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#1 2019-11-08 17:53:34

louis
Member
From: UK
Registered: 2008-03-24
Posts: 7,208

Musk's Stunning Tweets - 10 million tons to Mars!!

A stunning series of tweets from Elon Musk!

He's claiming that Space X will be able to launch Starship to LEO at $2million or $20 per kg.  He's claiming that the Starship will be able to go into orbit 3 times a day, so 1000 flights a year. He's claiming it will be possible using 100 Starships to transfer one million tons over 20 years to Mars Base Alpha. To create a Mars City will require 1000 Starships - so I guess 10 million tons to Mars over two decades.

https://twitter.com/elonmusk/status/1192547668817375232

https://twitter.com/teslaownersSV/statu … 3566363648

As always, one is astonished by the scale of his vision!

But I am also surprised by how much reliance Musk is placing on transferring mass to Mars as opposed to transferring manufacturing capability.

It seems to me he is putting the cart before the horse - he seems to be thinking in terms of creating a city, that will then attract colonists. Of course there is some logic to that. But on the other hand, it assumes the colonists with the right skills are there ready and waiting to be attracted.

I think he is underestimating the organisational task of dealing with such a huge undertaking. If we assume that half of the 10 million tons is going into residential accommodation at 10 tons per person, that would mean a population of 500,000. Will 500,000 well suited applicants come forward all of their own accord?  Or do you need to assess 10 million people to find the 500,000 with the right stuff? Are there 10 million people ready to move to Mars?

Then you need to match your 500,000 people to the tasks that need to be undertaken. You probably need a minimum 100,000 people educated to post grad level with lots of STEM skills because Mars with its life support systems and challenging environment is just that sort of place.

I have no problem with Musk's "Big Vision" but I do have a problem with "Big Vision" detached from reality!

But more than that I do query why we need 10 million tons of stuff transferred to Mars. I think ISRU on Mars can replace probably 97% of that stuff.

Things will be different on Mars. We can focus on the things that need to be done to create a self-sustaining civilisation rather than trying to replicate US lifestyles on Mars. So, there is no need for a family on Mars to have a couple of automobiles to get around, not least because there won't be many places to go to... smile  There's no need for people to have 50 pairs of shoes and 50 pairs of trousers. There's no need for lawnmowers...for newspapers...for all sorts of stuff. You are probably already reducing the "consumer" mass requirement by 80%.

For me the development phases need to be:

1. Establish a viable base on Mars, where humans can live.

2. This base will be permanently occupied but initially there won't be permanent residents...ie people will stay for 2,4,6 years to begin with.

3. The base will be focussed mostly on science, exploration and propellant production.

4.  Gradually build up the base's industrial and agricultural potential. This will involve liberal use of 3D printers, CNC machines and industrial robots. Eventually there will need to be a move from artificially lit agriculture to natural light farming.

5. Within a few decades the base should have a full industrial and agricultural infrastructure, albeit scaled down from Earth, which will allow it to use mainly ISRU to make machines, factories, PV solar arrays, residential and other habs, small domes for agriculture, clothing, rockets, transport vehicles, robot miners, and so on.

6. As the base develops, you can keep extending the safe period of residence and then work on ensuring safe procreation...once you can produce healthy babies on Mars, you have created humanity's second home.

Once you get to 5 I don't really see the need for millions of tons of mass to be transferred from Earth to Mars. Instead you can have a much more "organic" development process, whereby increasing numbers of people come to live on Mars and the settlement is able to expand to meet the needs of the new settlers.

That said, the possibility of a $20 per kg launch price to LEO is going to up huge possibilities for economic development on Mars as it would imply something like $80 per kg to Mars. If you can find easily accessible gold on Mars, you could be making $20 million profit on each kg returned to Earth. Fully automated robotic industrial processes on Mars could become economic because you have the economic advantage of no land costs, no licence costs, no taxation and v. low energy costs.


Let's Go to Mars...Google on: Fast Track to Mars blogspot.com

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#2 2019-11-09 03:43:34

Calliban
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From: Northern England, UK
Registered: 2019-08-18
Posts: 3,352

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

Excellent post Louis.  I agree with your assessment.  Most of the mass shipped to Mars should be capital equipment that allows other items to be produced using ISRU. We can divide pressurised areas into habitation space, industrial space and agricultural space.  The first two can be subterranean, using corralled surface materials to ballast internal pressure and provide insulation.  Agriculture is going to be quite difficult, as there is no escaping the need for either large amounts of electrical power with artificial lighting in underground growing systems or heating for surface polytunnels.  Either way a lot a area is needed and energy requirements are high.

I think $20/kg is ambitious.  It seems to require very rapid turnaround rates of rocket components.  The problem is that these are subject to thermal and mechanical fatigue.

A specific problem with colonising Mars is its distance.  It would take 2 years for Musk's Starship to make a round trip.  How cheap would an aeroplane flight be if it lasted for 2 years?

I wish he hadn't called it the 'Starship'.  That name has a very specific meaning and in this context it is a misnomer.

Last edited by Calliban (2019-11-09 03:51:11)


"Plan and prepare for every possibility, and you will never act. It is nobler to have courage as we stumble into half the things we fear than to analyse every possible obstacle and begin nothing. Great things are achieved by embracing great dangers."

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#3 2019-11-09 10:28:30

SpaceNut
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Posts: 28,747

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

The issue is dependant on starship landings of 2 cargo where the success rate is 100% landed perfectly versus 50% for the pair should there be any mishap for the mission landings. Then if there is any landing mishaps at all for the crewed is devistating to man for mars missions. Remember shuttle had only 2 fails in 100 missions and that was deemed to much for man to keep using. Mars will be treated at an even higher level of need for success.

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#4 2019-11-09 10:43:48

GW Johnson
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From: McGregor, Texas USA
Registered: 2011-12-04
Posts: 5,423
Website

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

One should be very careful about running and projecting numbers for a Starship/Super Heavy transportation system.  Those characteristics are still in great flux until the flight test program is nearer completion than starting.  That offered caution includes Musk himself,  despite his apparent predilection otherwise.

