My latest essay at The New Atlantis has come out from behind the paywall.
33 thoughts on “How Starship Changes Everything”
Does Starship change how fast we can get to Mars?
If you can make rocket fuel on the Moon, does change how fast we can get to Mars?
What most useful in order to explore Mars is to get crew fast to Mars.
Can you launch something from Earth and get it to Mars surface [intact] within 3 months?
Can you launch something from the Moon and get it to Mars surface {intact] within 3 months?
How if coming from Venus to Earth, and from the time that you reach Earth’s hill sphere, can get it to Mars surface [intact] within 3 months?
90 day Earth-Mars transit time won’t be happening with Hohmann transfer, so you would need something like constant-boost – which is a whole different set of technologies.
It’s not an either/or choice here. Starship can manage a non-Hohmann ballistic orbit from Earth to Mars that takes 90 days or less by a) choosing its launch window correctly, and b) carrying less payload. Crewed flights will use shorter flight times, cargo longer. Simple as that. Hohmnann Transfer is not the same thing as a ballistic orbit. It’s the name for the lowest energy ballistic orbit.
Constant boost is normally even slower than Hohmann Transfer, because of the weak thrust of efficient ion engines. The only fast transit constant thrust proposal was VASIMR. To make a fast transit with it requires either an 800-ton nuclear reactor activated in LEO or a 600-ton solar array the size of two shopping mall parking lots. The first runs into politcal trouble (No Nukes Ever!), the second into the giggle factor (Look! It’s a MILE LONG spaceship! Tee-hee-hee…).
One thing about the fast ballistic transfer is, Musk has recently acknowledged this will require a two-pass braking event because of the high arrival velocity from interplanetary transfer.
I don’t know if you’ve seen anything on the Alfvenic Plasmoid Reconnection Thruster, but it’s extremely interesting.
Here’s the paper on it. From page 10, some numbers on a small thruster.
5. Thrust and the thrust to power ratio
Because the plasmoids are ejected at the Alfven velocity, the expression for the thrust (Lovberg 1971) becomes F = ρ A V^2, where A is the area of the plasmoid cross section. Notably, the thrust does then not depend on ρ, and it scales as the magnetic field squared (B^2). For example, for plasmoids with radius 10cm (as in Fig. 3(d)) and reconnecting field of B=800G, the calculated thrust is about 50N, taking into account a duty cycle of about 33% (i.e. the distance between two consecutive plasmoids is twice the plasmoid length). The input power is given by Pinj= Iinj Vinj, where Iinj = 2πrBφ/µ0. In general Iinj could vary from a few to a few hundred kA. In our simulations, Iinj is about 100 kA (equivalent to Bφ ∼ 500G), corresponding to about 10 MW of power.
For this unoptimized high-power case (with a trust of 50-100N), the ratio of thrust over power is thus about 5-10 mN/kW. We have not yet performed a systematic optimization, but tentatively the optimal parameter range for this new thruster will be ISP (specific impulse) from 2,000 to 50,000 sec, power from 0.1 to 10 MW and thrust from 1 to 100 Newtons. It would thus occupy a complementary part of parameter space with little overlap with existing thrusters.
It’s got the high specific impulse ranges of the VASMIR, with the added advantage that you could build it in your basement out of old TV parts. The disadvantage, shared with the VASMIR, is that it needs a high-output nuclear reactor or giant solar panels.
And of course one added disadvantage is that the VASMIR is easy to understand because it’s just squeezing and heating, whereas this one will leave rocket scientists staring blankly at their right hand as they try to figure out if their thumb represents the B field, the E field, or the resultant force, and they turn their brain into a pretzel as they try to intuitively understand knotted field lines.
“…mile long spaceship…”
WE BRAKE FOR NOBODY
LOL.
Look at the porkchop plot for transfer times. I’m not sure of the exact number, but if you can spare an extra 3km/s or so, there are launch windows that get you there in about 3 months.
These transfers arrive at Mars faster velocity and at different vector than planet Mars.
I think you have change vector at Earth distance and arrive close to vector of Mars and at velocity slower Mars with a patched conic make up the difference {similar to 7 month trip time to Mars].
I am wondering if vector change from leaving Moon is less than compared low Earth orbit.
Therr’s no reason you can’t make a hyperbolic EDL at Mars, if you have a sufficiently robust entry vehicle, and that was always SpaceX plan until just recently. Musk is now saying at least one aerocapture pass is necessary at Mars, which means the current version of Starship is insufficiently robust. Either way, there’s absolutely no reason to rendezvous with Mars, doubling trip time and increasing the fuel requirement.
That said, this opens up all sorts of interesting possibilities, including getting stranded in a highly eliptical Mars orbit or, if you really screw up, back in solar orbit and not headed for Earth. I’ll follow these developments with interest! Maybe a loaded tanker on Mars ready to fly a rescue mission? Or, since the crew will have years of supplies with them, a rescue mission launched from Earth? Let the speculation begin!
My advice to all and sundry is to stop thinking in OldSpace terms and listen to Musk’s evolving plans.
I am wondering whether lunar surface or LEO is better to get to
Mars. But leaving from Earth/Moon L-1 is obviously better as compared to LEO.
