Starship On-orbit refueling - Options and Discussion

[OTV Booster]

When the airstream hits them just right I bet they whistle.

A whistling rocket?...that could be very interesting during lift-offs(if you can hear it over the engines) and the banking maneuver during landings.

They put a siren on it figuring that would loosen some sphincters. I think they were right.

[OTV Booster]

You just need a sunshade, maybe an earth shade, to keep methalox boiloff low.

Heck, a flat plate shielded from direct sunshine, painted the Earthside white with 95% reflectivity and the space-side at 0 albedo, the equilibrium temperature is around 115-120K at 500km or so.

A cylinder is more complicated, but a deployable shade really helps.

"Sunshade," eh? Well I guess I never thought about it before, but technically paint does cast a shadow on the surface immediately below it... 

A deployable system is doable (certainly for SpaceX), but R&D tends to be slow. I expect they prefer a static system.

"The best part is no part. The best process is no process."

Deployment can't fail if there's no deployment.

Need to wait and see how surface application works out but if necessary a relatively simple additional rectangular shade to block earthshine only could be a minimum viable hack. Keep the nose towards the sun.


Foam, paint and MLRI would be best if it works out.

[Twark_Main]

"Sunshade," eh? Well I guess I never thought about it before, but technically paint does cast a shadow on the surface immediately below it... 

I know you're kidding, but just for the sake of completeness, the paint conducts heat because it's in physical contact with the insulation, while a sunshade doesn't.

Can't conduct heat it doesn't absorb.    This is why solar white paint is still quite beneficial.

While we're on the subject of completeness, sunshades have radiative heat transfer because they're within the sight hemisphere of the insulation.

[dchenevert]

I'm curious about launch windows for orbitil refueling.
assumption: a depot is launched then  number of tankers are launched, with some amount of "haste". all launches are from starbase.

after the depot is launched, there is* a one orbit (about 90 minute) delay until the first tanker can be launched--the depot won't be overhead until then.

by "is*" I mean: it's easier for the 1st tanker if the depot passes "right overhead". But that's not trivial--the earth has turned in the mean time, so starbase is 90 minutes "to the east" of where it should be, at the speed of earth's spin.

how would this work in practice? I can think of a few options but I'd just be stumbling around mentally.

2nd question: how long do you think a "depot" load will last? for example, 50% boil-off in 3 days? 3 months? Answer can be really sloppy ("musks promises plus murphy's law plus some physics plus some engineering. and a little red bull")

3rd question: could a depot self-transport to LRHO? how much fuel would be left at that point? asking for a friend

[dchenevert]

in my fanboi mind, I could see an uncrewed HLS starship take off from starbase, and wait in orbit for dragon to bring the crew anad dock with it--then land on the moon, take off, return to earth orbit, and wait for a dragon to remove the crew.

if the solar system were littered with fully fueled depots -> probably not far from possible.
what is the actual math on this, assuming budget for multiple depots and dozens of tanker launches? I think this is a delta-V question but if not, go ahead and talk down to me about delta-v, I deserve it

By the way, just saying: I figger a successful artemis III will take about 6 depot launches, what with practice and dry-run-lnding-on-moon-with-uncrewed-hls-test-article. and gross estimate, 90 tanker launches.

[DanClemmensen]

in my fanboi mind, I could see an uncrewed HLS starship take off from starbase, and wait in orbit for dragon to bring the crew anad dock with it--then land on the moon, take off, return to earth orbit, and wait for a dragon to remove the crew.

if the solar system were littered with fully fueled depots -> probably not far from possible.
what is the actual math on this, assuming budget for multiple depots and dozens of tanker launches? I think this is a delta-V question but if not, go ahead and talk down to me about delta-v, I deserve it

By the way, just saying: I figger a successful artemis III will take about 6 depot launches, what with practice and dry-run-lnding-on-moon-with-uncrewed-hls-test-article. and gross estimate, 90 tanker launches.

This has been discussed at great length, with many variants, and discussing several obscure but solvable problems, all at:
    https://forum.nasaspaceflight.com/index.php?topic=59662.0

[rsdavis9]

In general and simplistically.
Keep your fuel at lowest orbit with highest velocity and that maximizes oberth effect which is just maximizing energy of fuel.