Be well aware that the number of Starship flights to LEO is not the same as the number of Starships than could go to Mars,  the moon,  or anywhere else.  Numbers vary among estimators,  but my own assessment posted over at "exrocketman" says there are 6 tankers required to refill a Starship in circular LEO,  and 7 are required for the elliptical parking orbit they have proposed for lunar flights. 

What that means is for circular LEO basing,  the total number of flights is 7,  in order to get one flight to Mars (or elsewhere).  From the elliptical departure orbit,  the total number of flights is 8,  to get one flight to the moon (or elsewhere).  I didn't see that reflected in the numbers bandied about in the post that started this thread.   

If you think in terms of 100 Starships on Mars,  you will have to launch 700 to 800 of them into LEO.  If you want 1000 Starships on Mars,  you will have to launch 7000 to 8000 of them into LEO.  While these are reusable vehicles,  and one of them can fly many times,  you can see how the number of launches (and the required launch rates) gets very ridiculous,  the bigger your dreams are.

And as I have repeatedly warned,  the final product that will be Starship/Super Heavy will be quite different from what is projected today,  precisely because of (1) experimental flight test results that are always and inherently unexpected,  and (2) problems whose solutions are not yet incorporated into the design (such as landing pad bearing loads vs soil strength,  and the instability of high center of gravity versus landing pad footprint size).

Also bear in mind that you can send a large payload and fly slow,  or you can fly a little faster,  but carry a small payload.  You CANNOT fly fast with a large payload,  rocket vehicle physics of mass ratio does not allow that.  It averages 9 months one-way to Mars on a min-energy Hohmann transfer at max payload.  You can fly a little faster with significantly less payload:  maybe 6-7 months one-way.  Talking about 3 months to Mars with this kind of vehicle is just nonsense.

Just trying to inject a little reality into these wild flights of fancy. 

GW


GW Johnson
McGregor,  Texas

"There is nothing as expensive as a dead crew,  especially one dead from a bad management decision"

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#5 2019-11-09 10:52:31

louis
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From: UK
Registered: 2008-03-24
Posts: 7,208

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

I think Musk must be thinking of direct flights.  How much more energy is required for a direct flight? I've never been able to nail that down exactly. But it's probably less of a challenge than we might expect. It's possible that Musk might have mid-flight refuelling in mind...that could be a possibility.  But, yes, $20 per kg to LEO is incredibly ambitious. If he can get it down to $200 per kg he will be doing very well!

Yes, I've never been a fan of the name "Starship" either! But I think it has traction with the public.

In terms of simple build techniques, I think using cut and cover would work well on Mars - dig a trench and then build over that bricked arches.  It has been shown on Earth that Mars regolith bricks can be created through simple compression.

NASA have also been experimenting with 3D printing techniques for building structures.

We need an industrial plan for Mars that (a) sources required raw materials through surface mining (b) processes and purifies the raw materials (c) produces industrial chemicals and gases (d) builds the required industrial units - scaled down versions of factories with the necessary machines and (e) manufactures the required items. We need materials like glass, plastics, steel, basalt, fibre glass, bamboo...

Calliban wrote:

Excellent post Louis.  I agree with your assessment.  Most of the mass shipped to Mars should be capital equipment that allows other items to be produced using ISRU. We can divide pressurised areas into habitation space, industrial space and agricultural space.  The first two can be subterranean, using corralled surface materials to ballast internal pressure and provide insulation.  Agriculture is going to be quite difficult, as there is no escaping the need for either large amounts of electrical power with artificial lighting in underground growing systems or heating for surface polytunnels.  Either way a lot a area is needed and energy requirements are high.

I think $20/kg is ambitious.  It seems to require very rapid turnaround rates of rocket components.  The problem is that these are subject to thermal and mechanical fatigue.

A specific problem with colonising Mars is its distance.  It would take 2 years for Musk's Starship to make a round trip.  How cheap would an aeroplane flight be if it lasted for 2 years?

I wish he hadn't called it the 'Starship'.  That name has a very specific meaning and in this context it is a misnomer.


Let's Go to Mars...Google on: Fast Track to Mars blogspot.com

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#6 2019-11-09 11:40:36

SpaceNut
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

A direct flight allows for no refueling in LEO as thats a launch and go...

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#7 2019-11-09 15:05:37

louis
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From: UK
Registered: 2008-03-24
Posts: 7,208

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

I meant a non-Hohmann transfer!

SpaceNut wrote:

A direct flight allows for no refueling in LEO as thats a launch and go...


Let's Go to Mars...Google on: Fast Track to Mars blogspot.com

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#8 2019-11-09 16:08:09

GW Johnson
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From: McGregor, Texas USA
Registered: 2011-12-04
Posts: 5,423
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Re: Musk's Stunning Tweets - 10 million tons to Mars!!

In the context of post 5 above and this thread,  I have no clue what "direct flight" means. 

To get from Earth to Mars,  the trajectory with the smallest delta-vee requirement is the "Hohmann min energy transfer".  This is an ellipse with its perihelion at Earth's orbit (when Earth is there),  and its aphelion at Mars's orbit on the opposite side of the sun (when Mars is there). 

Up to a point,  all faster (non-Hohmann) trajectories are also ellipses with their perihelions at Earth's orbit,  but their aphelions farther out than Mars's orbit,  and on the opposite side of the sun. 

The nature of such eccentric ellipses is that the faster velocities are at and near perihelion,  and that you are using less of the semi-perimeter of the ellipse as your transfer trajectory path,  because it sticks out past Mars's orbit. 
That's why these have shorter flight times.  They also have substantially-higher delta-vee requirements to get from Earth onto the trajectory at perihelion.  That's just physics.  There is no way to cheat.