As far as leaving from Lunar surface and 90 day Earth-Mars transit
time, was thinking of leaving lunar surface when the Moon is full, sending rocket thru the Earth/Moon L-2, and continuing going thru the Earth/Sun L-2.
If you are at L-2 you further from the Sun than Earth, and going along slightly longer orbit pathway around the Sun.
Or traveling about 10 million km distance longer in a year, a bit more 1%. 1% = .2978 km/sec faster than Earth’s velocity.
And you do same thing starting from Earth/Moon L-1. You first go to LEO distance and do rocket burn, and would going around
10 km/sec when about 200 km/sec above Earth. And add 7 km/sec delta – V, plus getting oberth effect which adds to rocket’s velocity, and flyby the full Moon and reach the L-2 point distance. As compared starting a 7.8 km/sec in LEO orbit or as compared to having rocket with 7 km/sec delta-v at lunar surface and flying past Earth/Sun L-2 at time the Moon is full.
In terms of a Starship with 6.9 delta-v in LEO, if uses 6.8 of delta-v at LEO, would be fastest thing which every left Earth?
And say Starship only carries 10 tons of payload, and get more delta-v, and flies past full moon and thru and past L-2?
Meant that my reply was here:
I am wondering whether lunar surface or LEO is better to get to Mars.
But leaving from Earth/Moon L-1 is obviously better as compared to LEO.
As far as leaving from Lunar surface and 90 day Earth-Mars transit time,
was thinking of leaving lunar surface when the Moon is full, sending the rocket thru the Earth/Moon L-2,
and continuing going thru the Earth/Sun L-2.
If you are at L-2 you are further from the Sun than Earth, and so going along slightly longer orbit pathway around the Sun.
Or traveling about 10 million km distance longer in a year, a bit more 1%.
1% = .2978 km/sec faster than Earth’s velocity.
And you do same thing starting from Earth/Moon L-1. You first go dow to LEO distance and do rocket burn,
and you would going around 10 km/sec when about 200 km/sec above Earth.
And adding 7 km/sec delta – V from rocket, plus getting oberth effect which adds to rocket’s velocity,
with this velocity flyby the full Moon and reach the L-2 point distance.
As compared starting a 7.8 km/sec in LEO orbit or as compared to having rocket with 7 km/sec delta-v at lunar surface and flying past Earth/Sun L-2 at time the Moon is full.
In terms of a Starship with 6.9 delta-v in LEO, if uses 6.8 of delta-v at LEO, would it be fastest thing which every left Earth?
And say Starship only carries 10 tons of payload, and get more delta-v, and flies past full moon and thru and past L-2?
“Does Starship change how fast we get to mars?”
Yes. We haven’t been there yet, so if we get there, it’s faster than before.
“If you can make rocket fuel on the Moon, does it change how fast we can get to mars?”
Yes. Less gravity well on the moon makes mass cheaper to loft into space. We can use more fuel when it costs less to deliver.
Your last sentence is the key. If Starship makes it possible to put fuel in LEO for $50/lb, at what point does lunar fuel become competitive? And how much infrastructure has to exist on the Moon before that point is reached? And how much will that cost for amortization purposes? Even Mars ISRU works only because of scale and distance.
“Your last sentence is the key. If Starship makes it possible to put fuel in LEO for $50/lb, at what point does lunar fuel become competitive?”
Lunar fuel where?
Before starship was an idea, lunar fuel wasn’t competitive in LEO. Where most competitive is lunar water and lunar LOX on the lunar surface. Next Lunar LOX in Low Lunar orbit, then Lunar LOX in Mars orbit or Venus’s orbit or High Earth orbits.
If lunar water is not very mineable, one might import LH2 to Lunar surface, and/or Low Lunar orbit.
Due to higher costs, I thought one had to export Lunar LOX to low lunar orbit.
Due spaceX lower costs, one should be able to get lower Lunar electrical costs. and not require Lunar export of LOX, but one might export lunar water.
Mars and lunar electrical costs are critical factor of whether one can have towns on Mars. I see no reason Lunar electrical costs should be more expensive than Mars electrical cost- other than perhaps having more market for electrical power on Mars and this allows space power satellites for Mars.
If the moon has mineable water, it seems the Moon will have cheapest electricity beyond Earth surface, unless there is large enough market elsewhere in space. If Venus or Earth orbit can import enough water, it could have cheaper electrical power. If/when Mars has towns, it seems it could export Mars water cheaper in the near term to Venus orbit.
“If Starship makes it possible to put fuel in LEO for $50/lb”
What price does lunar rocket fuel have to be to ship stuff to Mars.
This seems related to can moon get stuff faster to Mars as compared to from LEO.
And it seems where the Moon is in its orbit is related to shipping stuff to Mars {fast}.
The whole idea of mining lunar water is related to small scale lunar mining. Or if doing large scale where you are getting some He3, then you are getting lots of He4 and H2 {and other gases} and huge amounts of Iron- and Thorium and etc. But we are going ignore large scale and focus on relatively small-scale lunar water mining.
Let’s say one mine water to sell at less than $100 per kg.
