Moon Starship

[AC in NC]

I wanted to better understand a tanker scheme so I started looking around and eventually bit the bullet to dive into GMAT.

With caveats about various warts, no particular warranties re: orbits, and an obvious penchant for excessivity ... here was what I ended up with after goofing around with it for while.  Felt it had pretty decent fidelity and dovetails with a lot of the numbers others have previously derived and cite regularly.  Helped a lot to visualize the numbers others have used in their examples.

The scheme I modelled was to start at a 200km orbit (wart) for "Tanker 0".  When able to fully fuel a "Tanker 1", it would raise with a 180 sec burn to achieve about +1 km/s.  Repeat for Tankers 2 and 3.  Tanker 4 did a short burn to stay inside the moon at about 280K km.


Pretty interesting tool.

[ZachF]

https://twitter.com/elonmusk/status/1183860560418357248

Big challenge for Starship refueling on the moon is finding sources of carbon. Probably some pretty big deposits in craters from meteorites. Same goes for hydrogen & oxygen, also in (shadowed) craters.

What about eventually just building a hydrolox stage for the moon if you want to ISRU?

-Water is harvested on moon, split into hydrogen and oxygen.
-Starship lands, drops cargo, etc.
-Starship gets lifted onto hydrolox stage.
-Hydrolox stage boosts Starship back to Earth, returns to moon and lands.

?

[envy887]

Where are we getting the numbers for the amount of fuel delivered per trip? Are we just using the cargo capacities numbers and assuming thats all the fuel it will transfer? How much fuel will be left in the tanks after it reaches orbit? The tanks are never emptied to get it to orbit. How much excess fuel and oxidiser will be transfered too? Assuming they transfer all but whats in the header tanks.

The propellant launches should bring up an amount that is reasonably close to the max cargo payload of the vehicle. There are a number of reasons for it to be slightly more or slightly less, but these mostly cancel each other IMO.

[Twark_Main]

...The scheme I modelled was to start at a 200km orbit (wart) for "Tanker 0".  When able to fully fuel a "Tanker 1", it would raise with a 180 sec burn to achieve about +1 km/s.  Repeat for Tankers 2 and 3.  Tanker 4 did a short burn to stay inside the moon at about 280K km.

Sorry, this is a bit unclear to me. Trying to get on the same page.

• This system uses five separate tankers (numbered 0 to 4) in orbit simultaneously, correct? Or are we looking at a single tanker that visits multiple orbits?

• What do you mean by "when able to fully fuel a 'Tanker 1?'" I'm stumped because it seems impossible for Tanker 0 to fully fuel Tanker 1 with just a single rendezvous, so AIUI Tanker 0 would never perform its first orbit raise burn, because that particular Boolean condition ("Can Tanker 0 fully refuel Tanker 1 now?") would never become true, even when Tanker 0 is full. Would it be more accurate to say "when Tanker 0 is full" instead? Or am I missing it entirely?

• How many launches are required to fully fuel Tanker 4?

[AC in NC]

...The scheme I modelled was to start at a 200km orbit (wart) for "Tanker 0".  When able to fully fuel a "Tanker 1", it would raise with a 180 sec burn to achieve about +1 km/s.  Repeat for Tankers 2 and 3.  Tanker 4 did a short burn to stay inside the moon at about 280K km.

Sorry, this is a bit unclear to me. Trying to get on the same page.

• This system uses five separate tankers (numbered 0 to 4) in orbit simultaneously, correct? Or are we looking at a single tanker that visits multiple orbits?

• What do you mean by "when able to fully fuel a 'Tanker 1?'" It's impossible for Tanker 0 to fully fuel Tanker 1 with just a single rendezvous, so AIUI Tanker 0 would never perform its first orbit raise burn, because that particular Boolean condition ("Can Tanker 0 fully refuel Tanker 1 now?") would never become satisfied, even if Tanker 0 were full. Do you mean "when Tanker 0 is full" instead?

• Out of curiosity, how many launches are required to fully fuel Tanker 4 under this scheme?

Thanks for asking.  Yeah there was some semantic imprecision in the way I sketched out the scheme.

