Developing the cis-Lunar economy and infrastructure

SpaceNut · 2012-03-01 21:04:24

Then there is the solar panel problem as they can not be extended when taking the dive towards the atmosphere to slow into orbit. Which would mean bigger capacity batteries...

RGClark · 2012-03-03 14:28:58

RGClark wrote:

Hop, I wanted to see what kind of payload we could get to Mars for a sample return mission using one of the current 20 mT payload launchers or the 53 mT payload Falcon Heavy. If you use aerobraking both for landing at Mars and for Earth orbit insertion, what would be the total delta-V to go from LEO to the Martian surface and back to LEO?
From this diagram I get a surprisingly low 10.2 km/s:
Delta-v budget.
Delta-vs between Earth, Moon and Mars.
<image>
http://en.wikipedia.org/wiki/Delta-v_bu … n_and_Mars
LEO to GTO: 2.5 km/s
GTO to Earth C3: .7 km/s
Earth C3 to Mars transfer: .6 km/s
Now notice for the delta-v's after this leading into Mars they all have red arrows indicating this part of the trip can be done by aerobraking. So this portion leading into Mars orbit and landing on the surface is only 3.8 km/s.
Then for the return trip:
Mars(surface) to low Mars orbit: 4.1 km/s
low Mars orbit to Phobos transfer: .9 km/s
Phobos transfer to Deimos transfer: .3 km/s
Deimos transfer to Mars C3: .2 km/s
Mars C3 to Mars transfer: .9 km/s
Now the delta-v's after this leading into Earth all have red arrows indicating this part of the trip can be done by aerobraking. So the return part of the trip can amount to only 6.4 km/s, for a total of 10.2 km/s for the round trip.
As for the heat shield for these Mars return velocities notice that the SpaceX Dragon's PICA-X heat shield was designed to withstand such velocities. It reportedly weighs only half of Apollo era heat shields which would put it at about 8% of the landed mass.

Using a 465.5 s Isp for the LH2/LOX engines and one of the later Centaurs with a ca. 20 mT propellant load and ca. 2 mT dry mass then with a single stage you get a payload of .39 mT: 465.5*9.8ln(1 + 20/(2 + .39)) = 10,206 m/s. The problem is this payload mass also has to account for the heat shield mass for the aerobraking/aerocapture.

We can do better than this using staging, but the problem then is the smaller LH2/LOX stages do not have as good a mass/ratio. For instance see the gross mass/dry mass ratio of the Ariane upper stages here:

http://www.astronautix.com/props/loxlh2.htm

We may suppose that with using lightweight composites we can get a smaller stage with a high mass ratio, say 10 mT propellant load and 1 mT dry mass. Then with staging of two of these we can get a payload of 1.3 mT:

465.5*9.8ln(1 + 10/(1 + 11 + 1.3)) + 465.5*9.8ln(1 + 10/(1 + 1.3)) = 10,206 m/s.

Bob Clark

RGClark · 2012-04-07 09:54:43

I mentioned I had a question for Hop in this thread:

Robotic Mining.
http://newmars.com/forums/viewtopic.php … 92#p111392

It involves orbital mechanics of spaceflight. There are a few different components to the question. First, if our Mars rocket departed from the Moon or a Lagrange point propellant depot fully fueled towards Earth at, say, 11 km or more, so it's moving at speeds beyond Earth's escape velocity, then in just passing by the Earth it should pick up additional speed equal to Earth's escape velocity about 11 km/s. So at least temporarily it should have a speed of 22 km/s. But the problem is that it still be slowed down by the Earth as it proceeds to Mars, so it will lose some of this speed. How much speed will it lose?

What I want to do is leave Earth's vicinity at such high speed so that you don't have the long travel times of the Hohmann orbit, and in fact so that the trajectory approximates a straight-line path and if you do it at closest approach of Mars then the travel time could be say 60,000,000 km/22 km/s = 2,700,000 s, about 31 days. (You would have the problem of aerocapture at such highly elevated speeds but I'll leave that to another discussion.) So another question I have is at what high speed would you need so that the path is approximately straight-line?