It seems if lunar water is as cheap as $1 per kg, you talk about lunar settlements {towns on the Moon- like towns on Mars}. So, say, within range of $10 to $100 per kg.
And say electrical power is $1 per Kw hour
So split water is $50 of power per 1 kg of hydrogen and 8 kg of O2 and sell Hydrogen gas at $200 per kg, and LH2 at $225 per kg and LOX at $150 per kg. And also say can sell CO2 at $50 per kg.
So, say Starship brings 100 tons to Moon, and use Starship which can land on Mars. Got Starship with no rocket fuel on the Moon, but can buy lunar rocket, and could go to Mars vs Starship with cargo and/or passengers
at LEO which can refuel at $50 per kg.
Launching a Starship to LEO might soon have fuel as the main price determinant, but we’ll have to see if that gets swamped by the costs of early operations at lunar distances. If a thousand people are employed on the ground, running three shifts monitoring everything that’s going on up there, and each flight spans three weeks, the labor costs will exceed the fuel costs.
I think people are forgetting the scale of this, and are visualizing lunox from a pre-Starship perspective. Each refuel is 1500 metric tons of methane and oxygen. So the fuel and oxydizer load for a hundred such ships is 150,000 metric tons. That’s something the size of a supertanker in low earth orbit. What scale of lunar operation would be necessary to fill that up? By contrast, making it on Earth and sending it up 150 tons at a time is doable with the infrastructure we have now, or will have shortly (basically the large number of Starship launch pads).
Another issue I have, in addition to how much lunar ice really exists, is, proposals to process lunar regolith in bulk for resource sound like processing seawater in bulk for resources. What’s the scale of that operation? I;m in favor of lunar industrial parks, but we don’t have one now.
And, George, that labor cost component is characteristic of airline operations. But it’s not 1965, or even 1995 now. Rather than standing armies of people watching radar screens and shouting warnings to Starship pilots, everything will have to be completely automated. Remember, they’ll soon be launching through a layer of small comsats numbering in excess of 100,000. Like a sky permanently filled with Canada geese the size of refrigerators.
“SpaceX’s INSANE Portable Nuclear Reactor Will Change EVERYTHING!”
SpaceX began reusing those first-stage boosters in 2017, and by May 2021 had flown one of them ten times, meeting the original goal the company had set for reusability.
That aged quickly. With its latest Starlink launch, SpaceX just landed a booster for the 11th time! ^_^
Great article.
It flew it’s 11th mission today
When the US launched the first nuclear powered aircraft carrier, there were many unknown unknowns….
One of the most significant was the MASSIVE reduction in equipment corrosion (deck and aircraft) due to the lack of exhaust gasses…The Navy saved so much in not having corrosion damage to repair, that in the life of the Enterprise, the cost was covered.
I think Space-X will result in a similar Ah-Ha! moment. About what? Unknown unknowns….
If in LEO, it takes about 4.3 km/sec to reach Mars.
And 2.5 km/sec added would be some earth orbit.
And 2.5 km/sec from lunar surface is also gets you some Earth/Moon orbit.
If you had cannon on the lunar surface which gave 4 km/sec, you could get to Mars or Venus. If had the cannon in LEO, it could also get Mars or Venus because traveling at about 7.8 km/sec in LEO orbit and would get the Oberth effect.
With Moon or LEO, one has going to right direction to do this, and using hohmann transfer which is the least delta-v and longest distance to get somewhere in space.
And it seems from Lunar surface, one hit the Earth with the cannon and from LEO one can hit the lunar surface. And cannon shell from the Moon can hit something in LEO a lot faster- particularly if hitting something in LEO at different inclination. But if hitting it in order to hit it at lowest velocity difference, it would be +3 km/sec difference or trying to hit with Hohmann type trajectory with whatever it is, it going in same direction/vector.
Now with rocket in LEO, one match the lunar shell’s velocity- you have go fast enough and go in the right direction. Not sure if LEO cannon shell and Lunar cannon shell can be make to “dock” at a fairly low velocity. Seems easier for lunar shell to miss the cannon, and LEO cannon shoots at the lunar cannon shell. Or it’s easy to make the lunar shell to go faster relative to LEO cannon, at different vector. But can make Lunar cannon travel faster in same vector than +4 km sec?
Would New Moon or Full Moon be faster and/or would moon in perigee and apogee be faster?
It hit Earth, you have reduce the Moon’s orbital velocity, which is about 1 km/sec, or increasing it, moves it away from Earth: https://en.wikipedia.org/wiki/Lunar_phase https://en.wikipedia.org/wiki/Lunar_phase#/media/File:Moon_phases_en.jpg
Give a picture of it. In picture Moon going counterclockwise as viewing above north pole. And cannon in LEO likewise going counterclockwise. It seem it would be easier if LEO cannon was going clockwise {leave moon when full?}, but say going counterclockwise, so guess shot at Earth {and miss the rocky part] when Moon is New. So in 1000 seconds cannon shell going 5 km/sec relative to Moon going 1 km/sec and 5000 km from it. And at 50,000 km Earth gravity is around Moon’s gravity, and Sun’s gravity is insignificant. So say nearside Earth at Zenith and going at 45 degree. Or at pole going same direction. Or 45 degree off Zenith or Earth/Moon L-1 but within 1000 second that angle is reduced, in regards to Earth {one thing is you going above Lunar escape velocity]. And at 10,000 seconds, you are getting point of Earth’s gravity is as strong as the Moon. In terms of gravity loss say .8 m/s/s or .8 km/sec at vector of 45 degrees, and Earth weaker but increase gravity is roughly at 90 degrees and increases the slower you are going in a lunar orbit. Or could say Moon “trying” to pull cannon shell it’s direction [and losing that battle]. Anyhow in 10,000 seconds one going more directly towards Earth. Though 45 angle could not the proper angle to miss Earth as one wants to exactly. Or just guessing and using “round numbers”. Though one also ask do you want to leave when exactly at “new moon phase” it could better closer to “waxing crescent”.