I wanted to sort of conceptualize what set of tankers might be appropriate based on a remark speedevil made.  Something roughly to the effect of you can boost from LEO by 2km/s and retain half your fuel.  And I thought he suggested spacing tanker uphill by that amount meaning at least 1 and LEO+2km/s and 1 at LEO +4km/s.

So I just messed around with GMAT to conceptualize that.  I chose a 200km LEO orbit although I suspect that's quite a poor choice. IANARS so I'm just sort of spitballing.  From that 200km orbit (7.8 km/s) you can only get a single tanker uphill spaced at +2km/s.  The 2nd one goes earth escape for very close to it.  So I decided to limit my spacing to 180 sec burns uphilll which leads to a little less that +1km/s spacing.

So that's how I got to 5.  1 at 200 km.  3 uphill at 8.8 km/s, 9.7 km/s, and 10.6 km/s.  After that, there's only a small bit of deltaV that can be added and stay in any reasonable refueling orbit close to earth.  So #5 is at 10.9 km/s.

So the first part of the exercise was just to visualize that spacing and altitude of possible orbits.  Again, no representation at all if those make any sense.

So the way I'd envision these used is as follows based on having an (A) Stripped Down Orbit Tanker; and a (B) Reusable Tanker.

1)  Launch (A) to LEO with whatever residual prop remains but which should be greater (perhaps meaningfully) than the nominal 150t payload.  Let's say 300t residual (because it makes calculations easy).

2)  Launch (B) eight times with 150t each to top off (A) at 1200t + 300t extra

3)  Raise (A) to the next orbit with 900t residual prop at 8.8km/s

4)  Repeat 1-3 to bring another 900t residual prop to top off (A1) and leaving 600t in (A2) at 8.8km/s

5)  Raise (A1) to the next orbit with 900t residual prop at 9.7km/s

6)  Repeat 1-5 leaving (A1) with 900t at 10.6km, (A2) with 600t at 9.7km/s, and (A3) with 300t at 8.8km/s

7)  .... etc... 

That's the general idea.

When you need to use it, you can bring a normal Starship up, tank up enough to get to one of the tankers that's appropriate, top off there and go.

I'll run a spreadsheet later to calculate exactly how many trips it takes to fill the 5th tanker at altitude of 280,000 km

[freddo411]

What about eventually just building a hydrolox stage for the moon if you want to ISRU?

-Water is harvested on moon, split into hydrogen and oxygen.
-Starship lands, drops cargo, etc.
-Starship gets lifted onto hydrolox stage.
-Hydrolox stage boosts Starship back to Earth, returns to moon and lands.

?

Lifting Starship into hydrolox stage ... uh, I don't think that's feasible in the near future.   That would be quite a big stage.

Starship trips to the moon can benefit from using lunar ISRU LOX.   

[Twark_Main]

Ok so this is just spitballing, and not intended to be a provably optimal refueling scheme.


ISTM a sensible approach to first find the mathematically optimal refueling scheme, then work backwards to define the various parking orbits.

[AC in NC]

Ok so this is just spitballing, and not intended to be a provably optimal refueling scheme.

ISTM a sensible approach to first find the mathematically optimal refueling scheme, then work backwards to define the various parking orbits.

Heavens no!!!    The sensible approach is way above my skill set.  I just wanted to play around with some numbers (and the tool) to visualize how cheap an even sub-optimal scheme good be.

 

[RotoSequence]

How many tons of carbon would you need to bring to the moon in order to lift off with 50 tons of payload, without finding any carbon sources on the moon?

[Ace]

How many tons of carbon would you need to bring to the moon in order to lift off with 50 tons of payload, without finding any carbon sources on the moon?

Arguably, the best way to bring "extra" carbon with you to the Moon would be in the form of CH4.

Returning from the Moon to Earth would take about 2.6 km/s of delta-V, which corresponds to 171.5 tons of fuel with a cargo mass of 50 t. That's about 37 t of CH4.

Just going to low lunar orbit (LLO) would be a little less: 1.72 km/s dV, or 100 t of total fuel, 22 t of CH4.

[raketa]

https://twitter.com/elonmusk/status/1183860560418357248

Big challenge for Starship refueling on the moon is finding sources of carbon. Probably some pretty big deposits in craters from meteorites. Same goes for hydrogen & oxygen, also in (shadowed) craters.

