Starship On-orbit refueling - Options and Discussion

[Greg Hullender]

There's also the quantum of refueling to consider.  the 2.0km/sec gives you half full starships in the HEO so one full starship to do the boost at perigee.   [technically 2.2km/sec - 3650km/sec * ln(1320/720) ]

Anything else and the logistics get weird. 

3km/sec requires a mass ratio of  2.27 which uses 62% of the fuel.  You'd have to boost two refuelers to the HEO and that'd have to all be coordinated over 2 day orbits.  Yikes.

Okay, I get about the same numbers you do, assuming I'm understanding you correctly. Correct me if I'm wrong, but in your scenario a pair of depots (wet mass, 1200 mt; dry mass, 100 mt) are fueled in LEO and then boosted to HEO such that each is exactly half depleted (8 hour orbit, 2.3 kps ∆v, 27,600 km altitude of apogee). One tops up the other and then returns to Earth. The other waits to fully fuel a mission. (I'm using Raptor's target specific impulse of 3750 m/s.)

Your objection to using higher orbits is that it would take three depots, not just two. Is that right?

[Greg Hullender]

Not to mention an additional $30M in fuel.  (1200t * $25k/ton = $30M).  That's an optimistic estimate.  It takes half of that to make a Mars trip, so wherever it is you want to go, it better be worth the extra 1-2 km/sec (it likely isn't).

Far be it from me to defend the HEO depot idea, but, unless I'm seriously confused (always a possibility), the total ∆v for the mission isn't changed. It just moves some of it from the mission vehicle to the depot. The rocket equation can't be fooled.

The original motivation for it, as I recall, was that it appears that an HLS Starship that refueled in LEO before heading to NRHO and then to the moon would find itself just a little short of enough fuel to get back to NRHO. Refueling at NRHO is off the table because NASA won't allow the vehicle to be refueled twice. Hence the idea of a depot in HEEO. In this case, the determining factor is what's the most elliptical orbit an empty HLS can get into on a single load of fuel. That's when you put the depot.

I have to believe, though, that there are better knobs to twist to make this happen just using a single depot in LEO.

[Brigantine]

a pair of depots (wet mass, 1200 mt; dry mass, 100 mt)

I had thought the depots would have considerably more than 1200t of fuel capacity. Doesn't a standard ship with a payload section already have 1200t of fuel? And the depots are supposed to be stretched?


Also, re precession, the Near-Earth-Asteroid proposal uses a 7,800 x 113,300 km re-fueling orbit (and subsequently uses 242 m/s to lower perigee before the departure burn). Does that reduce the rate of precession? Either by design or incidentally.
[and more generally, what resources do you use to calculate precession?]

[InterestedEngineer]

There's also the quantum of refueling to consider.  the 2.0km/sec gives you half full starships in the HEO so one full starship to do the boost at perigee.   [technically 2.2km/sec - 3650km/sec * ln(1320/720) ]

Anything else and the logistics get weird. 

3km/sec requires a mass ratio of  2.27 which uses 62% of the fuel.  You'd have to boost two refuelers to the HEO and that'd have to all be coordinated over 2 day orbits.  Yikes.

Okay, I get about the same numbers you do, assuming I'm understanding you correctly. Correct me if I'm wrong, but in your scenario a pair of depots (wet mass, 1200 mt; dry mass, 100 mt) are fueled in LEO and then boosted to HEO such that each is exactly half depleted (8 hour orbit, 2.3 kps ∆v, 27,600 km altitude of apogee). One tops up the other and then returns to Earth. The other waits to fully fuel a mission. (I'm using Raptor's target specific impulse of 3750 m/s.)

Your objection to using higher orbits is that it would take three depots, not just two. Is that right?

The second ship *is* the mission.

So only two ships have to boost up to HEO.

Try counting the costs at $25 (or $50) per kilo for each of the scenarios you propose

[Greg Hullender]

Also, re precession, the Near-Earth-Asteroid proposal uses a 7,800 x 113,300 km re-fueling orbit (and subsequently uses 242 m/s to lower perigee before the departure burn). Does that reduce the rate of precession? Either by design or incidentally.
[and more generally, what resources do you use to calculate precession?]

I just used the formulas from the Wikipedia article on Nodal Precession and made an Excel spreadsheet.

For your NEA proposal, I figure an orbital period of 1.7 days and a precession time of 14.5 years. And are you sure you meant 242 m/s not 24.2 ms/s? Otherwise, I think you'll impact the planet.

You should double-check my math, of course. :-)

[Brigantine]

Noted. Thanks.

are you sure you meant 242 m/s not 24.2 ms/s? Otherwise, I think you'll impact the planet.

It says 0.242 km/s HEO lowering, I haven't done the maths... I wouldn't be too surprised if there is some degree of plane change included in that figure

[InterestedEngineer]

Noted. Thanks.

are you sure you meant 242 m/s not 24.2 ms/s? Otherwise, I think you'll impact the planet.