[TheRadicalModerate]

I'm curious about launch windows for orbitil refueling.
assumption: a depot is launched then  number of tankers are launched, with some amount of "haste". all launches are from starbase.

after the depot is launched, there is* a one orbit (about 90 minute) delay until the first tanker can be launched--the depot won't be overhead until then.

by "is*" I mean: it's easier for the 1st tanker if the depot passes "right overhead". But that's not trivial--the earth has turned in the mean time, so starbase is 90 minutes "to the east" of where it should be, at the speed of earth's spin.

how would this work in practice? I can think of a few options but I'd just be stumbling around mentally.

The traditional solution to this is not to try to do a launch directly to rendezvous.  As long as you launch into the same orbital plane, you can phase the orbit to chase down the depot.

Note that the depot itself never has any launch time dependencies.  It's gonna sit in orbit for its entire lifetime.  I think the question you're really asking is how to get as many tankers to the depot as quickly as possible.

Note that any launch site with a latitude < the inclination of the depot will get  two cracks at the orbital plane per day.  The closer the inclination is to the latitude, the closer together those two launch opportunities will be.  When latitude = inclination, there's only one opportunity per day.  When latitude > inclination, there are zero opportunities.

How long phasing takes is a real issue, so there are probably blocks of days where a tanker launch is impractical.  That said, if the depot is within, say, ±15º when the launch site matches the plane, it's not going to take more than a day to do the rendezvous.

Note that a depot with just a small difference between its inclination and the launch site has to be operationally capable of making two launches very close together.  BC Pads A and B may be able to do this, but this may be a reason to put the depot at a higher inclination than is most efficient (which would be where inclination = latitude).  There's also the issue of accessing the depot's orbit from both BC and KSC, which means that the inclination has to be higher than KSC's latitude (~28.5º), even though BC latitude (26.0º) would yield a bit more prop to orbit.

2nd question: how long do you think a "depot" load will last? for example, 50% boil-off in 3 days? 3 months? Answer can be really sloppy ("musks promises plus murphy's law plus some physics plus some engineering. and a little red bull")

Impossible to say.  What we can say is that sun- and earth-shades together are very complicated, so boiloff is most annoying during the VLEO depot prop accumulation phase.  (Things are easier in deep space.)  However, depots can launch empty, so their "payload" can be tens of tonnes of solar arrays and cryomachinery.  Throw enough power and mass at the problem and you can get to zero boiloff.  However, if your depot has to travel up to a higher apogee to refuel stuff in HEEO, then that mass can be a problem.

3rd question: could a depot self-transport to LRHO? how much fuel would be left at that point? asking for a friend

Yes.  Indeed, if SpaceX ever hopes to refuel HLS-style Ships in lunar orbit (not necessarily NRHO), this will be essential.  How you do this depends a lot on the size of the depot, and the operational architecture you're using in lunar orbit.  In a lot of cases, it takes two depots (or two trips by one depot) to provide enough prop to do the full NRHO-LS-NRHO trip in the plan of record.  However, there are a zillion caveats to that.

[DanClemmensen]

What we can say is that sun- and earth-shades together are very complicated, so boiloff is most annoying during the VLEO depot prop accumulation phase.  (Things are easier in deep space.)  However, depots can launch empty, so their "payload" can be tens of tonnes of solar arrays and cryomachinery.  Throw enough power and mass at the problem and you can get to zero boiloff.  However, if your depot has to travel up to a higher apogee to refuel stuff in HEEO, then that mass can be a problem.

Maybe create a specialized sunshade ship that stays in the low orbit. Depot docks to it there and only emerges to dock to Tanker before going back to hide in the shade. When full, Depot emerges and ascends to the higher orbit. When not in use between Depot filling campaigns, Sunshade ascends a bit to reduce drag.

[Greg Hullender]

Keep the nose towards the sun.

But you cannot point the nose towards the sun! It has to point in direction of orbit for two reasons: First, it minimizes MMOD risk, since 90% of impacts will be on the nose. Second, it reduces atmospheric drag by hugely reducing the surface area.