What about eventually just building a hydrolox stage for the moon if you want to ISRU?

-Water is harvested on moon, split into hydrogen and oxygen.
-Starship lands, drops cargo, etc.
-Starship gets lifted onto hydrolox stage.
-Hydrolox stage boosts Starship back to Earth, returns to moon and lands.

?

elon is not interested in Moon exploration..We not get from Spacex some additional big modification of SS for Moon.

[ZachF]

What about eventually just building a hydrolox stage for the moon if you want to ISRU?

-Water is harvested on moon, split into hydrogen and oxygen.
-Starship lands, drops cargo, etc.
-Starship gets lifted onto hydrolox stage.
-Hydrolox stage boosts Starship back to Earth, returns to moon and lands.

?

Lifting Starship into hydrolox stage ... uh, I don't think that's feasible in the near future.   That would be quite a big stage.

Starship trips to the moon can benefit from using lunar ISRU LOX.

It would not be that big if it's lifting a nearly empty Starship (~200tonnes). It would actually be much smaller than Starship itself.

A hydrolox stage that could lift a 200 tonne starship (120t empty + 30t landing fuel + 50t return cargo) into TLI and boost back to the lunar surface would only weigh ~200 tonnes fully fueled. It would probably fit inside a Starship's cargo bay too if it was folded up.

[freddo411]

What about eventually just building a hydrolox stage for the moon if you want to ISRU?

-Water is harvested on moon, split into hydrogen and oxygen.
-Starship lands, drops cargo, etc.
-Starship gets lifted onto hydrolox stage.
-Hydrolox stage boosts Starship back to Earth, returns to moon and lands.

?

Lifting Starship into hydrolox stage ... uh, I don't think that's feasible in the near future.   That would be quite a big stage.

Starship trips to the moon can benefit from using lunar ISRU LOX.

It would not be that big if it's lifting a nearly empty Starship (~200tonnes). It would actually be much smaller than Starship itself.

A hydrolox stage that could lift a 200 tonne starship (120t empty + 30t landing fuel + 50t return cargo) into TLI and boost back to the lunar surface would only weigh ~200 tonnes fully fueled. It would probably fit inside a Starship's cargo bay too if it was folded up.

Yes, it is possible to build such a lander/stage.   The centaur upper stage is about 10% the mass, and it is 12 m by 3 meters radius -- so the stage would be pretty large, (roughly 24m x 7m r)     A dry mass of about 20 tonnes is a pretty big lunar spacecraft, at least by today's standards.     Also, you'd need a crane to load the SS on the lander stage.

I'm doubtful that this approach would turn out to be easier than figuring out how to make CH3 and LOX for the starship on the moon.

I'd be delighted to be proven wrong by seeing this fly in real life.

[armchairfan]

Yes, it is possible to build such a lander/stage.   The centaur upper stage is about 10% the mass, and it is 12 m by 3 meters radius -- so the stage would be pretty large, (roughly 24m x 7m r)     A dry mass of about 20 tonnes is a pretty big lunar spacecraft, at least by today's standards.     Also, you'd need a crane to load the SS on the lander stage.

I'm doubtful that this approach would turn out to be easier than figuring out how to make CH3 and LOX for the starship on the moon.

I'd be delighted to be proven wrong by seeing this fly in real life.

Rather than lift the SS, why not mate two smaller independent but identical hydrolox boosters to each side of the SS and then fuel them? That should be a much easier task. Doing the math, a single Blue Origin BE3U engine (~500 710MN thrust) would be a good engine for each booster though using it to land back on the Moon would make for a bodacious hoverslam. Edit: Blue's BE-7 would be a better landing engine.

BTW, such a 10t dry mass booster loaded with 35t propellant delivered to LEO could land itself on the moon. Incidentally, the LEO payload capacity of New Glenn is ... wait for it ... 45t. Hummm ...

All that's needed is to get Elon and Jeff B to cooperate. 

Edit 2: Use (higher) BE3U thrust instead of BE3. You're definitely going to want a separate BE-7 landing engine.

[ZachF]

What about eventually just building a hydrolox stage for the moon if you want to ISRU?