There comes a point where the perihelion velocity of the transfer trajectory becomes equal to solar escape speed.  At that point,  the trajectory becomes an open-ended parabola.  The delta-vee to get onto this (at perihelion from the Earth) is quite high indeed.  Your residual velocity after escape from Earth has to be correctly oriented,  and must equal solar escape speed,  as measured with respect to the sun,  not the Earth.

Beyond that point,  with perihelion velocity greater than solar escape speed,  the shape of the trajectory becomes a hyperbola.  Delta-vee requirements to get onto this kind of orbit path are higher still. 

Arrival at Mars is simple,  because it has an atmosphere so that direct aerobraking entry (all in one pass,  mind you!!!) without firing a rocket is possible.  You time your arrival such that Mars is there just as you reach that distance from the sun on your trajectory. 

You have a speed with respect to ("wrt") the sun which can be figured from your trajectory by conservation of mechanical energy,  at that distance from the sun.  The direction of that speed is tangent to your trajectory.  Mars has a velocity tangential to its orbit. 

Your velocity with respect to Mars is the vector difference of your velocity wrt sun minus Mars's velocity wrt the sun.  The faster your transfer trajectory,  the larger the magnitude of your velocity is,  measured wrt Mars. 

This is your velocity wrt Mars "at infinity".  As you close with the planet,  its gravity speeds you up further,  so that your speed at entry interface (about 140 km altitude),  is yet higher still.  This rapidly becomes untenable in terms of heat shield design,  long before (!!!) your trajectory shifts from elliptic to parabolic.  And I do mean long before!!!  That is the inherent energy limit of chemical propulsion. 

Something similar is true for the Earth return voyage.  Your trajectory has a high perihelion velocity wrt the sun.  So also does the Earth have a velocity in its orbit wrt the sun.  Their vector difference is your velocity "at infinity" wrt the Earth. 

As you close with the Earth,  its gravity speeds you ever faster toward it.  Your entry interface speed (again,  near 140 km altitude),  can be very high indeed.  This becomes completely untenable for heat shield design,  and for peak entry gees,  and also for bouncing off the atmosphere above Earth escape speed,  on elliptical transfers well below the parabolic point.

As near as I can tell,  there is no alternative "direct flight" trajectory per post 5 above.  That's not a concept with any referent.  ALL of these I have just described are "direct flights" from Earth to Mars,  and/or from Mars to Earth.  Some are faster than others.  The min energy one is "Hohmann".  The rest are "non-Hohmann".  But that's all that there is.  And that's just physics. You can't cheat or deny it.

There is NO "in-flight refueling",  except just in Earth orbit,  with Spacex's concept for Starship/Super Heavy.  Here's why:

This vehicle concept can JUST BARELY reach orbit about the Earth,  with any significant payload,  and also with just enough propellant allowance (actually a small amount in comparison to the other items) still on board to deorbit and make its aerobrake/retropropulsive landing. That's true for the tankers as well as the cargo/passenger variants.  It must be refilled on-orbit to go anywhere else,  outside of Earth orbit!  Period.  You cannot argue with the physics!   

Payload to orbit is maximized if the orbit is circular LEO.  My estimate says 6 tankers are needed to refill fully,  when in circular LEO.

Payload to orbit is significantly less for a 400x1400 km elliptical orbit about the Earth.  That one also grazes the inner edge of the Van Allen radiation belts,  a fatal radiation exposure risk. However,  once refueled (with 7 tankers,  not 6 !!!),  departing from this elliptic orbit increases payload deliverable to the moon,  because the delta-vee to leave Earth is less,  with the higher perigee velocity of the elliptic Earth orbit. 

Less delta-vee is less required mass ratio,  thus allowing a bit more payload. But you had to pay for an extra tanker flight,  in order to get this effect.  Physics. You cannot argue with it!

Sorry,  if all of these things are different from what you would like.  That's just physics! 

Go run the numbers for yourselves.  I already did,  they're posted at "exrocketman". 

GW

Last edited by GW Johnson (2019-11-09 16:17:50)


GW Johnson
McGregor,  Texas

"There is nothing as expensive as a dead crew,  especially one dead from a bad management decision"

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#9 2019-11-09 17:15:01

louis
Member
From: UK
Registered: 2008-03-24
Posts: 7,208

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

I don't claim to understand trajectory dynamics, I simply understand (a) that Hohmann transfer is the most fuel-efficient way to get from Earth to Mars (and back) but (b) with enough fuel /power you can get from Earth to Mars more quickly than Hohmann transfer will get you (or assist you getting) there.

I've just never really found any good reference for how much more fuel/power you need to, say, reduce an average 7 months journey with Hohmann transfer to 3 months by any other trajectory. Is it three times, ten times or one hundred times more fuel/power?

GW Johnson wrote:

In the context of post 5 above and this thread,  I have no clue what "direct flight" means. 

To get from Earth to Mars,  the trajectory with the smallest delta-vee requirement is the "Hohmann min energy transfer".  This is an ellipse with its perihelion at Earth's orbit (when Earth is there),  and its aphelion at Mars's orbit on the opposite side of the sun (when Mars is there). 

Up to a point,  all faster (non-Hohmann) trajectories are also ellipses with their perihelions at Earth's orbit,  but their aphelions farther out than Mars's orbit,  and on the opposite side of the sun. 

The nature of such eccentric ellipses is that the faster velocities are at and near perihelion,  and that you are using less of the semi-perimeter of the ellipse as your transfer trajectory path,  because it sticks out past Mars's orbit. 
That's why these have shorter flight times.  They also have substantially-higher delta-vee requirements to get from Earth onto the trajectory at perihelion.  That's just physics.  There is no way to cheat.

There comes a point where the perihelion velocity of the transfer trajectory becomes equal to solar escape speed.  At that point,  the trajectory becomes an open-ended parabola.  The delta-vee to get onto this (at perihelion from the Earth) is quite high indeed.  Your residual velocity after escape from Earth has to be correctly oriented,  and must equal solar escape speed,  as measured with respect to the sun,  not the Earth.