And, if you have something like Solar White tiles, there's no real need to point at the sun. Absent the effect of the Earth's reradiated heat, the vehicle would get cold enough to freeze oxygen. It's the nearby Earth that really messes you up.

That's the challenge that has to be dealt with, given the forced geometry of the problem.

[InterestedEngineer]

2nd question: how long do you think a "depot" load will last? for example, 50% boil-off in 3 days? 3 months? Answer can be really sloppy ("musks promises plus murphy's law plus some physics plus some engineering. and a little red bull")

 assume we face the sun, so the sunshine never hits the tanks, so we only have to deal with earthshine.

Now the tank area facing the earth is about 270m² facing 100W of earthshine.  assuming none of this is radiated back out (it will be), that's a rate of heat absorption of 27kW.

It takes 4.6GJ to evaporate 10t of liquid methane.  4.6GJ/27kW = 2 days.   5t a day.

In reality it's going to be on the order of 2t a day because we radiate into the dark night half the orbit, the tiles do absorb heat without conducting it all back into the ship, there is a heat gradient into the ship and so some of the energy is lost due to Stefan-Boltzmann radiation (which the tiles are extraordinarily good at),  only part of the time is the ship facing full profile at the earth, etc.

This is with no white paint, no sunshades, no nothing, just a bog-standard Starship V3.

So a starship V3, this coming year, should be able to launch 100t of fuel into LEO and still have between zero and little less than half of it left after a month in LEO.

that's your kinda-sloppy-but-probably-close answer.
Note: Sorry it's 200W/m² average earth shine.  oops.  SO that's 54kW.

OTOH, stefan-boltzmann for the entire starship at the boiling point of methane (130K) is -10kW/m².

The tiles themselves will have a heat gradient on them and stay at 170K while conducting little heat to the inside, so half the ship is really at -20kW/m².   So that's 34kW/m² net just to have the entire starship at boiling point of methane (an equilibrium).

So I'll extend the range of possibilities to be 2-6t per day of boiled methane assuming the whole ship was methane..  I haven't done the LOX side yet.  Probably shoiuld, it's 4/5 the mass.

[InterestedEngineer]

2nd question: how long do you think a "depot" load will last? for example, 50% boil-off in 3 days? 3 months? Answer can be really sloppy ("musks promises plus murphy's law plus some physics plus some engineering. and a little red bull")

 assume we face the sun, so the sunshine never hits the tanks, so we only have to deal with earthshine.

Now the tank area facing the earth is about 270m² facing 100W of earthshine.  assuming none of this is radiated back out (it will be), that's a rate of heat absorption of 27kW.

It takes 4.6GJ to evaporate 10t of liquid methane.  4.6GJ/27kW = 2 days.   5t a day.

In reality it's going to be on the order of 2t a day because we radiate into the dark night half the orbit, the tiles do absorb heat without conducting it all back into the ship, there is a heat gradient into the ship and so some of the energy is lost due to Stefan-Boltzmann radiation (which the tiles are extraordinarily good at),  only part of the time is the ship facing full profile at the earth, etc.

This is with no white paint, no sunshades, no nothing, just a bog-standard Starship V3.

So a starship V3, this coming year, should be able to launch 100t of fuel into LEO and still have between zero and little less than half of it left after a month in LEO.

that's your kinda-sloppy-but-probably-close answer.
Note: Sorry it's 200W/m² average earth shine.  oops.  SO that's 54kW.

OTOH, stefan-boltzmann for the entire starship at the boiling point of methane (130K) is -10kW/m².

The tiles themselves will have a heat gradient on them and stay at 170K while conducting little heat to the inside, so half the ship is really at -20kW/m².   So that's 34kW/m² net just to have the entire starship at boiling point of methane (an equilibrium).

So I'll extend the range of possibilities to be 2-6t per day of boiled methane assuming the whole ship was methane..  I haven't done the LOX side yet.  Probably shoiuld, it's 4/5 the mass.

How embarrassing, the LOX side is on the order of 19t / day.

Oddly enough the net heat flux is on the order 100W/m² but alas the heat of vaporization of LOX is far less than that of Methane. 

Roughly, with irradiance being 200W/m² on the tile side and none on the metal side, boiling point being 90K,

insulation k of the tiles being 0.05 W/m⋅K, .05m thick => R =1.0 K⋅m²/W.