Now as move faster toward a gravity well {ie Earth] you get less “gravity assist” or spending less time gaining velocity from the gravity, it seems leaving at nearer “waxing crescent” is better.
Or at 100,000 km from Moon, you have pulled closer to the L-1 point and you are in L-1 “space/zone”, maybe about 20,000 km from point [and getting further from it].
When you write about the necessary reliability of satellites that make them expensive then the knock-on reliability of rockets that make them expensive; I thought about all the time spent analyzing computer models due to lack of ability to conduct routine testing in the actual integrated flight regime (as is done with certified airframes). Years are lost before flights due to the time spent running models over and over in the hope to get enough data to make up for lack of flying.
That difference is apparent in the approaches to Manned Dragon versus Boeing’s Starliner or the Orion. Regular Dragons have flown about 20 missions (not counting various test flights), and three of the Dragon capsules have now flown three missions each. Crew Dragons have flown four cargo missions and five manned missions, for a total of eighteen people. Starliner and Orion have flown a combined total of zero missions.
Elon just embiggened the upper stage with six raptor vacs for a total of 9 engines. Plus a tanks stretch.
And yes, before anyone asks, embiggen is a perfectly cromulent word.
Yo, Matt Groening isn’t exactly Shakespeare?
It is around here. Rand is a fan too.
I wonder if SpaceX will solo test fly the new version?
We should place bets on whether he did it for technical reasons, or just so there’d be 42 engines after all…
“I think people are forgetting the scale of this, and are visualizing lunox from a pre-Starship perspective. Each refuel is 1500 metric tons of methane and oxygen. So the fuel and oxydizer load for a hundred such ships is 150,000 metric tons. That’s something the size of a supertanker in low earth orbit. What scale of lunar operation would be necessary to fill that up?”
Well, if talking about starship which uses methane, and important factor is how much CO2 is there on the Moon.
With the Moon, it said there is billions of tons of water at lunar polar region, and then you have question of billions tons how much is the most mineable.
If mineable is 10% water, there could be no findable deposit which is 10% or more of lunar water.
Or could be million of tons of water which 10% or more lunar water.
But what about 5% water and 5% CO2 or 5% water and 10% CO2?
Frozen lunar CO2 could cheaper than getting CO2 from Mars atmosphere. Frozen CO2 could be regarded as resource of CO2 at high pressure- put in pressure pot, warm it in sunlight.
Though seems like if there frozen CO2 on lunar surface, it will be mixed to water and other volatiles.
Likewise, with frozen water.
Anyhow, like Mars, if bring LH2 one can make methane.
If brought 60 tons of LH2, you need 420 tons of CO2 which makes
160 ton of CH4 and 320 tons of water. Then split water and get more H2 and O2. And with H2, and CO2 can make more Methane.
1/9 of water is H2, 320 / 9 = 35.55 tons of H2, so a bit more than 80 tons of Methane with a bit more than 210 tons of CO2. And a bit more than 160 tons water {halving each time, but at this point trading 60 tons of H2 for +240 tons of Methane and +160 tons of water. Can bring 1/2 of passengers and rest payload is LH2. Passengers will need water to travel to Mars, but less than 160 tons.
Maybe send two starship to Moon, and one of starship goes to Mars.
The two make 2 times +240 tons of Methane and +160 tons of water. split 320 tons water give: 80 + 480 = 560 tons of methane with more water than you need for trip. And got enough LOX and sell the empty starship on lunar surface to the Lunatics.
If Lunatics get to point mining lots H2, they could export frozen/slush H2 to Mars with their bought starship. Maybe a Martian wants to buy a starship on surface of Mars. But probably sending 100 tons of “lunar samples” to Earth, would be more profitable in near term.
There’s no reason to carry H2 to Mars (which was part of the original Mars Direct plan). Mars has plenty of water, much of it in the form of a near surface permafrost layer. So the first two cargo ships will land near a known deposit, with drill rigs, power shovels, and robots, Sabatier reactors, and compressors (for the CO2 atmosphere), along with deployable solar panels and batteries. Their main task will be to see if they can make methane and oxygen, with which to refill the tanks of the two ships.
When two more cargo ships, and two crew ships arrive the following synod, there should be enough methalox to refuel them both in case they need to come home right away. If not, I think one crew ship will head for Earth after 30 days, and the other one will stay, so it’s crew can set up the beginnings of the first Mars Base, and have it ready when eight more ships arrive at the next synod.
