles I posted showing you're dead wrong on this opinion.

Again, the 7km/sec is from LEO orbit,but those are way beyond the Mars.

You can not do industrial activity on LEO, and even on medium earth orbit.

Due to tidal forces ( or the dramatic speed differences between close orbits) you can not put two object in close proximity without continuous propellant consumption, and example in LOE orbit the force in a tether between two objects separated by 100 km will be extremely high.
Additionally there is air drag.

IT will be in the range of 0.1 m/sec^2, if you calculate the force on 100 tons then the tether will be very heavy on its own.

Means there is no chance to use example solar furnaces.

So, if anyone want to make space industry the natural point of that is the L4/5 point.
It is cheap to bring back thing from the solar system, it is stable, and generally it require small and light structures to keep together a big and complicated structure.

Quickie said:
Wow, you never give up, don't you, despite the few articles I posted showing you're dead wrong on this opinion.

Bring equations.
Similar like the one that I scanned for you from my notebook : )

I suggest to start with deltaV requirement for LEO-L4/5-NEA, rocket force calculation, tidal force calculation for LEO objects,NEQ- L4/5 capture orbit calculation.
Or it can be interesting if you can calculate the moon mining delta Vs with required fuel mass as well. Just to compare .
Connections / reasons / orbit differences, masses , effect on exhaust speed and time, hohmann transfer and so on.
IF you really on the top I can be interested if you can calculate same NEA - L4/5 low energy capture orbit. That require sophisticated math .

Don't forget, it is not "social science". bring the math. Show solved equations .

Quickie · Sep 30, 2018

Anlsvrthng said:
I have no clue what you want to say here.

The rockets working on a simple principle: they accelerate M.exhaust mass to V. exhaust, and it accelerate the rocket weighting M.rocket by a.rocket.

means there is two balancing force, the engine "pushing"force against the rocket inertia.
So, you can either calculate the engine force by using the acceleration of reactive mass and the mass flow, OR you can use the speed change of the rocket and the mass of rocket ( this is actually beyond the high school level if anyone wants it precisely )
............

I have no clue what you want to say here.

The rockets working on a simple principle: they accelerate M.exhaust mass to V. exhaust, and it accelerate the rocket weighting M.rocket by a.rocket.

means there is two balancing force, the engine "pushing"force against the rocket inertia.
So, you can either calculate the engine force by using the acceleration of reactive mass and the mass flow, OR you can use the speed change of the rocket and the mass of rocket ( this is actually beyond the high school level if anyone wants it precisely )

No. You said the rocket thrust can be found from the Tsiolkovsky's Rocket Equation.

Prove it then.

I have no clue what you want to say here.

The rockets working on a simple principle: they accelerate M.exhaust mass to V. exhaust, and it accelerate the rocket weighting M.rocket by a.rocket.

means there is two balancing force, the engine "pushing"force against the rocket inertia.
So, you can either calculate the engine force by using the acceleration of reactive mass and the mass flow, OR you can use the speed change of the rocket and the mass of rocket ( this is actually beyond the high school level if anyone wants it precisely )

No. You said the rocket thrust can be found from the Tsiolkovsky's Rocket Equation.

You said this: "The rocket equitation doesn't need the trust, but it can be calculated with the help of it."

Prove it then.

The earth escape velocity is 11 186 meters / seconds.

It calculated from the surface of earth ( sea level) , and means that if you accelerate on object horizontally on a rail ( with 0 air friction) by this mach it will enter a heliocentric orbit .

Now, to enter low level orbit takes 7800 meter/sec from sea level. By accelerating this much you enter a few hundred km height orbit around earth.

The thing is the escape velocity has nothing to do with the Delta-V needed for orbit transfer!

Even here, you are wrong again! on the object going into heliocentric orbit. Going at escape velocity on the surface of earth will send the object to infinity of space!

So, if you came back from the solar system then 3386 km/sec to go back to LEO.

Wrong again! The Delta V required depends on which part of the solar system the object is coming back from.

However this is nonsense from the startin point of the topics, considering if you bring back mass from moon, then you have to d othe next to came back leo:
(moon surface-orbit)+(orbit-earth transfer)+(earth transfer-LEO)
The (moon surface-orbit) on its own 2.4 km/sec.

Similar amount of energy is required coming back from NEA to LEO, too! See below!