We get an equilibrium equation with Stefan Boltzmann 199K on the tiles => -88W/m²

And the gradient into the tank is 110K, which gives us a flux of 110W, so that's pretty close to equilibrium.

110W/m² over 425m² is 47kW.   for LOX the latent heat is 213kJ/kg at 1  bar, so the boil off rate is 47/213 = .22kg/sec or 19t/day.

So the lox in a Starship V3 will last about 4 days in LEO.  Considerably worse.  Gonna need shades.

[Vultur]

Heh, I did the LOX with the same heat numbers you used and got 17.3 tons/day, which is decently close.

I'll leave it since I already wrote it out....

First guess (please correct this)

Oxygen heat of vaporization is 6.82 kJ/mol, molar mass of O2 is 32 g/mol, so about 213 kJ/kg. That's assuming the oxygen is already at boiling point and only needs the heat of vaporization.

So using your same 200W/m^2 x 270 m^2 = 54kW then that is ~ 200 grams per second, 12kg/minute, 720kg/hr, 17.3 tons per day.

That seems significantly worse than the methane, but I may have messed up the numbers.

(BTW, when you calculated the 4.6GJ per 10 tons, are you assuming the methane is already at boiling point or is down at O2 boiling point?)

[InterestedEngineer]

Heh, I did the LOX with the same heat numbers you used and got 17.3 tons/day, which is decently close.

I'll leave it since I already wrote it out....

First guess (please correct this)

Oxygen heat of vaporization is 6.82 kJ/mol, molar mass of O2 is 32 g/mol, so about 213 kJ/kg. That's assuming the oxygen is already at boiling point and only needs the heat of vaporization.

So using your same 200W/m^2 x 270 m^2 = 54kW then that is ~ 200 grams per second, 12kg/minute, 720kg/hr, 17.3 tons per day.

That seems significantly worse than the methane, but I may have messed up the numbers.

(BTW, when you calculated the 4.6GJ per 10 tons, are you assuming the methane is already at boiling point or is down at O2 boiling point?)

You are ignoring stefan-boltzmann, which rids us of a little under half the heat.  (heat conduction through the tiles leaves us an equilibrium equation  with the inside boiling point and the outside radiating away and a temperature gradient)

Which means if our numbers are the same, I or you must have done something wrong...  and it's probably the 270m² in yours.   Just to check, the tank is about 30m long and 9m in diameter.

area = circumference x length = 9*3.14 * 30 = 850m²

The half that is subject to earth shine is the same half that's tiled so that's 425² (as opposed to your 270)

So you 2x error in one area offsets the 0.5x area in ignoring Stefan-Boltzmann.



And yes I ignored the specific heat of the methane.  Equilibrium will be boiling point.

[Vultur]

Heh, I did the LOX with the same heat numbers you used and got 17.3 tons/day, which is decently close.

I'll leave it since I already wrote it out....

First guess (please correct this)

Oxygen heat of vaporization is 6.82 kJ/mol, molar mass of O2 is 32 g/mol, so about 213 kJ/kg. That's assuming the oxygen is already at boiling point and only needs the heat of vaporization.

So using your same 200W/m^2 x 270 m^2 = 54kW then that is ~ 200 grams per second, 12kg/minute, 720kg/hr, 17.3 tons per day.

That seems significantly worse than the methane, but I may have messed up the numbers.

(BTW, when you calculated the 4.6GJ per 10 tons, are you assuming the methane is already at boiling point or is down at O2 boiling point?)

You are ignoring stefan-boltzmann, which rids us of a little under half the heat.  (heat conduction through the tiles leaves us an equilibrium equation  with the inside boiling point and the outside radiating away and a temperature gradient)

Which means if our numbers are the same, I or you must have done something wrong...  and it's probably the 270m² in yours.   Just to check, the tank is about 30m long and 9m in diameter.

area = circumference x length = 9*3.14 * 30 = 850m²

The half that is subject to earth shine is the same half that's tiled so that's 425² (as opposed to your 270)

So you 2x error in one area offsets the 0.5x area in ignoring Stefan-Boltzmann.

Ok, got it.