Wasn’t the Falcon 9 cheaper than the alternatives even when flown expendably? Musk could have stopped right there and still been considered a genius innovator.
What companies are making plug and play methods to de-orbit all the new objects in space that had less emphasis placed on reliability when they were constructed?
Cheaper than comparable rockets, but not cheaper than alternatives. Soyuz 2.1a price is maybe $17.5mln (it’s hard to tell), whereas expendable Falcon 9 is at least $50mln. Semi-reusability cuts the Falcon price in half (again, it’s hard to tell). And then Inida’s PSLV may be even cheaper (who knows?). Expendable Falcon 9’s big selling point is, it’s comparable to Angara A5, at half the price.
Does Starship change how fast we can get to Mars?
If you can make rocket fuel on the Moon, does change how fast we can get to Mars?
What most useful in order to explore Mars is to get crew fast to Mars.
Can you launch something from Earth and get it to Mars surface [intact] within 3 months?
Can you launch something from the Moon and get it to Mars surface {intact] within 3 months?
How if coming from Venus to Earth, and from the time that you reach Earth’s hill sphere, can get it to Mars surface [intact] within 3 months?
90 day Earth-Mars transit time won’t be happening with Hohmann transfer, so you would need something like constant-boost – which is a whole different set of technologies.
It’s not an either/or choice here. Starship can manage a non-Hohmann ballistic orbit from Earth to Mars that takes 90 days or less by a) choosing its launch window correctly, and b) carrying less payload. Crewed flights will use shorter flight times, cargo longer. Simple as that. Hohmnann Transfer is not the same thing as a ballistic orbit. It’s the name for the lowest energy ballistic orbit.
Constant boost is normally even slower than Hohmann Transfer, because of the weak thrust of efficient ion engines. The only fast transit constant thrust proposal was VASIMR. To make a fast transit with it requires either an 800-ton nuclear reactor activated in LEO or a 600-ton solar array the size of two shopping mall parking lots. The first runs into politcal trouble (No Nukes Ever!), the second into the giggle factor (Look! It’s a MILE LONG spaceship! Tee-hee-hee…).
One thing about the fast ballistic transfer is, Musk has recently acknowledged this will require a two-pass braking event because of the high arrival velocity from interplanetary transfer.
I don’t know if you’ve seen anything on the Alfvenic Plasmoid Reconnection Thruster, but it’s extremely interesting.
Youtube video
Here’s the paper on it. From page 10, some numbers on a small thruster.
It’s got the high specific impulse ranges of the VASMIR, with the added advantage that you could build it in your basement out of old TV parts. The disadvantage, shared with the VASMIR, is that it needs a high-output nuclear reactor or giant solar panels.
And of course one added disadvantage is that the VASMIR is easy to understand because it’s just squeezing and heating, whereas this one will leave rocket scientists staring blankly at their right hand as they try to figure out if their thumb represents the B field, the E field, or the resultant force, and they turn their brain into a pretzel as they try to intuitively understand knotted field lines.
“…mile long spaceship…”
WE BRAKE FOR NOBODY
LOL.
Look at the porkchop plot for transfer times. I’m not sure of the exact number, but if you can spare an extra 3km/s or so, there are launch windows that get you there in about 3 months.
https://marspedia.org/Earth-Mars_Transfer_Trajectory
These transfers arrive at Mars faster velocity and at different vector than planet Mars.
I think you have change vector at Earth distance and arrive close to vector of Mars and at velocity slower Mars with a patched conic make up the difference {similar to 7 month trip time to Mars].
I am wondering if vector change from leaving Moon is less than compared low Earth orbit.
Therr’s no reason you can’t make a hyperbolic EDL at Mars, if you have a sufficiently robust entry vehicle, and that was always SpaceX plan until just recently. Musk is now saying at least one aerocapture pass is necessary at Mars, which means the current version of Starship is insufficiently robust. Either way, there’s absolutely no reason to rendezvous with Mars, doubling trip time and increasing the fuel requirement.
That said, this opens up all sorts of interesting possibilities, including getting stranded in a highly eliptical Mars orbit or, if you really screw up, back in solar orbit and not headed for Earth. I’ll follow these developments with interest! Maybe a loaded tanker on Mars ready to fly a rescue mission? Or, since the crew will have years of supplies with them, a rescue mission launched from Earth? Let the speculation begin!
My advice to all and sundry is to stop thinking in OldSpace terms and listen to Musk’s evolving plans.
I am wondering whether lunar surface or LEO is better to get to
Mars. But leaving from Earth/Moon L-1 is obviously better as compared to LEO.
As far as leaving from Lunar surface and 90 day Earth-Mars transit
time, was thinking of leaving lunar surface when the Moon is full, sending rocket thru the Earth/Moon L-2, and continuing going thru the Earth/Sun L-2.
If you are at L-2 you further from the Sun than Earth, and going along slightly longer orbit pathway around the Sun.
Or traveling about 10 million km distance longer in a year, a bit more 1%. 1% = .2978 km/sec faster than Earth’s velocity.