Beyond that point,  with perihelion velocity greater than solar escape speed,  the shape of the trajectory becomes a hyperbola.  Delta-vee requirements to get onto this kind of orbit path are higher still. 

Arrival at Mars is simple,  because it has an atmosphere so that direct aerobraking entry (all in one pass,  mind you!!!) without firing a rocket is possible.  You time your arrival such that Mars is there just as you reach that distance from the sun on your trajectory. 

You have a speed with respect to ("wrt") the sun which can be figured from your trajectory by conservation of mechanical energy,  at that distance from the sun.  The direction of that speed is tangent to your trajectory.  Mars has a velocity tangential to its orbit. 

Your velocity with respect to Mars is the vector difference of your velocity wrt sun minus Mars's velocity wrt the sun.  The faster your transfer trajectory,  the larger the magnitude of your velocity is,  measured wrt Mars. 

This is your velocity wrt Mars "at infinity".  As you close with the planet,  its gravity speeds you up further,  so that your speed at entry interface (about 140 km altitude),  is yet higher still.  This rapidly becomes untenable in terms of heat shield design,  long before (!!!) your trajectory shifts from elliptic to parabolic.  And I do mean long before!!!  That is the inherent energy limit of chemical propulsion. 

Something similar is true for the Earth return voyage.  Your trajectory has a high perihelion velocity wrt the sun.  So also does the Earth have a velocity in its orbit wrt the sun.  Their vector difference is your velocity "at infinity" wrt the Earth. 

As you close with the Earth,  its gravity speeds you ever faster toward it.  Your entry interface speed (again,  near 140 km altitude),  can be very high indeed.  This becomes completely untenable for heat shield design,  and for peak entry gees,  and also for bouncing off the atmosphere above Earth escape speed,  on elliptical transfers well below the parabolic point.

As near as I can tell,  there is no alternative "direct flight" trajectory per post 5 above.  That's not a concept with any referent.  ALL of these I have just described are "direct flights" from Earth to Mars,  and/or from Mars to Earth.  Some are faster than others.  The min energy one is "Hohmann".  The rest are "non-Hohmann".  But that's all that there is.  And that's just physics. You can't cheat or deny it.

There is NO "in-flight refueling",  except just in Earth orbit,  with Spacex's concept for Starship/Super Heavy.  Here's why:

This vehicle concept can JUST BARELY reach orbit about the Earth,  with any significant payload,  and also with just enough propellant allowance (actually a small amount in comparison to the other items) still on board to deorbit and make its aerobrake/retropropulsive landing. That's true for the tankers as well as the cargo/passenger variants.  It must be refilled on-orbit to go anywhere else,  outside of Earth orbit!  Period.  You cannot argue with the physics!   

Payload to orbit is maximized if the orbit is circular LEO.  My estimate says 6 tankers are needed to refill fully,  when in circular LEO.

Payload to orbit is significantly less for a 400x1400 km elliptical orbit about the Earth.  That one also grazes the inner edge of the Van Allen radiation belts,  a fatal radiation exposure risk. However,  once refueled (with 7 tankers,  not 6 !!!),  departing from this elliptic orbit increases payload deliverable to the moon,  because the delta-vee to leave Earth is less,  with the higher perigee velocity of the elliptic Earth orbit. 

Less delta-vee is less required mass ratio,  thus allowing a bit more payload. But you had to pay for an extra tanker flight,  in order to get this effect.  Physics. You cannot argue with it!

Sorry,  if all of these things are different from what you would like.  That's just physics! 

Go run the numbers for yourselves.  I already did,  they're posted at "exrocketman". 

GW


Let's Go to Mars...Google on: Fast Track to Mars blogspot.com

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#10 2019-11-09 19:27:27

SpaceNut
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From: New Hampshire
Registered: 2004-07-22
Posts: 28,747

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

Those dammed math equations of distance, mass change, fuel allotment for burn period, to speed...(isp) or exit velocity thrust....which is momentum once the engines fire and all fuel is used over a period of time.....

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#11 2019-11-09 19:48:04

louis
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From: UK
Registered: 2008-03-24
Posts: 7,208

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

So do you have an estimate for the additional power required to turn an average 7 month Hohmann transfer into a  3 month journey?

SpaceNut wrote:

Those dammed math equations of distance, mass change, fuel allotment for burn period, to speed...(isp) or exit velocity thrust....which is momentum once the engines fire and all fuel is used over a period of time.....


Let's Go to Mars...Google on: Fast Track to Mars blogspot.com

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#12 2019-11-09 20:50:02

SpaceNut
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From: New Hampshire
Registered: 2004-07-22
Posts: 28,747

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

Remember the ship that goes faster to mars is all that much harder to slow in the thin atmospher of mars and will produce a lot more heat for the shield to remove as the drag component of heating is related to the speed of the ship and the angle of attack. Going faster also means you will need more fuel on retro propulsion to slow the ship to a touch down once its in landing. Si the current 100 ton of fuel will need to cut into the payload mass to mars surface to allow for the ship size to stay the same while compensating for larger tanks for landing due to the increase speed.

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#13 2019-11-10 06:42:24

louis
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From: UK
Registered: 2008-03-24
Posts: 7,208

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

Yes, I meant including all factors including deceleration...I am just wondering what the proportion of propellant/fuel to payload/rocket mass would be to, say, achieve a 3 month journey time.

SpaceNut wrote:

Remember the ship that goes faster to mars is all that much harder to slow in the thin atmospher of mars and will produce a lot more heat for the shield to remove as the drag component of heating is related to the speed of the ship and the angle of attack. Going faster also means you will need more fuel on retro propulsion to slow the ship to a touch down once its in landing. Si the current 100 ton of fuel will need to cut into the payload mass to mars surface to allow for the ship size to stay the same while compensating for larger tanks for landing due to the increase speed.