[Greg Hullender]

3rd question: could a depot self-transport to LRHO? how much fuel would be left at that point? asking for a friend

I figure you can get it to lunar orbit with 64% of the fuel provided you fuel up two depots in LEO, launch them into a final-fueling orbit up to the point where there's just enough fuel between them to fill one of them up, transfer the fuel to one, leaving the other almost empty. A perigee, the full one does just enough of a burn to get to orbit at the distance of the moon and then circularizes the orbit when it gets there. In other words, this gets it to the moon's orbit but not actually close to the moon itself. if it's going to be in some sort of distant halo orbit that's probably not going to cost a lot more.

Assumptions: the depots refuel at an altitude of 281 km (from some SpaceX doc a while back), Raptor 3 has ISP of 3.7278 km/s, depot dry mass is 88.5 metric tons, a fully fueled depot holds 1588 tons of propellant. The FFO is at about 33,400 km with a period of 9 hours 47 minutes. It arrives at the moon with just over 1,000 tons of propellant.

These numbers haven't been updated in a while, so I'll happily rerun this with more current numbers.

[TheRadicalModerate]

Assume we face the sun, so the sunshine never hits the tanks, so we only have to deal with earthshine.

Note that this requires some amount of attitude control to oppose tidal forces.  It's small but probably not trivial.

Now the tank area facing the earth is about 270m² facing 100W of earthshine.  assuming none of this is radiated back out (it will be), that's a rate of heat absorption of 27kW.

Just taking the Google AI answer at face value, IR re-radiation from Earth is 150-350W/m², average 230W/m².  Average Earth albedo is about 0.3,¹ which would translate to ~400W/m².  So your average luminance from Earth is 630W/m².

OTOH, stefan-boltzmann for the entire starship at the boiling point of methane (130K) is -10kW/m².

The tiles themselves will have a heat gradient on them and stay at 170K while conducting little heat to the inside, so half the ship is really at -20kW/m².   So that's 34kW/m² net just to have the entire starship at boiling point of methane (an equilibrium).

I'm on shaky ground here, but I think you're doing this wrong.  Yes, the skin itself will be at 90K-130K, but you can't really calculate the emittance without knowing the thermal resistance of the insulation (SOFI, MLI, who knows?), which should give the gradient, and eventually the outer surface temperature.  The net result should give a better result than what you're estimating, i.e., the outside of the insulation should be hotter than you estimated.

________
¹I'm unclear if that number is for the whole Earth, or just the illuminated side.  If the former, then my back-of-napkin should be about right.  If the latter, divide by roughly 2.

[Twark_Main]

Assume we face the sun, so the sunshine never hits the tanks, so we only have to deal with earthshine.

Note that this requires some amount of attitude control to oppose tidal forces.  It's small but probably not trivial.

This is one reason why I favor having an ISS-class CMG. Total mass <1 t.

The other reason, of course, is better Solar storm protection during Mars cruise. The radiation is directional but moves around a lot, so you can get better shielding if your autopilot can "surf" the solar storm while the crew shelters in the storm shelter. This lets you use an asymmetrically shielded storm shelter, which gives better protection for the same shielding mass.

TBPINP, but it seems this is a part that easily pays for itself in terms of adding both efficiency and redundancy.

Now the tank area facing the earth is about 270m² facing 100W of earthshine.  assuming none of this is radiated back out (it will be), that's a rate of heat absorption of 27kW.

Just taking the Google AI answer at face value, IR re-radiation from Earth is 150-350W/m², average 230W/m².  Average Earth albedo is about 0.3,¹ which would translate to ~400W/m².  So your average luminance from Earth is 630W/m².

OTOH, stefan-boltzmann for the entire starship at the boiling point of methane (130K) is -10kW/m².

The tiles themselves will have a heat gradient on them and stay at 170K while conducting little heat to the inside, so half the ship is really at -20kW/m².   So that's 34kW/m² net just to have the entire starship at boiling point of methane (an equilibrium).

I'm on shaky ground here, but I think you're doing this wrong.  Yes, the skin itself will be at 90K-130K, but you can't really calculate the emittance without knowing the thermal resistance of the insulation (SOFI, MLI, who knows?), which should give the gradient, and eventually the outer surface temperature.  The net result should give a better result than what you're estimating, i.e., the outside of the insulation should be hotter than you estimated.