And you do same thing starting from Earth/Moon L-1. You first go to LEO distance and do rocket burn, and would going around
10 km/sec when about 200 km/sec above Earth. And add 7 km/sec delta – V, plus getting oberth effect which adds to rocket’s velocity, and flyby the full Moon and reach the L-2 point distance. As compared starting a 7.8 km/sec in LEO orbit or as compared to having rocket with 7 km/sec delta-v at lunar surface and flying past Earth/Sun L-2 at time the Moon is full.
In terms of a Starship with 6.9 delta-v in LEO, if uses 6.8 of delta-v at LEO, would be fastest thing which every left Earth?
And say Starship only carries 10 tons of payload, and get more delta-v, and flies past full moon and thru and past L-2?
Meant that my reply was here:
I am wondering whether lunar surface or LEO is better to get to Mars.
But leaving from Earth/Moon L-1 is obviously better as compared to LEO.
As far as leaving from Lunar surface and 90 day Earth-Mars transit time,
was thinking of leaving lunar surface when the Moon is full, sending the rocket thru the Earth/Moon L-2,
and continuing going thru the Earth/Sun L-2.
If you are at L-2 you are further from the Sun than Earth, and so going along slightly longer orbit pathway around the Sun.
Or traveling about 10 million km distance longer in a year, a bit more 1%.
1% = .2978 km/sec faster than Earth’s velocity.
And you do same thing starting from Earth/Moon L-1. You first go dow to LEO distance and do rocket burn,
and you would going around 10 km/sec when about 200 km/sec above Earth.
And adding 7 km/sec delta – V from rocket, plus getting oberth effect which adds to rocket’s velocity,
with this velocity flyby the full Moon and reach the L-2 point distance.
As compared starting a 7.8 km/sec in LEO orbit or as compared to having rocket with 7 km/sec delta-v at lunar surface and flying past Earth/Sun L-2 at time the Moon is full.
In terms of a Starship with 6.9 delta-v in LEO, if uses 6.8 of delta-v at LEO, would it be fastest thing which every left Earth?
And say Starship only carries 10 tons of payload, and get more delta-v, and flies past full moon and thru and past L-2?
“Does Starship change how fast we get to mars?”
Yes. We haven’t been there yet, so if we get there, it’s faster than before.
“If you can make rocket fuel on the Moon, does it change how fast we can get to mars?”
Yes. Less gravity well on the moon makes mass cheaper to loft into space. We can use more fuel when it costs less to deliver.
Your last sentence is the key. If Starship makes it possible to put fuel in LEO for $50/lb, at what point does lunar fuel become competitive? And how much infrastructure has to exist on the Moon before that point is reached? And how much will that cost for amortization purposes? Even Mars ISRU works only because of scale and distance.
“Your last sentence is the key. If Starship makes it possible to put fuel in LEO for $50/lb, at what point does lunar fuel become competitive?”
Lunar fuel where?
Before starship was an idea, lunar fuel wasn’t competitive in LEO. Where most competitive is lunar water and lunar LOX on the lunar surface. Next Lunar LOX in Low Lunar orbit, then Lunar LOX in Mars orbit or Venus’s orbit or High Earth orbits.
If lunar water is not very mineable, one might import LH2 to Lunar surface, and/or Low Lunar orbit.
Due to higher costs, I thought one had to export Lunar LOX to low lunar orbit.
Due spaceX lower costs, one should be able to get lower Lunar electrical costs. and not require Lunar export of LOX, but one might export lunar water.
Mars and lunar electrical costs are critical factor of whether one can have towns on Mars. I see no reason Lunar electrical costs should be more expensive than Mars electrical cost- other than perhaps having more market for electrical power on Mars and this allows space power satellites for Mars.
If the moon has mineable water, it seems the Moon will have cheapest electricity beyond Earth surface, unless there is large enough market elsewhere in space. If Venus or Earth orbit can import enough water, it could have cheaper electrical power. If/when Mars has towns, it seems it could export Mars water cheaper in the near term to Venus orbit.
“If Starship makes it possible to put fuel in LEO for $50/lb”
What price does lunar rocket fuel have to be to ship stuff to Mars.
This seems related to can moon get stuff faster to Mars as compared to from LEO.
And it seems where the Moon is in its orbit is related to shipping stuff to Mars {fast}.
The whole idea of mining lunar water is related to small scale lunar mining. Or if doing large scale where you are getting some He3, then you are getting lots of He4 and H2 {and other gases} and huge amounts of Iron- and Thorium and etc. But we are going ignore large scale and focus on relatively small-scale lunar water mining.
Let’s say one mine water to sell at less than $100 per kg.
It seems if lunar water is as cheap as $1 per kg, you talk about lunar settlements {towns on the Moon- like towns on Mars}. So, say, within range of $10 to $100 per kg.
And say electrical power is $1 per Kw hour
So split water is $50 of power per 1 kg of hydrogen and 8 kg of O2 and sell Hydrogen gas at $200 per kg, and LH2 at $225 per kg and LOX at $150 per kg. And also say can sell CO2 at $50 per kg.
So, say Starship brings 100 tons to Moon, and use Starship which can land on Mars. Got Starship with no rocket fuel on the Moon, but can buy lunar rocket, and could go to Mars vs Starship with cargo and/or passengers
at LEO which can refuel at $50 per kg.