-Water is harvested on moon, split into hydrogen and oxygen.
-Starship lands, drops cargo, etc.
-Starship gets lifted onto hydrolox stage.
-Hydrolox stage boosts Starship back to Earth, returns to moon and lands.

?

Lifting Starship into hydrolox stage ... uh, I don't think that's feasible in the near future.   That would be quite a big stage.

Starship trips to the moon can benefit from using lunar ISRU LOX.

It would not be that big if it's lifting a nearly empty Starship (~200tonnes). It would actually be much smaller than Starship itself.

A hydrolox stage that could lift a 200 tonne starship (120t empty + 30t landing fuel + 50t return cargo) into TLI and boost back to the lunar surface would only weigh ~200 tonnes fully fueled. It would probably fit inside a Starship's cargo bay too if it was folded up.

Yes, it is possible to build such a lander/stage.   The centaur upper stage is about 10% the mass, and it is 12 m by 3 meters radius -- so the stage would be pretty large, (roughly 24m x 7m r)     A dry mass of about 20 tonnes is a pretty big lunar spacecraft, at least by today's standards.     Also, you'd need a crane to load the SS on the lander stage.

I'm doubtful that this approach would turn out to be easier than figuring out how to make CH3 and LOX for the starship on the moon.

I'd be delighted to be proven wrong by seeing this fly in real life.

It wouldn't be nearly that large. 200 tonnes of hydrolox fuel is about 600m^3, Starship's cargo bay is 1000m^3.

[AC in NC]

Ok so this is just spitballing, and not intended to be a provably optimal refueling scheme.

ISTM a sensible approach to first find the mathematically optimal refueling scheme, then work backwards to define the various parking orbits.

So I worked up some numbers off an Unreasonably Cheap Orbital Tanker @ $10M per launch and Unreasonably Cheap Reusable Tanker @ $1M per launch.  There's a bit of built-in error because the orbital burn and residual propellant numbers I had run have a modest flaw.  But these get you in the ballpark subject obviously to the premised costs.


        11 Tankers    $110M
        88 Reusable   $ 88M
        99 Launches   $198M

      5700 T @   5K km (8.8kms)     in 6 Tankers (2 empty, 3 1500 T, 1 1200 T)
      3000 T @  16K km (9.7kms)     in 3 Tankers (1 empty, 2 1500 T          )
      2400 T @  85K km (10.6kms)    in 2 Tankers (         3 1500 T, 1 900  T)
    11,100 T @ $18K/T or $9/kg


The 5700 T could be made into 4500 T @ 9.7 kms or 3600 T @ 10.6kms
The 3000 T could be made into 2400 T @ 10.6kms

For a total of 8400 T at 10.6kms for the same price but I think it would be better to leave the Tankers distributed for keeping each level topped off so that a mission can come up, min tank for a quick rendezvous with the appropriate tanker.

[armchairfan]

It wouldn't be nearly that large. 200 tonnes of hydrolox fuel is about 600m^3, Starship's cargo bay is 1000m^3.

Here's a scale figure for two hypothetical Blue Origin boosters (BE-3U, BE-7U) with SS. The boosters are 6x12m (340m^3) based on a simple scaling of the space shuttle external tank (I drew it before your post). Each booster lifts off with the BE-3U and lands with the BE-7. Ground support equipment must invert the booster on the ground and mate it with the SS. That might not be that hard for a 10t booster in lunar gravity.

As previously noted, the boosters
- mass 10t each
- can deliver themselves to the moon if fueled with 35t of propellant each in LEO; thus each fueled booster can be delivered to LEO with a New Glenn
- as a pair can deliver 200t SS to TEI and return the lunar surface

So not too much rocket lego.

Edit: Obviously this doesn't make sense until lunar hydrolox production starts. No carbon (for methane) is needed and there's apparently plenty of water ice near the lunar south pole.

[rakaydos]

It wouldn't be nearly that large. 200 tonnes of hydrolox fuel is about 600m^3, Starship's cargo bay is 1000m^3.