Coming back from the moon surface to LEO requires only 2.74 km/sec! Surprised?
From moon surface to LLO only 1.87 km/sec.

Coming back from NEA to LEO need more energy than that! See below.

The NEA-LEO looks like
(solar orbit-earth capture) + (earth capture-LEO) .
The first leg in this case is 0.382 km/sec.

So, is it make sense to mine on moon?

Please, that figure 0.382 km/s comes from your baloney rocketry math.

You can't change the Delta-V required with some wizardry math.

If you continue with this phony math thing. I'M GOING TO REPORT YOU!

Unless the NEA come very near to Earth during the loop around the Sun, or it's just a flyby mission, the Delta-V to latch onto/from the NEA heliocentric orbit, (i.e. LEO-to-NEO and NEO-to-LEO) is the same as shown in this link
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,

And it starts from Delta-V of 3.76 km/s!

You can not do industrial activity on LEO, and even on medium earth orbit.

Due to tidal forces ( or the dramatic speed differences between close orbits) you can not put two object in close proximity without continuous propellant consumption, and example in LOE orbit the force in a tether between two objects separated by 100 km will be extremely high.
Additionally there is air drag.

IT will be in the range of 0.1 m/sec^2, if you calculate the force on 100 tons then the tether will be very heavy on its own.

Means there is no chance to use example solar furnaces.

So, if anyone want to make space industry the natural point of that is the L4/5 point.
It is cheap to bring back thing from the solar system, it is stable, and generally it require small and light structures to keep together a big and complicated structure.

I didn't say anything about doing industrial activity at LEO.

And are you intentionally ignorant about the advantages of Lagrange Points in relation to its short distance to the moon. Everyone already knows the Lagrange Points are good staging point for space activity. You so conveniently forget that the LPts are only a relatively short distance from the moon surface, requiring only 2.5 km/s.

On the other hand, the Delta-V for LPts-to-NEA is similar to the values for LEO-to-NEA with the majority of them actually more. Also see below.

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Bring equations.
Similar like the one that I scanned for you from my notebook : )

I suggest to start with deltaV requirement for LEO-L4/5-NEA, rocket force calculation, tidal force calculation for LEO objects,NEQ- L4/5 capture orbit calculation.
Or it can be interesting if you can calculate the moon mining delta Vs with required fuel mass as well. Just to compare .
Connections / reasons / orbit differences, masses , effect on exhaust speed and time, hohmann transfer and so on.

You have to be kidding me. Why would anyone waste their time with you with all your false pretense of knowing the subject at hand.

Furthermore, they wouldn't want to find themselves bearing with all the new baloney claims with every new replies of yours, like this one.

"" The earth escape velocity is 11 186 meters / seconds.
It calculated from the surface of earth ( sea level) , and means that if you accelerate on object horizontally on a rail ( with 0 air friction) by this mach it will enter a heliocentric orbit . ""

My answer to this is in the above.

I already have enough of your ploy.

Most of the info's and data are already available on the internet.

IF you really on the top I can be interested if you can calculate same NEA - L4/5 low energy capture orbit.

I've already answered this in the above.

Obviously, you really don't know anything about this subject. Do you?

There is no such thing as a NEA - L4/5 low energy capture orbit!

It's only low energy for asteroids coming very near to Earth orbit, which is a rare occurrence. Most of them do not come that near, as you can see from the Delta-V values in the link above.

It requires as much energy to go from L4/5-to-NEA as it is from NEA-to-L4/5 , as it is the usual case with orbital mechanics.

No amount of baloney magical math calculation of yours is going to change that! With which you have tried to bamboozle people here, from the start!

There's really no free lunch in space exploration, too.

Alright, I have already given you enough time.

Anlsvrthng · Sep 30, 2018

Arguing loudly doesn't' make you right.

Please bring calculations.
IT is not complicated, only high school math.

Quickie said:
No. You said the rocket thrust can be found from the Tsiolkovsky's Rocket Equation.

Prove it then.

I explained if by two different way.

I am happy to explain again, but please help in it by telling me what is not clear for you.

Quickie said:
Coming back from the moon surface to LEO requires only 2.74 km/sec! Surprised?

Not only me, it surprise the NASA as well.
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Let me highlight few me line from this file :

For comparison, delta-v for transferring from low-Earth orbit to rendezvous
with the Moon and Mars:
Moon: 6.0 km/s
Mars: 6.3 km/s

2.74 is the escape velocity from the surface of moon (the landing is only 1.5 km/sec, my bad).