Launching a Starship to LEO might soon have fuel as the main price determinant, but we’ll have to see if that gets swamped by the costs of early operations at lunar distances. If a thousand people are employed on the ground, running three shifts monitoring everything that’s going on up there, and each flight spans three weeks, the labor costs will exceed the fuel costs.
I think people are forgetting the scale of this, and are visualizing lunox from a pre-Starship perspective. Each refuel is 1500 metric tons of methane and oxygen. So the fuel and oxydizer load for a hundred such ships is 150,000 metric tons. That’s something the size of a supertanker in low earth orbit. What scale of lunar operation would be necessary to fill that up? By contrast, making it on Earth and sending it up 150 tons at a time is doable with the infrastructure we have now, or will have shortly (basically the large number of Starship launch pads).
Another issue I have, in addition to how much lunar ice really exists, is, proposals to process lunar regolith in bulk for resource sound like processing seawater in bulk for resources. What’s the scale of that operation? I;m in favor of lunar industrial parks, but we don’t have one now.
And, George, that labor cost component is characteristic of airline operations. But it’s not 1965, or even 1995 now. Rather than standing armies of people watching radar screens and shouting warnings to Starship pilots, everything will have to be completely automated. Remember, they’ll soon be launching through a layer of small comsats numbering in excess of 100,000. Like a sky permanently filled with Canada geese the size of refrigerators.
“SpaceX’s INSANE Portable Nuclear Reactor Will Change EVERYTHING!”
https://www.youtube.com/watch?v=2RRmtSIgqgo
That aged quickly. With its latest Starlink launch, SpaceX just landed a booster for the 11th time! ^_^
Great article.
It flew it’s 11th mission today
When the US launched the first nuclear powered aircraft carrier, there were many unknown unknowns….
One of the most significant was the MASSIVE reduction in equipment corrosion (deck and aircraft) due to the lack of exhaust gasses…The Navy saved so much in not having corrosion damage to repair, that in the life of the Enterprise, the cost was covered.
I think Space-X will result in a similar Ah-Ha! moment. About what? Unknown unknowns….
If in LEO, it takes about 4.3 km/sec to reach Mars.
And 2.5 km/sec added would be some earth orbit.
And 2.5 km/sec from lunar surface is also gets you some Earth/Moon orbit.
If you had cannon on the lunar surface which gave 4 km/sec, you could get to Mars or Venus. If had the cannon in LEO, it could also get Mars or Venus because traveling at about 7.8 km/sec in LEO orbit and would get the Oberth effect.
With Moon or LEO, one has going to right direction to do this, and using hohmann transfer which is the least delta-v and longest distance to get somewhere in space.
And it seems from Lunar surface, one hit the Earth with the cannon and from LEO one can hit the lunar surface. And cannon shell from the Moon can hit something in LEO a lot faster- particularly if hitting something in LEO at different inclination. But if hitting it in order to hit it at lowest velocity difference, it would be +3 km/sec difference or trying to hit with Hohmann type trajectory with whatever it is, it going in same direction/vector.
Now with rocket in LEO, one match the lunar shell’s velocity- you have go fast enough and go in the right direction. Not sure if LEO cannon shell and Lunar cannon shell can be make to “dock” at a fairly low velocity. Seems easier for lunar shell to miss the cannon, and LEO cannon shoots at the lunar cannon shell. Or it’s easy to make the lunar shell to go faster relative to LEO cannon, at different vector. But can make Lunar cannon travel faster in same vector than +4 km sec?
Would New Moon or Full Moon be faster and/or would moon in perigee and apogee be faster?
It hit Earth, you have reduce the Moon’s orbital velocity, which is about 1 km/sec, or increasing it, moves it away from Earth:
https://en.wikipedia.org/wiki/Lunar_phase
https://en.wikipedia.org/wiki/Lunar_phase#/media/File:Moon_phases_en.jpg
Give a picture of it. In picture Moon going counterclockwise as viewing above north pole. And cannon in LEO likewise going counterclockwise. It seem it would be easier if LEO cannon was going clockwise {leave moon when full?}, but say going counterclockwise, so guess shot at Earth {and miss the rocky part] when Moon is New. So in 1000 seconds cannon shell going 5 km/sec relative to Moon going 1 km/sec and 5000 km from it. And at 50,000 km Earth gravity is around Moon’s gravity, and Sun’s gravity is insignificant. So say nearside Earth at Zenith and going at 45 degree. Or at pole going same direction. Or 45 degree off Zenith or Earth/Moon L-1 but within 1000 second that angle is reduced, in regards to Earth {one thing is you going above Lunar escape velocity]. And at 10,000 seconds, you are getting point of Earth’s gravity is as strong as the Moon. In terms of gravity loss say .8 m/s/s or .8 km/sec at vector of 45 degrees, and Earth weaker but increase gravity is roughly at 90 degrees and increases the slower you are going in a lunar orbit. Or could say Moon “trying” to pull cannon shell it’s direction [and losing that battle]. Anyhow in 10,000 seconds one going more directly towards Earth. Though 45 angle could not the proper angle to miss Earth as one wants to exactly. Or just guessing and using “round numbers”. Though one also ask do you want to leave when exactly at “new moon phase” it could better closer to “waxing crescent”.