Let's Go to Mars...Google on: Fast Track to Mars blogspot.com

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#14 2019-11-10 08:50:16

tahanson43206
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Registered: 2018-04-27
Posts: 16,754

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

For Louis re #13 and topic in general ...

An option available using Newtonian physics is momentum transfer of mass travelling from Earth (or anywhere) to Phobos.

The engineering is beyond human capability at present, but were a solution to be found, the momentum given to a vehicle for a fast transfer to Mars would be given to Phobos, which needs help staying in orbit, instead of to heating the atmosphere of Mars, which is the only technique humans can manage at present.

(th)

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#15 2019-11-10 08:57:36

Calliban
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From: Northern England, UK
Registered: 2019-08-18
Posts: 3,352

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

tahanson43206 wrote:

For Louis re #13 and topic in general ...

An option available using Newtonian physics is momentum transfer of mass travelling from Earth (or anywhere) to Phobos.

The engineering is beyond human capability at present, but were a solution to be found, the momentum given to a vehicle for a fast transfer to Mars would be given to Phobos, which needs help staying in orbit, instead of to heating the atmosphere of Mars, which is the only technique humans can manage at present.

(th)

How about a very long piece of elastic?

:-)


"Plan and prepare for every possibility, and you will never act. It is nobler to have courage as we stumble into half the things we fear than to analyse every possible obstacle and begin nothing. Great things are achieved by embracing great dangers."

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#16 2019-11-10 10:14:11

Oldfart1939
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Registered: 2016-11-26
Posts: 2,366

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

Robert Zubrin has these figures published in both Mars Direct and Entering Space.

The combined culprits preventing the 3 month Mars journey are (1) Rocket Equation, and (2) the Isp of Methane/Oxygen fuel combination (325 to maybe 380 sec). In order to achieve shorter transit times, we need to look at Nuclear Thermal with double to triple the Isp (850-900 sec.). As GW has repeatedly pointed out--Physics cannot be defeated through wishful thinking.

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#17 2019-11-10 10:21:13

SpaceNut
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From: New Hampshire
Registered: 2004-07-22
Posts: 28,747

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

louis wrote:

Yes, I meant including all factors including deceleration...I am just wondering what the proportion of propellant/fuel to payload/rocket mass would be to, say, achieve a 3 month journey time.

SpaceNut wrote:

Remember the ship that goes faster to mars is all that much harder to slow in the thin atmospher of mars and will produce a lot more heat for the shield to remove as the drag component of heating is related to the speed of the ship and the angle of attack. Going faster also means you will need more fuel on retro propulsion to slow the ship to a touch down once its in landing. Si the current 100 ton of fuel will need to cut into the payload mass to mars surface to allow for the ship size to stay the same while compensating for larger tanks for landing due to the increase speed.

Work the equation backups for the new on orbit to mars arrival speed to be able to solve for the new thrust needed to slow the ship for retro propulsion. First to solve is the guide path aerobraking speed at exit of mach even if the mass of the ship, fuel, and payload leave at the same consolidated value to give you a first pass number. This is the engine thrust that must be achieved at altitude to be able to slow the ship down.

https://www.reddit.com/r/spacex/comment … on_system/

Mars departure speed is more like 4.3 kms for a 6 month journey to mars and you are looking to get there is half the time which would mean a new arrival speed of 8.6 or higher depending on distance which could be as high as 9.50 kms. We can solve for the new value of fuel mass change later for the earth departure.

http://design.ae.utexas.edu/mission_pla … riteup.pdf
DV vs C3 from Low Earth Orbit

There is a simulator program link which has been posted by another member here by rgclark I think for the glide path.

Fast transits and aerocapture topic

If we were trying for high orbit breaking we would just pass and bounce off to hopefully slow in time for another pass as the orbitors do.
https://space.stackexchange.com/questio … ut-landing

pMPR9.gif

Of course months or years looping to slow down is not what we would want for a manned mission.

https://ntrs.nasa.gov/archive/nasa/casi … 209887.pdf
ENABLING EXPLORATION MISSIONS NOW: APPLICATIONS OF ON-ORBIT STAGING

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#18 2019-11-10 11:02:25

tahanson43206
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Registered: 2018-04-27
Posts: 16,754

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

For Calliban re #15 ...

First, thanks for taking up this suggestion.  I appreciate that the opening post is on the light hearted side.

The proposal ** is ** so far beyond the capability of currently living engineers that is almost not worth considering

However, the principle at play has been known for thousands of years, since the first early humanoid tried catching fish by dropping a baited line in the water.

Modern fly catching equipment is a state-of-the art manifestation of the principle.

A very light weight (somewhat elastic) line is cast a great distance.

A fish which is more than strong enough to break the line snags the lure, and makes a dash for freedom.

The angler patiently lets out line while maintaining a small but steady drag (today using ratchet mechanisms in the reel, supplemented by skillfully applied pressure from the angler).

Eventually, if all goes well, the fish tires or changes direction.  In either case, the angler draws in on the line, until eventually the fish fetches up next to the angler, where supplemental capture methods are often employed to complete the transaction.

In the case of Phobos and Mr. Musk's millions of tons of payload, what I have in mind is a metal cable laid out on a line that allows the approaching object to engage with the "lure".  The "reel" in this case is yet to be invented, but a visual image might be of a reel of cable on an axle oriented so that the line will pay out in the direction of the moving object, with the ability to change orientation as the target moves across the horizon.

Just as with the operation of a fly fishing rod and reel, the line would be paid out at a pace to allow a gentle tug on the target, well within the tension limits of the cable.  Over time, and given a cable of sufficient length, any object can be slowed and then drawn back to Phobos.

Likewise, as discussed at some length earlier in this forum, a cable attached to Phobos can allow a lander to move gently down into the atmosphere of Mars, so that at an appropriate point, the lander can be release to fly to a designated landing site.  In the case of Mars, the lander would need rockets for the final landing phase, to eliminate the disadvantage of high landing speed and long runways necessitated by the low density of the atmosphere.