If you launch a spacecraft with this speed toward earth then it will pass by the earth, and will go to the infinity.
Doing aero breaking require 1.3 km/sec more , going to LEO require earth escape velocity - LEO speed delta V.

Lagrange require similar amount.

So, I explain again what the NASA file say :
2018 AV2 require 3.758 km/sec
Moon require 6 km/sec.

6 is bigger than 3.758.
NASA say that.
Please argue with NASA.And bring math : )

Jura The idiot · Oct 1, 2018

now I read
China Media Group launches the first 4K Ultra HD channel today
2018-10-01 11:10 GMT+8
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China Media Group launched it's first domestic superstar TV channel – CCTV 4K Ultra HD channel on Monday, October 1. The channel will broadcast 18 hours per day from 6 a.m. to midnight starting October 2.

The channel will continue to provide viewers with many 4K programs including documentaries, sports events, TV series and cartoons.

At that time, the first batch of cable TV network users in 13 municipalities and provinces such as Beijing, Shanghai and Guangdong are expected to have a good time.

4K Ultra HD TV is a rapidly developing, emerging technology. Compared with existing high-definition and standard TV systems, 4K Ultra HD TV is a technical upgrade.

By providing high-quality 4K Ultra HD programs and enhancing the viewing experience, China Media Group stars a new journey of upgrading its production and broadcasting system.

The follow-up construction will be carried out in stages from 2019 to 2021, when the celebration of the 70th anniversary of the founding of the People's Republic of China in 2019 will adopt 4K Ultra HD TV broadcasting.

The launch of the 4K channel marks a solid step for China Media Group to create a new world-class mainstream media target with strong leadership, communication and influence.

Quickie · Oct 1, 2018

Anlsvrthng said:
Arguing loudly doesn't' make you right.

Please bring calculations.
IT is not complicated, only high school math.

I explained if by two different way.

I am happy to explain again, but please help in it by telling me what is not clear for you.

Not only me, it surprise the NASA as well.
Please, Log in or Register to view URLs content!

Let me highlight few me line from this file :

For comparison, delta-v for transferring from low-Earth orbit to rendezvous
with the Moon and Mars:
Moon: 6.0 km/s
Mars: 6.3 km/s

2.74 is the escape velocity from the surface of moon (the landing is only 1.5 km/sec, my bad).

If you launch a spacecraft with this speed toward earth then it will pass by the earth, and will go to the infinity.
Doing aero breaking require 1.3 km/sec more , going to LEO require earth escape velocity - LEO speed delta V.

Lagrange require similar amount.

So, I explain again what the NASA file say :
2018 AV2 require 3.758 km/sec
Moon require 6 km/sec.

6 is bigger than 3.758.
NASA say that.
Please argue with NASA.And bring math : )

Arguing loudly doesn't' make you right.

They are facts.

Please bring calculations.
IT is not complicated, only high school math.

No need to for a general discussion. All the data is already available out there. Furthermore, it's not like you are any good at it.

As proved by another of your many baloney rocketry math like this.

"going to LEO require earth escape velocity - LEO speed delta V."

Are you under the influence of drugs or what? You're at it again.

If that works, the whole world will turn upside down!

If I'm the moderator, I would have banned you long time ago already.

I explained if by two different way.

I am happy to explain again, but please help in it by telling me what is not clear for you.

You're pretending to not understand, or you're so challenged to even understand a simple question?

Ok, one last time. You have the Tsiolkovsky Rocket Equation:

" delta V [final] =V[exhaust speed] *ln(launch_mass/empty_mass)"

Derive the Thrust (i.e. the Force in their usual dimensions of mass, length and time) from it.

You said it can be done. I said NO.

So, I explain again what the NASA file say :
2018 AV2 require 3.758 km/sec
Moon require 6 km/sec.

6 is bigger than 3.758.
NASA say that.
Please argue with NASA.And bring math

Those are within a very small percentile of the rest.
I thought this has been discussed before. You don't seem to have learned anything from it.

You bring back processed material from the moon surface which can be 1 percent of the raw
material on the moon.

You would have to bring back unprocessed raw material from the asteroid which is 100 times
heavier than processed material, meaning that the launch weight of the spacecraft (with
heat shields and all) would have to be 100 times heavier i.e 100 times the Delta-V
required.