________
¹I'm unclear if that number is for the whole Earth, or just the illuminated side.  If the former, then my back-of-napkin should be about right.  If the latter, divide by roughly 2.

Yes, correct. Easy way to solve is to take the thermal conductivity equation and the radiant balance equation and then "goal seek" the surface temperature until the two equations give the same thermal flux.

Yet another reason to have a CMG. You can (without burning propellant) roll the Starship so the black insulated heat shield aims at the brightest + hottest part of the Earth. This also reduces brightness as a side-effect, or as an intentional goal.

TUFI is a documented material, you can find the thermal conductivity (0.843 W/m-K +/- 0.0421) and emissivity (87.3% +/- 4.36%).

Source: https://tpsx.arc.nasa.gov/Material?id=8

Solar reflectivity should be around 10%, can't find a good source at the moment.

[InterestedEngineer]

Assume we face the sun, so the sunshine never hits the tanks, so we only have to deal with earthshine.

Note that this requires some amount of attitude control to oppose tidal forces.  It's small but probably not trivial.

Well, the boiled methalox has to go *somewhere*, attitude control seems like a good place for it.

[InterestedEngineer]

Now the tank area facing the earth is about 270m² facing 100W of earthshine.  assuming none of this is radiated back out (it will be), that's a rate of heat absorption of 27kW.

Just taking the Google AI answer at face value, IR re-radiation from Earth is 150-350W/m², average 230W/m².  Average Earth albedo is about 0.3,¹ which would translate to ~400W/m².  So your average luminance from Earth is 630W/m².

OTOH, stefan-boltzmann for the entire starship at the boiling point of methane (130K) is -10kW/m².

The tiles themselves will have a heat gradient on them and stay at 170K while conducting little heat to the inside, so half the ship is really at -20kW/m².   So that's 34kW/m² net just to have the entire starship at boiling point of methane (an equilibrium).

I'm on shaky ground here, but I think you're doing this wrong.  Yes, the skin itself will be at 90K-130K, but you can't really calculate the emittance without knowing the thermal resistance of the insulation (SOFI, MLI, who knows?), which should give the gradient, and eventually the outer surface temperature.  The net result should give a better result than what you're estimating, i.e., the outside of the insulation should be hotter than you estimated.

________
¹I'm unclear if that number is for the whole Earth, or just the illuminated side.  If the former, then my back-of-napkin should be about right.  If the latter, divide by roughly 2.

Yes, correct. Easy way to solve is to take the thermal conductivity equation and the radiant balance equation and then "goal seek" the surface temperature until the two equations give the same thermal flux.

Yet another reason to have a CMG. You can (without burning propellant) roll the Starship so the black insulated heat shield aims at the brightest + hottest part of the Earth. This also reduces brightness as a side-effect, or as an intentional goal.

TUFI is a documented material, you can find the thermal conductivity (0.843 W/m-K +/- 0.0421) and emissivity (87.3% +/- 4.36%).

Source: https://tpsx.arc.nasa.gov/Material?id=8

Solar reflectivity should be around 10%, can't find a good source at the moment.

That's what I did, I took the equilibrium and "goal seeked" it.  You can make a table with the two going in opposite directions.  You can write a Python script for that.  modern LLMs can write a python script (but I always verify the result).

I had already assumed the heat shield faced earth shine - using the boiled methalox for that kind of positioning.  Note you just point the nose at the sun ALL THE TIME and I'm pretty sure that's the optimal orientation, and then you rotate around the axis to keep the heat shield pointed at the earth.  which is at its worst when 90 degrees from the sun (at the terminator).  When you are on the opposite side of the sun the nose is facing the earth, when you are on the side of the sun the tail is facing the earth.  You angle it just enough to hide the bare steel section, no more.  That's why I went with an average of 200W/m^2.

As far as TUFI, I used the Shuttle tiles, which was 0.05 W/m-K at room temperature - is TUFI really that much worse?  Maybe you used the high temperature thermal conductivity? (which is also worse for the shuttle tiles).  I also pushed it out to 5cm to include the saffil layer underneath.