(th)

Last edited by tahanson43206 (2019-11-10 11:03:16)

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#19 2019-11-10 11:50:01

GW Johnson
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From: McGregor, Texas USA
Registered: 2011-12-04
Posts: 5,423
Website

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

There is no 7 month Hohmann transfer to Mars,  Louis. 

For min energy Hohmann transfer ellipse,  whose perihelion is at Earth's orbit when Earth is there,  and whose aphelion is at Mars's orbit,  across the sun when Mars is there,  the worst case transfer time is when Earth is farthest from the sun on one side, and Mars is farthest from the sun on the other.  That's the longest-axis transfer ellipse.

The best case transfer time is when Earth is nearest the sun on one side,  and Mars is nearest the sun on the other.  That's the shortest-axis transfer ellipse.  The average distances give you an average transfer ellipse length that is in-between.  For any of these cases,  the transfer time is half the period of the transfer orbit,  which is in turn determined by its axis length.  That's just physics!

It does NOT matter what "source" you want to quote,  the orbital mechanics numbers DO NOT LIE!  Run the numbers for yourself,  if you do not believe me.  Running the numbers is what I did.  Here they are:  worst case 283.4 days.  Average case 258.6 days.  Best case 234.5 days.  Using 30 days as one month,  these range from 7.83 to 9.44 months,  with the average being 8.63. 

Thus there is no "7 month Hohmann transfer".  The shortest is rare,  and just a tad under 8 months. Most of the time it's just about 9 months,  give or take a couple of weeks.

As for repeat-pass aerocapture,  the first pass is THE critical one!  It must decrease your entry interface velocity from above Mars escape to significantly below Mars escape,  or else you will NOT make any subsequent decelerating passes,  you will instead be "lost in space".  Permanently. 

With Hohmann min energy transfer,  allowing for the gravitational pull of Mars accelerating you toward entry,  entry interface speed is 7.5 km/s give or take about 0.5 km/s.  Mars escape speed is 5 km/s.  You have to come out of that aerobraking at a speed near 4 km/s in order not to spend many months in a single long ellipse to the second pass.  (Mars circular low orbit speed is about 3.5 km/s.)

So you have to lose AT LEAST 3 to 4 km/s velocity by hypersonic drag (in a very thin atmosphere) on that very first pass!  For faster transfer trajectories,  the interface velocity and deceleration requirement is higher still,  and it gets that way VERY,  VERY quickly as you try to fly faster.

Thin atmosphere indeed!  That means you dive very deep into that atmosphere to find enough "air" to brake with.  The faster your entry,  the deeper you must dive, the worse the heat shield problem,  the higher the deceleration gee peak,  and the closer you must come to hitting the surface. 

There are some very practical limitations to just how fast you can fly if you intend to use aerobraking instead of propulsion to effect capture at Mars.  That is true for one-pass direct entry as well as multi-pass aerobraking. 

Sorry,  it's just physics.  You cannot wish it away. 

7.5 km/s entry speed is comparable to entry on Earth from LEO (8 km/s).  That's about a 3-4 gee ride,  usually.  The hypersonics take place at comparable speeds and densities,  it's just that the altitudes are far lower on Mars. 

For Apollo coming back from the moon,  entry was a tad under 11 km/s,  and it was a peak 11 gee ride.  See how fast that builds up?  So does the peak heating and total heat absorption to be dealt with,  especially since plasma radiation heating dominates above 10 km/s at entry,  and it is worse than convective heating. You get into trouble VERY quickly! 

I haven't yet set up a spreadsheet to run the numbers for faster trajectories.  But I probably will,  soon.  I may be wrong,  but my hunch is the notion of a 3-month trip to Mars with "Starship" is just utter nonsense!  The speeds at Mars entry will be way too high,  even if the mass ratio works out feasible at zero useful payload.   Which I also doubt.

All that being said,  the numbers I have worked out for "Starship" and posted on "exrocketman" as the "reverse-engineering of the 2019 version",  are based on a vehicle inert mass of 120 metric tons,  and a max propellant load of 1200 metric tons,  fitted with 3 vacuum Raptor engines and 3 sea level Raptor engines.  Note how different the first flight test prototype is:  200 metric tons inert,  and only 3 engines (sea level Raptor). 

By the way,  Spacex thinks currently that they can vector the sea level engines,  but not so much the vacuum engines.  No room to swing those bells.  Which may force them to use the lower-Isp sea level engines to land on Mars.  My numbers assume they are using the vacuum engines to land on Mars.  My payload-to-Mars estimates may be over-optimistic.  Vacuum Isp=355 s for the sea level engines,  versus 381 s for the vacuum engines.  That increases landing propellant required,  which has to come directly out of deliverable payload.

Note also that they have yet to announce in public that they have achieved in tests the full 4400 psia chamber pressure of that Raptor engine design.  Getting the full quoted Isp numbers depends utterly and crucially upon operating routinely at that 4400 psia chamber pressure.  That's just rocket engine/nozzle physics.   You cannot wave your hands at that and get it to go away,  either.

This thing is still a very long way away from any sort of readiness to fly to Earth orbit,  much less the moon or Mars!   And when it does fly,  remember that first and foremost it is a two-stage transport to Earth orbit.  THAT is its primary function!  You have to "misuse" the second stage of an orbital TSTO transport (by on-orbit refilling) to leave Earth orbit at all to go anywhere else.

Now,  as for using Phobos or slingshot things,  that is not currently a family of technologies we can use yet.  Rockets and heat shields are.  That is why I usually don't address those other ideas. 