This is just using Earth flyby. Could we in addition also use a Venus flyby? You would need an orbital arrangement where both Venus and Mars are near the Earth at the same time. Say you are now traveling at 22 km/s towards Venus, minus the amount you're slowed by leaving the Earth. You can likewise pick up about 11 km/s additional speed by just passing by Venus on the way to Mars, perhaps arranging it so that the path is bent by Venus to aim the craft towards Mars. So you could conceivably be traveling now at 33 km/s, again though I need to know how much speed you would lose in leaving Venus. You would also have to factor in the additional time it takes to get to Venus and the longer straight-line distance to Mars from Venus. Also, in being within Venus's orbit around the Sun, the greater gravitational effects of the Sun will have a greater effect to curve the trajectory.

Finally, could we use repeatedly the gravitational boosts of Earth and Venus? Suppose we are now at 33 km/s, more or less, after leaving Venus but we arrange it so our path is bent completely around to head back towards Earth. Could we once more get an additional 11 km/s to bring our velocity to 44 km/s? Could we do this repeatedly to get arbitrarily high speeds?

Bob Clark

RobS · 2012-04-07 21:47:46

Back to the question what does one need to start a lunar base to mine lunar ices. Michael Duke, when he was working in Houston, made a proposal using the Space Shuttle that would still work--would work better--with Falcon Heavy:

Duke's system involved solar-powered ion engines to lift cargo from low earth orbit to a lagrange point, one of the points where the gravities of the Earth and Moon balance each other (Duke uses L2, which lies beyond the moon).

An eight-tonne solar electric propulsion system, he estimates, can be launched with sixteen tonnes of cargo and transport the latter to the lagrange point over a six-month period. (If the solar-electric vehicle returned to low earth orbit for reuse, the next twenty-four tonne launch presumably could involve eighteen or nineteen tonnes of payload and five or six tonnes of propellant for the ion engine.)

The sixteen tonnes carried to the lagrange point in turn could include eight tonnes of payload for the lunar surface and a reusable eight-tonne lunar-based vehicle (including the hydrogen and oxygen fuel) to transport it there.

One could replace this system with a Falcon 9, which can land a bit more than 8 metric tonnes on the lunar surface. Or if one developed a small solar thermal rocket (where a solar mirror heats a beryllium or graphite block to high temperature and hydrogen is run through it and heated to produce an exhaust velocity of 9 km/sec; about the same as a nuclear thermal rocket), I figure you could get something like 30 tonnes of the 53 into low lunar orbit.

Duke's first launch would put an eight-tonne fuel making plant near the lunar North or South Pole, including a one-tonne nuclear reactor or solar power system able to make 25 kilowatts of power. The plant would ingest ice-laden regolith, heat it, extract the water, convert the water to hydrogen and oxygen, liquefy them, and store them back in the reusable lunar-based vehicle. (Note sheets of ice are hard as rock at that temperature, but a regolith-frost mixture might not be so hard and could be excavated).

The lunar-based vehicle would be filled with sixteen tonnes of hydrogen and oxygen fuel in a few months, enough to launch itself back to the lagrange point with eight tonnes of fuel left over to bring another eight tonnes of cargo to the moon.

A second launch from earth would land a second lunar-based vehicle on the moon with an eight-tonne habitat and supplies. That lunar-based vehicle would also be refueled in a few six months. Then one of the lunar-based vehicles would be launched to the lagrange point to await a crew.

The third launch from Earth would lift an eight-tonne crew vehicle and a sixteen-tonne stage to propel it quickly through earth's Van Allen radiation belts to the lagrange point. It would dock to the waiting lunar-based vehicle, which would land the crew vehicle on the surface and be refueled with the hydrogen and oxygen in the other lunar-based vehicle.

The crew would breathe oxygen and drink water made from lunar ice. Once the crew completed routine maintenance of the fuel-making plant and explored the area, the lunar-based vehicle would launch the crew vehicle back to the lagrange point, where a small engine firing would send it back to the earth's atmosphere for aerobraking and rendezvous with the International Space Station.

Duke's proposal is perhaps the simplest and most elegant proposal for sending astronauts back to the moon that has been make in recent years. Three launches would establish a reusable transportation system to the moon.

He estimated that an eight-tonne fuel-making system on the moon would allow one or two manned flights to the moon per year. The lunar-based vehicles would be designed for ten uses before a solar electric vehicle would have to bring them back to low earth orbit for refurbishment (if a facility for such refurbishment were built).