Here's a scale figure for two hypothetical Blue Origin boosters (BE-3U, BE-7U) with SS. The boosters are 6x12m (340m^3) based on a simple scaling of the space shuttle external tank (I drew it before your post). Each booster lifts off with the BE-3U and lands with the BE-7. Ground support equipment must invert the booster on the ground and mate it with the SS. That might not be that hard for a 10t booster in lunar gravity.

As previously noted, the boosters
- mass 10t each
- can deliver themselves to the moon if fueled with 35t of propellant each in LEO; thus each fueled booster can be delivered to LEO with a New Glenn
- as a pair can deliver 200t SS to TEI and return the lunar surface

So not too much rocket lego.

Edit: Obviously this doesn't make sense until lunar hydrolox production starts. No carbon (for methane) is needed and there's apparently plenty of water ice near the lunar south pole.

So why are we bringing 120 tons of sheet metal and methalox engine to the moon? just transfer the cargo to the hydrolox landers and let them handle last mile delivery.

[Twark_Main]

Ok so this is just spitballing, and not intended to be a provably optimal refueling scheme.

ISTM a sensible approach to first find the mathematically optimal refueling scheme, then work backwards to define the various parking orbits.

So I worked up some numbers off an Unreasonably Cheap Orbital Tanker @ $10M per launch and Unreasonably Cheap Reusable Tanker @ $1M per launch.  There's a bit of built-in error because the orbital burn and residual propellant numbers I had run have a modest flaw.  But these get you in the ballpark subject obviously to the premised costs.


        11 Tankers    $110M
        88 Reusable   $ 88M
        99 Launches   $198M

      5700 T @   5K km (8.8kms)     in 6 Tankers (2 empty, 3 1500 T, 1 1200 T)
      3000 T @  16K km (9.7kms)     in 3 Tankers (1 empty, 2 1500 T          )
      2400 T @  85K km (10.6kms)    in 2 Tankers (         3 1500 T, 1 900  T)
    11,100 T @ $18K/T or $9/kg


The 5700 T could be made into 4500 T @ 9.7 kms or 3600 T @ 10.6kms
The 3000 T could be made into 2400 T @ 10.6kms

For a total of 8400 T at 10.6kms for the same price but I think it would be better to leave the Tankers distributed for keeping each level topped off so that a mission can come up, min tank for a quick rendezvous with the appropriate tanker.

So how many launches just for filling Tanker 4? 85,000 km apogee corresponds to Tanker 3.

I'm working up (what I think is) an optimal scheme. Unfortunately it depends on the payload mass and desired delta-v, so pre-positioned depots are less favorable. But since HEEO depots would need to "point at" at the target, I think mission-specific depots are more practical anyway.

For their to be a "quick rendezvous with an appropriate tanker" in the right orbit, you'd need not just a single 11-Tanker constellation, but a whole array of them! One 11-Tanker ladder pre-positioned in every different departure direction.

So I think the best course is to just bite the bullet and accept that HEEO retankings will be serviced by mission-specific tanker missions, not some general-purpose fleet of prepositioned tankers.

Brain dump follows, apologies for the half-formed thoughts:

Broadly speaking, the optimal retanking scheme is to fully top off Starship in LEO using Tanker 0, then have Starship rendezvous with Tanker 1 in a HEEO having an apogee specially chosen such that A) the tanker transfers 100% of its available fuel to Starship, and B) the tanker completely fills Starship's tanks. Tanker 1 was refilled in LEO by Tanker 0. If Tanker 1 can reach the aforementioned HEEO in a single burn from LEO, then that's your optimal solution right there! If not... well I'm still working on that part. But it's some form of recursion where you retank Tanker 2 using the same trick as we used to refuel Starship. Think Towers of Hanoi.

A useful heuristic: in general you want to minimize the total tanker dry mass you wastefully drag uphill (or more accurately, total impulse = dry mass * deltav from LEO). In fact the problem is very nearly equivalent, in the same way that optimizing the Oberth effect is mathematically equivalent to minimizing the total kinetic + potential energy of the rocket's exhaust.

I can't be the first person to work this. Someone (presumably a real mathematician) must have solved this problem already. Anyone got the citation handy?

[Ludus]

There’s about a 3.8 oxygen to methane mass ratio anyway so if the lunar base can just supply lox from water ice it’s already gone much of the way.