GW

Last edited by GW Johnson (2019-11-10 11:59:40)


GW Johnson
McGregor,  Texas

"There is nothing as expensive as a dead crew,  especially one dead from a bad management decision"

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#20 2019-11-10 15:09:58

Oldfart1939
Member
Registered: 2016-11-26
Posts: 2,366

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

The table in The Case for Mars, Table 4.2, lists the departure V and associated transit times for free return trajectories to Mars; Zubrin lists 2 departure velocities as being suitable; V = 3.34 km/sec and a 250 day transit time for unmanned cargo flights, and a 5.08 km/sec departure velocity for manned spacecraft with a transit time of 180 days. A 6.93 km/sec velocity is dangerous for aeroentry on Mars, with a 140 day transit time. The 7.93 km/sec departure velocity yields a 130 day transit time with an impossible aeroentry at Mars, but with a free return to Earth. After 4 years in space nearing Jupiter's orbit.

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#21 2019-11-10 15:31:59

Calliban
Member
From: Northern England, UK
Registered: 2019-08-18
Posts: 3,352

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

Using hydrogen-oxygen propellant, increasing dV from 3.4 to 7.93km/s, almost doubles the mass requirement.  It therefore doubles the payload cost.  Until humanity perfects a propulsion system that combines high exhaust velocity with high thrust; it will make sense to minimise dV at the expense of flight time.


"Plan and prepare for every possibility, and you will never act. It is nobler to have courage as we stumble into half the things we fear than to analyse every possible obstacle and begin nothing. Great things are achieved by embracing great dangers."

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#22 2019-11-10 15:52:40

GW Johnson
Member
From: McGregor, Texas USA
Registered: 2011-12-04
Posts: 5,423
Website

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

Oldfart1939: 

Looks like Zubrin has already done what I propose to do,  regarding higher-energy transfer orbits. 

I am showing about 7.5 km/s at Mars entry interface from a min-energy Hohmann transfer.  From Earth LEO,  it is 7.9-to-8.0 km/s.  I think it's really up around 12+ km/s that it gets very dangerous. 

In the old days,  they (NASA) quoted 12-17 km/s entry at Earth coming back from Mars,  for direct (one-pass) aerobraking entry.  That was REALLY challenging,  but supposedly the heat shield on Dragon was originally designed for that. 

Calliban:

There already is a high-thrust high-Isp propulsion technology available whose feasibility was demonstrated about 6 decades ago.  But no one seems to want to use it.  That is nuclear pulse propulsion. 

It gets better performance in very large ship masses.  For a 10,000 ton ship,  it's in the 10,000 to 20,000 sec Isp range,  at ship acceleration levels hard to hold down to 2-4 gees.  Side effects are some radioactive fallout if you want to surface launch from Earth,  and nuclear EMP effects if you fire the thing anywhere within the Van Allen belts.

It's based on 1955-vintage fission device technology.  Shaped charge effects very far from an omni-directional blast weapon.  Fractional-kiloton to few-kiloton yield sizes.  Thousands used at 1 to 2 carefully-ranged and carefully oriented blasts per second. 

We can do better today with nuclear devices.  But it has been forgotten-to-death since 1965.  Makes for a poor small exploration ship design,  but unsurpassed-great for a giant colonization vessel.

GW

Last edited by GW Johnson (2019-11-10 15:53:18)


GW Johnson
McGregor,  Texas

"There is nothing as expensive as a dead crew,  especially one dead from a bad management decision"

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#23 2019-11-10 16:05:03

GW Johnson
Member
From: McGregor, Texas USA
Registered: 2011-12-04
Posts: 5,423
Website

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

Now,  flight rates with Starship/SuperHeavy are still entirely speculative,  as of today.  So are the basic characteristics of the vehicle.  That is because not even the first experimental prototypes have flown.  And the design has changed drastically from year to year,  from first reveal in 2017,  to this year.

OK,  assume for the sake of argument that 4-5 flights per week are feasible.  That's about 200+ flights per year.  For Mars,  6 of every 7 are tankers,  the other being a Mars-bound ship.  For one in 7 of 200+,  that's 30 ships to Mars a year.

Now,  many payload numbers per ship to Mars have been bandied about.  I point out that absolutely NOTHING is yet known as to what the real number might be.  The low is likely near 100 tons.  The highest I have seen is my own estimate of 200 tons.  Let's be optimistic,  and call it 200 tons per ship.  At 30 ships/year,  that's 6000 tons to Mars per year,  max.

So at 6000 tons per year max,  how many years does it take to ship 1 million tons to Mars?  Something like 160 years?  1600 years for 10 million tons?

I think somebody's been smoking a little too much of something.

GW


GW Johnson
McGregor,  Texas

"There is nothing as expensive as a dead crew,  especially one dead from a bad management decision"

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#24 2019-11-10 17:55:54

louis
Member
From: UK
Registered: 2008-03-24
Posts: 7,208

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

Well there's a lot of info out there but not always clear... This article states:

"The total journey time from Earth to Mars takes between 150-300 days depending on the speed of the launch, the alignment of Earth and Mars, and the length of the journey the spacecraft takes to reach its target. It really just depends on how much fuel you’re willing to burn to get there. More fuel, shorter travel time."

https://www.universetoday.com/14841/how … t-to-mars/

Now if that article is correct it's between 5 months and 10 months, which I think is why I was thinking of an average of about 7 months.

The article also claims that:  "The successful Mariner 7 only required 131 days to make the journey."

I am in no position to judge the accuracy of your statements against theirs. Perhaps your figures apply only to Starships configured in line with Musk's presentations...?

GW Johnson wrote:

There is no 7 month Hohmann transfer to Mars,  Louis. 

For min energy Hohmann transfer ellipse,  whose perihelion is at Earth's orbit when Earth is there,  and whose aphelion is at Mars's orbit,  across the sun when Mars is there,  the worst case transfer time is when Earth is farthest from the sun on one side, and Mars is farthest from the sun on the other.  That's the longest-axis transfer ellipse.