RGClark · 2012-04-08 01:41:49

RobS wrote:

...
Duke's proposal is perhaps the simplest and most elegant proposal for sending astronauts back to the moon that has been make in recent years. Three launches would establish a reusable transportation system to the moon.
He estimated that an eight-tonne fuel-making system on the moon would allow one or two manned flights to the moon per year. The lunar-based vehicles would be designed for ten uses before a solar electric vehicle would have to bring them back to low earth orbit for refurbishment (if a facility for such refurbishment were built).

Thanks for that. Do you have a link to the full paper?

Bob Clark

RobS · 2012-04-08 07:54:17

Some time ago, David Portee had this great site about Moon and Mars plans, and there was a summary of his plan there, maybe with a link to a longer version. But Portee's website suddenly went away several years ago, which is a real shame; it was really great. The only link I had was to the description there, and it is gone. I may have printed it out, though. If I did, maybe I can scan the paper copy and pdf it. I'll be home tonight and I'll take a look in my files then. The text above I cut and pasted from an article I wrote that space.com published about 6 or 8 years ago.

RobS · 2012-04-08 08:07:29

I should add that Duke's plan used xenon propellant and would have demanded the world's entire annual production of the rare gas, which is more expensive than gold per ounce! A plan that is practical only if you use the space shuttle and launch costs are $20,000 per kilogram. Solar panels are now much more efficient--his plan was made in the early or mid 90s--so a vehicle of the same size would now work with argon, I suppose (which requires more electricity to ionize than xenon). Research on solar thermal rockets has also advanced, since then. A 50 or 100 pound thrust solar rocket, thrusting for 15 minutes every perigee, using hydrogen, can get stuff to the moon (or almost to escape velocity) in a month or two. There is some new research on the web I saw that proposes direct sunlight straight onto the hydrogen through some sort of glass window that could achieve a specific impusle of 1200 seconds. Nuclear thermal can get about 900, so this is much better (that's an exhaust velocity of almost 12 kilometers per second). The down side, of course, is that you need a huge mirror--maybe 70 or 80 meters in diameter--to produce the 50 or 100 pounds (250-500 newtons) of thrust. You can't use this for manned missions to Mars. You could use it to move the bulk of your mass to EML1, then a small rocket would send up the crew and a kick stage, which would send the vehicle back toward earth for trans-Mars injection deep in Earth's gravity well. You could use the solar thermal rocket in interplanetary cruise to shorten the transit time to Mars, but you'd have to use it to slow you down at the other end as well (because if you make the trip too short, you can't aerobrake; the Martian atmosphere isn't thick enough). You could also use a solar thermal rocket cargo flights to low lunar orbit; it'd be very efficient for that purpose. It'd also be great for manned and unmanned missions to Mercury, where the more intense sunlight would make the engine more powerful. Basically, the sun's gravity increases and makes a flight to Mercury harder in proportion to the increase in solar intensity, so they balance out. It might even be possible to use solar thermal to slow a manned vehicle into a Mercury orbit (the crew, of course, would have to land at the poles, an environment similar to the lunar poles, but with even more volatiles, and volatiles that will tell us a lot about the volcanic history of Mercury). I suspect solar thermal is pretty easy to develop, too; someone just needs to give it a try. It's a "poor man's nuke."

RobS · 2012-04-08 23:26:10

I've made a pdf of my printout of the description of Duke's plan. I see no way to attach it to a message, but if anyone wants a copy, email me at rstockma@depaul.edu and I can send it.

RGClark · 2012-04-09 09:59:26

RGClark wrote:

...
It involves orbital mechanics of spaceflight. There are a few different components to the question. First, if our Mars rocket departed from the Moon or a Lagrange point propellant depot fully fueled towards Earth at, say, 11 km or more, so it's moving at speeds beyond Earth's escape velocity, then in just passing by the Earth it should pick up additional speed equal to Earth's escape velocity about 11 km/s. So at least temporarily it should have a speed of 22 km/s. But the problem is that it still be slowed down by the Earth as it proceeds to Mars, so it will lose some of this speed. How much speed will it lose?
What I want to do is leave Earth's vicinity at such high speed so that you don't have the long travel times of the Hohmann orbit, and in fact so that the trajectory approximates a straight-line path and if you do it at closest approach of Mars then the travel time could be say 60,000,000 km/22 km/s = 2,700,000 s, about 31 days. (You would have the problem of aerocapture at such highly elevated speeds but I'll leave that to another discussion.) So another question I have is at what high speed would you need so that the path is approximately straight-line?
This is just using Earth flyby. Could we in addition also use a Venus flyby? You would need an orbital arrangement where both Venus and Mars are near the Earth at the same time. Say you are now traveling at 22 km/s towards Venus, minus the amount you're slowed by leaving the Earth. You can likewise pick up about 11 km/s additional speed by just passing by Venus on the way to Mars, perhaps arranging it so that the path is bent by Venus to aim the craft towards Mars. So you could conceivably be traveling now at 33 km/s, again though I need to know how much speed you would lose in leaving Venus. You would also have to factor in the additional time it takes to get to Venus and the longer straight-line distance to Mars from Venus. Also, in being within Venus's orbit around the Sun, the greater gravitational effects of the Sun will have a greater effect to curve the trajectory.
Finally, could we use repeatedly the gravitational boosts of Earth and Venus? Suppose we are now at 33 km/s, more or less, after leaving Venus but we arrange it so our path is bent completely around to head back towards Earth. Could we once more get an additional 11 km/s to bring our velocity to 44 km/s? Could we do this repeatedly to get arbitrarily high speeds?

I'm having trouble disentangling the gravitational slingshot effect and the Oberth effect.
By the Oberth effect I can get greater velocity if I apply my rocket burn when I'm closest to the planet. Plugging in some speeds into the equation on the Wikipedia page I am able to get an additional boost about that of Earth's or Venus' escape speed if I make the rocket burn high, say, 10 km/s or above. The problem is this page seems to be suggesting to get the gravity boost, I need to apply a rocket burn but I wanted to get the gravity boost without having to apply an additional rocket burn.
On the other hand the Wikipage on the gravitational slingshot effect suggests I can get an additional boost without having to supply an additional burn. But the problem here is I want to get these additional boosts while my craft is already moving at high speed, to boost even higher, but if I'm going too fast I won't swing around the planet but instead go right by it without getting the slingshot effect.
But this slingshot effect is potentially quite large though according to the Wikipedia page. It can be as high as twice the speed of the planet around the Sun. Since for the Earth this is about 30 km/s, this means you can get a boost of about 60 km/s (!) How fast can you be going beforehand and still get swung around by the planet to still get the boost?

Bob Clark

Last edited by RGClark (2012-04-09 10:00:21)

RobS · 2012-04-09 13:53:21

Dear Bob: You'll need to find some equations somewhere to determine how big the slingshot effect is. If you are moving fast already, you enter and leave a gravity well quickly and that gives the planet less time to bend your trajectory. A weak gravitational field can't bend your path practically at all.

To send Messenger to Mercury, NASA had to use an Earth flyby and two Venus flybys. To send Cassini to Saturn they used a Venus flyby and an Earth flyby, if I recall. And in The Promise of Space, Arthur Clarke once commented that the only way to drop a space probe into the sun was to send it to Jupiter, because it was big enough to cancel out the probe's circular motion around the sun.

I have often wondered why NASA didn't use a gravitational slingshot past the moon to send a spacecraft to Mars. The moon has a gravitational well and you'd think it could give a space probe the extra half kilometer or so per second needed to send a spacecraft to Mars. But apparently not.

RGClark · 2012-04-10 14:32:45

RGClark wrote:

...
But this slingshot effect is potentially quite large though according to the Wikipedia page. It can be as high as twice the speed of the planet around the Sun. Since for the Earth this is about 30 km/s, this means you can get a boost of about 60 km/s (!) How fast can you be going beforehand and still get swung around by the planet to still get the boost?

The gravitational slingshot won't work from Earth since the spacecraft even if you launch from the Moon is still moving in the same direction as Earth with respect to the Sun.
It might work from Venus. I remember reading some of the plans to reduce the return time from a Mars mission is to do a swingby of Venus. As I recall though, the reduction in time was not that dramatic as to reduce the trip time to days instead of months, so likely the same would be true for using a Venus swingby for the outbound trip.
I think we could use the Oberth effect though.