Now as move faster toward a gravity well {ie Earth] you get less “gravity assist” or spending less time gaining velocity from the gravity, it seems leaving at nearer “waxing crescent” is better.
Or at 100,000 km from Moon, you have pulled closer to the L-1 point and you are in L-1 “space/zone”, maybe about 20,000 km from point [and getting further from it].
When you write about the necessary reliability of satellites that make them expensive then the knock-on reliability of rockets that make them expensive; I thought about all the time spent analyzing computer models due to lack of ability to conduct routine testing in the actual integrated flight regime (as is done with certified airframes). Years are lost before flights due to the time spent running models over and over in the hope to get enough data to make up for lack of flying.
That difference is apparent in the approaches to Manned Dragon versus Boeing’s Starliner or the Orion. Regular Dragons have flown about 20 missions (not counting various test flights), and three of the Dragon capsules have now flown three missions each. Crew Dragons have flown four cargo missions and five manned missions, for a total of eighteen people. Starliner and Orion have flown a combined total of zero missions.
Elon just embiggened the upper stage with six raptor vacs for a total of 9 engines. Plus a tanks stretch.
And yes, before anyone asks, embiggen is a perfectly cromulent word.
Yo, Matt Groening isn’t exactly Shakespeare?
It is around here. Rand is a fan too.
I wonder if SpaceX will solo test fly the new version?
We should place bets on whether he did it for technical reasons, or just so there’d be 42 engines after all…
“I think people are forgetting the scale of this, and are visualizing lunox from a pre-Starship perspective. Each refuel is 1500 metric tons of methane and oxygen. So the fuel and oxydizer load for a hundred such ships is 150,000 metric tons. That’s something the size of a supertanker in low earth orbit. What scale of lunar operation would be necessary to fill that up?”
Well, if talking about starship which uses methane, and important factor is how much CO2 is there on the Moon.
With the Moon, it said there is billions of tons of water at lunar polar region, and then you have question of billions tons how much is the most mineable.
If mineable is 10% water, there could be no findable deposit which is 10% or more of lunar water.
Or could be million of tons of water which 10% or more lunar water.
But what about 5% water and 5% CO2 or 5% water and 10% CO2?
Frozen lunar CO2 could cheaper than getting CO2 from Mars atmosphere. Frozen CO2 could be regarded as resource of CO2 at high pressure- put in pressure pot, warm it in sunlight.
Though seems like if there frozen CO2 on lunar surface, it will be mixed to water and other volatiles.
Likewise, with frozen water.
Anyhow, like Mars, if bring LH2 one can make methane.
If brought 60 tons of LH2, you need 420 tons of CO2 which makes
160 ton of CH4 and 320 tons of water. Then split water and get more H2 and O2. And with H2, and CO2 can make more Methane.
1/9 of water is H2, 320 / 9 = 35.55 tons of H2, so a bit more than 80 tons of Methane with a bit more than 210 tons of CO2. And a bit more than 160 tons water {halving each time, but at this point trading 60 tons of H2 for +240 tons of Methane and +160 tons of water. Can bring 1/2 of passengers and rest payload is LH2. Passengers will need water to travel to Mars, but less than 160 tons.
Maybe send two starship to Moon, and one of starship goes to Mars.
The two make 2 times +240 tons of Methane and +160 tons of water. split 320 tons water give: 80 + 480 = 560 tons of methane with more water than you need for trip. And got enough LOX and sell the empty starship on lunar surface to the Lunatics.
If Lunatics get to point mining lots H2, they could export frozen/slush H2 to Mars with their bought starship. Maybe a Martian wants to buy a starship on surface of Mars. But probably sending 100 tons of “lunar samples” to Earth, would be more profitable in near term.
There’s no reason to carry H2 to Mars (which was part of the original Mars Direct plan). Mars has plenty of water, much of it in the form of a near surface permafrost layer. So the first two cargo ships will land near a known deposit, with drill rigs, power shovels, and robots, Sabatier reactors, and compressors (for the CO2 atmosphere), along with deployable solar panels and batteries. Their main task will be to see if they can make methane and oxygen, with which to refill the tanks of the two ships.
When two more cargo ships, and two crew ships arrive the following synod, there should be enough methalox to refuel them both in case they need to come home right away. If not, I think one crew ship will head for Earth after 30 days, and the other one will stay, so it’s crew can set up the beginnings of the first Mars Base, and have it ready when eight more ships arrive at the next synod.
Wasn’t the Falcon 9 cheaper than the alternatives even when flown expendably? Musk could have stopped right there and still been considered a genius innovator.
What companies are making plug and play methods to de-orbit all the new objects in space that had less emphasis placed on reliability when they were constructed?
Cheaper than comparable rockets, but not cheaper than alternatives. Soyuz 2.1a price is maybe $17.5mln (it’s hard to tell), whereas expendable Falcon 9 is at least $50mln. Semi-reusability cuts the Falcon price in half (again, it’s hard to tell). And then Inida’s PSLV may be even cheaper (who knows?). Expendable Falcon 9’s big selling point is, it’s comparable to Angara A5, at half the price.