The best case transfer time is when Earth is nearest the sun on one side,  and Mars is nearest the sun on the other.  That's the shortest-axis transfer ellipse.  The average distances give you an average transfer ellipse length that is in-between.  For any of these cases,  the transfer time is half the period of the transfer orbit,  which is in turn determined by its axis length.  That's just physics!

It does NOT matter what "source" you want to quote,  the orbital mechanics numbers DO NOT LIE!  Run the numbers for yourself,  if you do not believe me.  Running the numbers is what I did.  Here they are:  worst case 283.4 days.  Average case 258.6 days.  Best case 234.5 days.  Using 30 days as one month,  these range from 7.83 to 9.44 months,  with the average being 8.63. 

Thus there is no "7 month Hohmann transfer".  The shortest is rare,  and just a tad under 8 months. Most of the time it's just about 9 months,  give or take a couple of weeks.

As for repeat-pass aerocapture,  the first pass is THE critical one!  It must decrease your entry interface velocity from above Mars escape to significantly below Mars escape,  or else you will NOT make any subsequent decelerating passes,  you will instead be "lost in space".  Permanently. 

With Hohmann min energy transfer,  allowing for the gravitational pull of Mars accelerating you toward entry,  entry interface speed is 7.5 km/s give or take about 0.5 km/s.  Mars escape speed is 5 km/s.  You have to come out of that aerobraking at a speed near 4 km/s in order not to spend many months in a single long ellipse to the second pass.  (Mars circular low orbit speed is about 3.5 km/s.)

So you have to lose AT LEAST 3 to 4 km/s velocity by hypersonic drag (in a very thin atmosphere) on that very first pass!  For faster transfer trajectories,  the interface velocity and deceleration requirement is higher still,  and it gets that way VERY,  VERY quickly as you try to fly faster.

Thin atmosphere indeed!  That means you dive very deep into that atmosphere to find enough "air" to brake with.  The faster your entry,  the deeper you must dive, the worse the heat shield problem,  the higher the deceleration gee peak,  and the closer you must come to hitting the surface. 

There are some very practical limitations to just how fast you can fly if you intend to use aerobraking instead of propulsion to effect capture at Mars.  That is true for one-pass direct entry as well as multi-pass aerobraking. 

Sorry,  it's just physics.  You cannot wish it away. 

7.5 km/s entry speed is comparable to entry on Earth from LEO (8 km/s).  That's about a 3-4 gee ride,  usually.  The hypersonics take place at comparable speeds and densities,  it's just that the altitudes are far lower on Mars. 

For Apollo coming back from the moon,  entry was a tad under 11 km/s,  and it was a peak 11 gee ride.  See how fast that builds up?  So does the peak heating and total heat absorption to be dealt with,  especially since plasma radiation heating dominates above 10 km/s at entry,  and it is worse than convective heating. You get into trouble VERY quickly! 

I haven't yet set up a spreadsheet to run the numbers for faster trajectories.  But I probably will,  soon.  I may be wrong,  but my hunch is the notion of a 3-month trip to Mars with "Starship" is just utter nonsense!  The speeds at Mars entry will be way too high,  even if the mass ratio works out feasible at zero useful payload.   Which I also doubt.

All that being said,  the numbers I have worked out for "Starship" and posted on "exrocketman" as the "reverse-engineering of the 2019 version",  are based on a vehicle inert mass of 120 metric tons,  and a max propellant load of 1200 metric tons,  fitted with 3 vacuum Raptor engines and 3 sea level Raptor engines.  Note how different the first flight test prototype is:  200 metric tons inert,  and only 3 engines (sea level Raptor). 

By the way,  Spacex thinks currently that they can vector the sea level engines,  but not so much the vacuum engines.  No room to swing those bells.  Which may force them to use the lower-Isp sea level engines to land on Mars.  My numbers assume they are using the vacuum engines to land on Mars.  My payload-to-Mars estimates may be over-optimistic.  Vacuum Isp=355 s for the sea level engines,  versus 381 s for the vacuum engines.  That increases landing propellant required,  which has to come directly out of deliverable payload.

Note also that they have yet to announce in public that they have achieved in tests the full 4400 psia chamber pressure of that Raptor engine design.  Getting the full quoted Isp numbers depends utterly and crucially upon operating routinely at that 4400 psia chamber pressure.  That's just rocket engine/nozzle physics.   You cannot wave your hands at that and get it to go away,  either.

This thing is still a very long way away from any sort of readiness to fly to Earth orbit,  much less the moon or Mars!   And when it does fly,  remember that first and foremost it is a two-stage transport to Earth orbit.  THAT is its primary function!  You have to "misuse" the second stage of an orbital TSTO transport (by on-orbit refilling) to leave Earth orbit at all to go anywhere else.

Now,  as for using Phobos or slingshot things,  that is not currently a family of technologies we can use yet.  Rockets and heat shields are.  That is why I usually don't address those other ideas. 

GW

Last edited by louis (2019-11-10 17:56:49)


Let's Go to Mars...Google on: Fast Track to Mars blogspot.com

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#25 2019-11-10 19:40:44

SpaceNut
Administrator
From: New Hampshire
Registered: 2004-07-22
Posts: 28,747

Re: Musk's Stunning Tweets - 10 million tons to Mars!!

I think that its articles like this one that twists the facts without the full picture.
https://www.space.com/24701-how-long-do … -mars.html

The fastest launched mission https://en.wikipedia.org/wiki/New_Horizons of which there was no refueling, no circular staging which is called a direct path launch.  Atlas V rocket directly into an Earth-and-solar escape trajectory with a speed of about 16.26 km/s (10.10 mi/s; 58,500 km/h; 36,400 mph).

It launched on Atlas V (551) AV-010 with a payload of 30.4 kg (67 lb).

January 19, 2006: Successful launch at 19:00 UTC after a brief delay due to cloud cover
April 7, 2006: The probe passed Mars' orbit 1.5 AU from Earth

Thats a mer 75 days of flying....

https://upload.wikimedia.org/wikipedia/ … e_appr.png

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