Bob Clark

RobS · 2012-04-11 12:38:21

No, I think the moon can provide some slingshot effect. It's moving around the earth at something like half a kilometer per second. Once a month it's moving around the sun half a kilometer per second faster than the Earth and once a month it's moving half a kilometer per second slower. It's basically outside the Earth's gravity well, too. But it has a pretty small gravity well, so I'm not sure it can swing a spaceship approaching the Earth around and slow it relative to the Earth significantly.

RGClark · 2012-04-12 07:45:16

RobS wrote:

Dear Bob: You'll need to find some equations somewhere to determine how big the slingshot effect is. If you are moving fast already, you enter and leave a gravity well quickly and that gives the planet less time to bend your trajectory. A weak gravitational field can't bend your path practically at all.
To send Messenger to Mercury, NASA had to use an Earth flyby and two Venus flybys. To send Cassini to Saturn they used a Venus flyby and an Earth flyby, if I recall. And in The Promise of Space, Arthur Clarke once commented that the only way to drop a space probe into the sun was to send it to Jupiter, because it was big enough to cancel out the probe's circular motion around the sun.
I have often wondered why NASA didn't use a gravitational slingshot past the moon to send a spacecraft to Mars. The moon has a gravitational well and you'd think it could give a space probe the extra half kilometer or so per second needed to send a spacecraft to Mars. But apparently not.

Thanks for that. You couldn't use the gravitational slingshot effect directly with respect to Earth when launching from Earth's vicinity, such as from the Moon or Lagrange points, but your examples suggest you might be able to use it if you first get to Venus or Mars. Then the idea of repeatedly using such flyby's to get to very high speeds might indeed work for a Mars cycler. This possibility was also raised on this page:

Gravitational Slingshot.
http://www.mathpages.com/home/kmath114/kmath114.htm

I like your idea of using a Moon flyby to get an extra 500 m/s delta-V for a Mars mission. On the "Mars Semi-Direct with Falcon" thread you mentioned that a shorter outbound transit time is made possible with a 4.3 km/s delta-V, rather than the Hohmann trajectory delta-V of 3.8 km/s. Then you could use the lunar flyby to get the extra 500 m/s for the shorter transit time.
I wonder if it may be possible to also use the Oberth effect with respect to the Moon at the same time as the lunar gravitational slingshot since they are distinct effects. With a lunar escape velocity of 2.4 km/s, you could get an extra delta-V close to this with the Oberth effect.

I also wonder if these effects can be used also for the return trip with respect to Mars' moons. For instance Phobos has a orbital speed of 2.1 km/s around Mars. Then that could subtract off from the 6.4 km/s to 7 km/s delta-V of the return trip.

Bob Clark

Last edited by RGClark (2012-04-12 07:51:29)

RobS · 2012-04-12 08:02:30

The issue of the "depth" of the gravity well needs to be remembered, though. Basically, the way the "gravity slingshot" works is that the object you are passing changes the direction of your flight. Relative to that object, though, the momentum does not change and you leave the object with the same relative speed you approached it. If you fly by Phobos, it doesn't have the gravity to bend your path significantly. Maybe each encounter can give you a couple meters per second and after a few years your spacecraft is in a highly elliptical orbit. Maybe Phobos can fling you to a few meters per second more than Martian escape velocity, eventually. But that won't get you to Earth, and it may take a looong time. I doubt Luna can bend one's trajectory much, either; otherwise, they would have used it already.

Can you use the gravity slingshot and the Oberth effect? I'm sure you can. The Earth just knows your direction and speed and pulls on you based on your distance.

RGClark · 2012-04-13 01:56:52

RobS wrote:

The issue of the "depth" of the gravity well needs to be remembered, though. Basically, the way the "gravity slingshot" works is that the object you are passing changes the direction of your flight. Relative to that object, though, the momentum does not change and you leave the object with the same relative speed you approached it. If you fly by Phobos, it doesn't have the gravity to bend your path significantly. Maybe each encounter can give you a couple meters per second and after a few years your spacecraft is in a highly elliptical orbit. Maybe Phobos can fling you to a few meters per second more than Martian escape velocity, eventually. But that won't get you to Earth, and it may take a looong time. I doubt Luna can bend one's trajectory much, either; otherwise, they would have used it already.
Can you use the gravity slingshot and the Oberth effect? I'm sure you can. The Earth just knows your direction and speed and pulls on you based on your distance.