Check the earlier posts for other reasons.

I said I've given you enough time.

In this case, I'm replying to you NOT for your sake, but
for the sake of the rest of the members only, just to see through your veiled intention, as
it is the case when you're in the economics thread and other threads.

Jura The idiot · Oct 2, 2018

now I read
Beidou navigation system serves Tibet
Xinhua| 2018-10-01 20:52:39
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Beidou, a domestically engineered satellite navigation system of China, started to serve its second largest provincial region of Tibet on Monday.

A Beidou-supported information platform made a debut in Lhasa, capital of Tibet Autonomous Region. It is expected to provide services in disaster alert and relief, emergency rescue, transportation, agriculture, forestry, water conservancy, as well as targeted poverty reduction and smart travel.

"We will accelerate the development of Beidou navigation industry and its application in Tibet," said Qin Zhenhua, chairman of the Tibet Xingchuan Beidou Satellite Navigation Platform Co., Ltd.

Tibet occupies about one eighth of China's land area with 3.17 million people. Beidou meets the urgent need from the sparsely populated region.

Named after the Chinese term for the Big Dipper constellation, Beidou is a rival to the U.S. GPS system, Russia's GLONASS and the European Union's Galileo as a global satellite navigation system.

China launched its first Beidou satellite in 2000.

Beidou, with 38 satellites, will provide danger alerts and navigation services for global users after the successful launch of two satellites on September 19.

Beidou was listed on a global satellite search and rescue implementation plan in February by COSPAS-SARSAT, an international satellite system for search and rescue.

Anlsvrthng · Oct 2, 2018

Quickie said:
You're pretending to not understand, or you're so challenged to even understand a simple question?

Ok, one last time. You have the Tsiolkovsky Rocket Equation:

" delta V [final] =V[exhaust speed] *ln(launch_mass/empty_mass)"

Derive the Thrust (i.e. the Force in their usual dimensions of mass, length and time) from it.

You said it can be done. I said NO.

You can calculate from it, either with the mass flow or with the flight time.

However both of the above is interesting ONLY if you take of from the surface of moon, or you try do do hohmann.

Still, no math. only talk.
I still waiting for your calculation how can it require less delta V to reach the moon surface and get back to L4/5 than to reach a NEA and get back.
Don't forget , the discussion started from that you said it is cheaper to go to the moon than to go to the NEAs.
I supported calculations and so on, but all that you have is same ride on the rocket equation : D

And the all space agency waiting for it as well. Maybe they calculating by the wrong way the delta V.

Quickie said:
Those are within a very small percentile of the rest.
I thought this has been discussed before. You don't seem to have learned anything from it.

You bring back processed material from the moon surface which can be 1 percent of the raw
material on the moon.

You would have to bring back unprocessed raw material from the asteroid which is 100 times
heavier than processed material, meaning that the launch weight of the spacecraft (with
heat shields and all) would have to be 100 times heavier i.e 100 times the Delta-V
required.

Check the earlier posts for other reasons.

I said I've given you enough time.

In this case, I'm replying to you NOT for your sake, but
for the sake of the rest of the members only, just to see through your veiled intention, as
it is the case when you're in the economics thread and other threads.

How you get energy for industrial process?
How you get radiation shielding for humans being outside the magnetic shield of earth?

A Salyut sized space station require shielding in the range of 60 tons as bare minimum for long term space living.
The surface of moon has too low gravity.
In the space they need centrifuge to keep the bones healthy , but that need damping mass. Lot of mass,again.

Forget the reactor as a source, a nuclear reactor mindbogglingly expensive even in space terms ,if all activity is to melt metals , or reduce oxides to make oxygen.
(and the reactor weight more than the same power solar panel in )
Not possible to melt the nickel-iron meteorite with solar furnace, possible no clean , oxide free metal on the surface, 14 days long night makes it impossible to use solar panels.

And it is just the beginning.
Not possible to take of from the surface with ion propulsion, it needs chemical rockets.
In space the dust from asteroid can be used as reaction mass, requiring only a grinder and maybe facility to make mesh , on the moon it require complicated chemical processes, with all required equipment, if you use that small amount of water that is on the poles, instead to put that into life support systems on humans living quarters.
But it doesn't help the lack of nitrogen.

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