Actually, the Moon can because that's what happens with a lunar free return trajectory. But you're right about Phobos. It's escape velocity is too low, only 11 m/s, not km/s, according to the wikipage.

Bob Clark

Terraformer · 2012-04-20 08:47:48

Copy of a post I made on space fellowship re. the mini-moons:

http://news.nationalgeographic.com/news … e-science/
http://www.moondaily.com/reports/Mini_m … h_999.html
http://news.discovery.com/space/how-man … 20406.html
Couple this with the volatiles supply at Luna and we might have a viable business. With a Lunar fuel infrastructure in place, the wholesale capture and return to the Lunar surface of ~100 tonne metal asteroid would be far more feasible. Extract the Platinum and other precious metals on the surface, send them to Terra...
...and crash the planetary market? How much can we sell without having to worry? With a Platinum price of $50k/kg, selling 10 tonnes of the stuff will gross $500M. Enough to make a base viable when coupled with other business?

RobS · 2012-04-20 09:09:43

You don't need to do any crashing. The moon must be peppered with nickel-iron meteoroids that have been there billions of years and have never rusted. Since the moon will have an infrastructure and some gravity, it may be easier to process existing nickel-iron debris there than to go to an asteroid. A nickel-iron asteroid might have no volatiles needed for processing and would have no gravity, which will complicate mining and separation.

Terraformer · 2012-04-20 09:59:07

That's not a problem if you return it wholesale...

Anyway, it's something to bear in mind if Lunar reserves aren't good enough. Besides, bringing them into a Lunar orbit would be good even without using their resources - they provide scientists with an easy to study on-site asteroid. Maybe.

Hmmm, how about tourism to these places...?

RobS · 2012-04-20 10:40:24

Return it wholesale . . . where? How? Cut it into 1-meter spheres and send them into the Australian outback? Remember, a mistake could cost a few trillion dollars, if the pieces hit something big. Send unprocessed chunks to earth in a cargo capsule? An expensive way to import iron. Process the stuff on-site (very difficult without gravity or volatiles) and just send the good stuff home? Crash the pieces into the moon 200 km from a base and build a road to the Meteor-crater sized hole?

Terraformer · 2012-04-20 11:54:34

No, return it wholesale to Luna, process it there, and then send the good stuff home in a capsule - I'm sure someone could build a capsule capable of returning say 4 times it's mass in Platinum from Luna, at maybe 2 tonnes for the capsule. It would be energetically more expensive than processing it onsite, but quite probably cheaper, and you don't have to expose your workers to the deep space radiation environment.

RGClark · 2012-05-31 10:05:15

RGClark wrote:

...
I like your idea of using a Moon flyby to get an extra 500 m/s delta-V for a Mars mission. On the "Mars Semi-Direct with Falcon" thread you mentioned that a shorter outbound transit time is made possible with a 4.3 km/s delta-V, rather than the Hohmann trajectory delta-V of 3.8 km/s. Then you could use the lunar flyby to get the extra 500 m/s for the shorter transit time.
I wonder if it may be possible to also use the Oberth effect with respect to the Moon at the same time as the lunar gravitational slingshot since they are distinct effects. With a lunar escape velocity of 2.4 km/s, you could get an extra delta-V close to this with the Oberth effect.

The Moon's velocity around the Earth is 1 km/s. Then since you could add twice the body's speed to the spacecraft you could conceivably get 2 km/s extra delta-v this way:

Gravitational Slingshot.
http://www.mathpages.com/home/kmath114/kmath114.htm

So is there some reason why this wouldn't work?

Bob Clark

RobS · 2012-05-31 11:47:06

I think we would need to ask someone who knows the physics really well. The velocity change is a function of the direction and speed of approach, which is in turn a function of where you started from. It may be impossible to launch from Earth orbit and get the 2 km delta-v because you are approaching the moon from the wrong direction.

RGClark · 2012-06-11 12:56:35

I really do not like the 6 to 8 month transit times proposed for manned Mars missions. So I wanted to explore generating high delta-v's to get approximately straight-line trajectories to reduce travel time. But I needed to know how high the speed needs to be to get this.

Found this reference after a web search for short Mars transit times:

Entry Velocities at Mars and Earth for Short Transit Times.

Abstract : Propulsion systems composed of a Shuttle External Tank, appropriately modified for the purpose, with a rocket engine that is either an SSME or a NERVA could inject a gross personnel payload of 100,000 lb on a trans-Mars trajectory from Space Station Freedom with aerobraking at Mars with transit times of less that 70 days. Such transit times reflect a significant reduction from the 200- plus days generally considered. The 100,000-lb payload would include the mass of a hypothetical aerobrake for aerocapture at Mars. The entry velocities at Mars compatible with such transit times are greater that 21 km/sec, to be compared with previously stated constraints of 8.5 to 9.5 km/sec for nominal Mars entry velocity. Limits of current aerobrake technology are not well enough defined to determine the feasibility of an aerobrake to handle Mars-entry velocities for short-transit-time trajectories. Return from Mars to Earth on a mirror image of 70-days outbound trajectory (consistent with a stay time of about 12 days) would require a Mars-departure velocity increment more than twice as great as that at Earth departure and would require a correspondingly more capable propulsion system. The return propulsion system would preferably be predeployed at Mars by one or more separate minimum-energy, 0.5-to-1.1-Mlb-gross-payload cargo flights with the same outbound propulsion systems as the personnel flight, before commitment of the personnel flight. Aerobraking entry velocity at Earth after such a transit time would be about 16 km/sec, to be compared with constraints set at 12.5 to 16 km/sec.

www.dtic.mil/dtic/tr/fulltext/u2/a272591.pdf

It gives the equations for the conic section flight paths you would get for high speed departures. For a departure from Earth orbit using an additional delta-v of ca. 8.8 km/s, the transit time to Mars would be about 70 days. See page 14, by the internal page numbering. However, the paper notes after a short stay of 12 days, to make a comparable short transit time of ca. 70 days back to Earth from Mars orbit would require a delta-v twice as large, ca. 18 km/s.

The paper gives both chemical and nuclear propulsion options. For the chemical propulsion it uses the SSME engine, and for the nuclear, NERVA. For the Earth departure, the chemical propulsion version uses a ET style tank at 1,600,000 lb. propellant load, 86,700 lb. vehicle dry mass, and 100,000 lb. payload.

Unfortunately, the paper does not give the structure that would allow a return trip to Earth at a 18 km/s delta-v. Presumably it would be refueled at Mars or Phobos, but it does not give the make-up of such a vehicle. For a delta-v this high it would have to be staged.

Some possibilities for the architecture:

First of all, we'll assume that there are propellant depots in Earth orbit and the vehicles carry along their own propellant production equipment to generate their own propellant on Mars or perhaps Phobos.

Then,
1.) The vehicle carries along an empty propellant tank to be refilled at Mars or Phobos for the return trip.
2.) The vehicle leaves Earth with two fully fueled stages. Only one stage is burned on the way to Mars. At Mars, this depleted stage is used to land on Mars. The unburned stage is left in Mars orbit to link up with the landing stage for the return trip.
3.)Two fully fueled stages leave Earth orbit, and again only one is burned but the vehicle lands now on Phobos. The depleted stage refuels on Phobos. The unburned stage lands on Mars. The two stages link up in space fully fueled for the return trip.

Bob Clark

(Edited for clarity.)

Last edited by RGClark (2012-06-11 23:44:26)

louis · 2012-06-11 18:26:18

RGClark wrote:

I really do not like the 6 to 8 month transit times proposed for manned Mars missions. So I wanted to explore generating high delta-v's to get approximately straight-line trajectories to reduce travel time. But I needed to know how high the speed needs to be to get this.
Found this reference after a web search for short Mars transit times:
Entry Velocities at Mars and Earth for Short Transit Times.
Abstract : Propulsion systems composed of a Shuttle External Tank, appropriately modified for the purpose, with a rocket engine that is either an SSME or a NERVA could inject a gross personnel payload of 100,000 lb on a trans-Mars trajectory from Space Station Freedom with aerobraking at Mars with transit times of less that 70 days. Such transit times reflect a significant reduction from the 200- plus days generally considered. The 100,000-lb payload would include the mass of a hypothetical aerobrake for aerocapture at Mars. The entry velocities at Mars compatible with such transit times are greater that 21 km/sec, to be compared with previously stated constraints of 8.5 to 9.5 km/sec for nominal Mars entry velocity. Limits of current aerobrake technology are not well enough defined to determine the feasibility of an aerobrake to handle Mars-entry velocities for short-transit-time trajectories. Return from Mars to Earth on a mirror image of 70-days outbound trajectory (consistent with a stay time of about 12 days) would require a Mars-departure velocity increment more than twice as great as that at Earth departure and would require a correspondingly more capable propulsion system. The return propulsion system would preferably be predeployed at Mars by one or more separate minimum-energy, 0.5-to-1.1-Mlb-gross-payload cargo flights with the same outbound propulsion systems as the personnel flight, before commitment of the personnel flight. Aerobraking entry velocity at Earth after such a transit time would be about 16 km/sec, to be compared with constraints set at 12.5 to 16 km/sec.
www.dtic.mil/dtic/tr/fulltext/u2/a272591.pdf
It gives the equations for the conic section flight paths you would get for high speed departures. For a departure from Earth orbit using an additional delta-v of ca. 8.8 km/s, the transit time to Mars would be about 70 days. See page 14, by the internal page numbering. However, the paper notes after a short stay of 12 days, to make a comparable short transit time of ca. 70 days back to Earth from Mars orbit would require a delta-v twice as large, ca. 18 km/s.
The paper gives both chemical and nuclear propulsion options. For the chemical propulsion it uses the SSME engine, and for the nuclear, NERVA. For the Earth departure, the chemical propulsion version uses a ET style tank at 1,600,000 lb. propellant load, 86,700 lb. vehicle dry mass, and 100,000 lb. payload.
Unfortunately, the paper does not give the structure that would allow a return trip to Earth at a 18 km/s delta-v. Presumably it would be refueled at Mars or Phobos, but it does not give the make-up of such a vehicle. For a delta-v this high it would have to be staged.
Some possibilities for the architecture:
First of all, we'll assume that there are propellant depots in Earth orbit and the vehicles carry along their own propellant production equipment to generate their own propellant on Mars or perhaps Phobos.
Then,
1.) The vehicle carries along an empty propellant tank to be refilled at Mars or Phobos for the return trip.
2.) The vehicle leaves Earth with two fully fueled stages. Only one stage is burned on the way to Mars. At Mars, the unfueled stage is used to land on Mars. The unburned stage is left in Mars to link up with the landing stage for the return trip.
3.)Two fully fueled stages leave Earth orbit, and again only one is burned but the vehicle lands now on Phobos. The unfueled stage refuels on Phobos. The unburned stage lands on Mars. The two stages link up in space fully fueled for the return trip.

Bob Clark

Excuse my ignorance...

How does the SSME match up with the Falcon series? Is there anything there which could get you to Mars that quickly?

RobS · 2012-06-11 21:14:00

Zubrin's book The Case for Mars has a chart on page 96 of the new edition:

Departure Vel Orbital Period Time to Earth Return Transit to Mars Mars Aeroentry
3.34 km/sec 1.5 years 3 years 250 days easy
5.08 km/sec 2.0 years 2 years 180 days acceptable
6.93 km/sec 3.0 years 3 years 140 days dangerous
7.97 km/sec 4.0 years 4 years 130 days impossible

The departure velocity is your velocity after you leave the Earth's gravitational field, NOT the delta-v. The orbital period is how long your space ship will take to make one orbit around the sun. Time to Earth return is how long before the spacecraft will return to the Earth by itself (a free return trajectory). Transit time is how long to go from Earth to Mars. Mars aeroentry is whether the atmosphere can stop you, based on current technology. If you want to go from Earth to Mars in 70 days, keep in mind that if your have an engine failure during trans-Mars injection, you are in trouble. The orbit you put your spacecraft on may miss Mars if the velocity is wrong, and if you are trying a transit of 70 days, your spacecraft will go well past the orbit of Jupiter before falling back to the orbit of the Earth, at which point everyone on board will have run out of consumables and died. Furthermore, right now we don't know how to use the thin Martian atmosphere to slow a vehicle moving at such a velocity. Zubrin, I think, assumes direct entry to landing. Aerobraking into orbit is easier, but I have my doubts it's possible with a 70-day trajectory.

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