It says 0.242 km/s HEO lowering, I haven't done the maths... I wouldn't be too surprised if there is some degree of plane change included in that figure

Why is perigee at 7,800 km?

[Brigantine]

Why is perigee at 7,800 km?

That's a great question, and the best hypothesis I could come up with is that it's to mitigate collision avoidance complications, including in the event of an "anomaly" during refueling. I imagine debris in a e.g. 300 x 113,000 km orbit has the potential to cause inconvenience for a considerable while, unlike LEO. Essentially Pe = 7,800 is an acceptable enough accidental graveyard orbit

If so, perhaps a suborbital Pe would be an acceptable alternative at lower cost... depending on timing and geography

[Further NEA - specific discussion: thread exists]
[Any comments on broader implications for HEO re-fueling in general: still on topic]

[tbellman]

The original motivation for it, as I recall, was that it appears that an HLS Starship that refueled in LEO before heading to NRHO and then to the moon would find itself just a little short of enough fuel to get back to NRHO. Refueling at NRHO is off the table because NASA won't allow the vehicle to be refueled twice. Hence the idea of a depot in HEEO. In this case, the determining factor is what's the most elliptical orbit an empty HLS can get into on a single load of fuel. That's when you put the depot.

I have to believe, though, that there are better knobs to twist to make this happen just using a single depot in LEO.

(My bolding.)
I have not heard any such requirement.  I suspect that you are misremembering and mixing a couple of things up:

• First, in the source selection statement for HLS Option A, Kathy Lueders wrote about the Starship propellant refilling that the high number of launches needed (up to 14 tankers, one depot, and one lander, all in just six months) meant a high risk of delays, but that it was tempered by doing them in Earth orbit (presumably LEO) rather than in lunar orbit "where an unexpected event would create a much higher risk to loss of mission".

But this is not an outright ban, it is just an additional risk.  And it is mostly a schedule risk: if refilling in NRHO fails, then it will take longer to send up a new tanker/depot there.  That could e.g. be long enough that all propellant in the lander ship has time to boil off, and the ship then becomes derelict.

This was also written in April 2021.  In the preceeding six months, SpaceX had launched 19 times.  Doing 16 launches in a similar length of time, and doing it in 2024, just to support a lunar landing, was not something NASA could consider a trivial task when they evaluated the HLS bids.

• Second, doing a second refilling in a high elliptic orbit, would entail passing through the Van Allen belts thrice instead of once (on the outgoing leg of the journey).  You probably want to avoid that if the ship is carrying living humans during those passages due to the radiation.  But:
A) A second refilling doesn't have to happen in such an orbit; e.g. at the Lunar Gateway in NRHO, during the trans-lunar coasting, or in low lunar orbit after taking off from the lunar surface, depending on the mission.
B) The HLS lander will not be carrying crew until after rendez-vous with Orion in NRHO, so unless the ship itself becomes significantly radioactive due to passing through the Van Allen belts multiple times, this doesn't matter.

[Barley]

The original motivation for it, as I recall, was that it appears that an HLS Starship that refueled in LEO before heading to NRHO and then to the moon would find itself just a little short of enough fuel to get back to NRHO. Refueling at NRHO is off the table because NASA won't allow the vehicle to be refueled twice. Hence the idea of a depot in HEEO. In this case, the determining factor is what's the most elliptical orbit an empty HLS can get into on a single load of fuel. That's when you put the depot.

You also have to look at the least elliptical orbit where a fully fuel HLS has to be to complete the mission.  The final refueling can be anywhere between that and the most elliptical orbit you can reach.  If there is no overlap, you cannot complete the mission subject to the single refueling restraint.

[Barley]

3km/sec requires a mass ratio of  2.27 which uses 62% of the fuel.  You'd have to boost two refuelers to the HEO and that'd have to all be co is ordinated over 2 day orbits.  Yikes.

IMHO the yikes is completely inappropriate.  It's not like navigating a sailboat, where wind and tide are at least somewhat unpredictable.  If you do the orbital math right everything ends up exactly where you need it.  The math is absolutely trivial compared to every other part of a rocket program, such as finite element analysis, let alone computational fluid dynamics.  I am often amazed by how much money and effort engineers will expend to avoid simple calculations.

Show the plan including all the 200t launches to LEO to make this work.   then tell me it's inappropriate.

Not to mention an additional $30M in fuel.  (1200t * $25k/ton = $30M).  That's an optimistic estimate.  It takes half of that to make a Mars trip, so wherever it is you want to go, it better be worth the extra 1-2 km/sec (it likely isn't).

To solve the problem you need to work backwards from the goal.  If you need the extra 1km/s it's priceless. "Yikes  lets not return them safely to Earth" is not going to work.

We don't know the mass of starship.  We don't know the cost to launch.  We don't know how hard or trivial refueling is.  We don't know about boiloff.  Far too early to take parts of the trade space off the table.  Easy refueling with low performance SS leads to a different plan than hard refueling with outstanding performance.

[Twark_Main]

You'd have to boost two refuelers to the HEO and that'd have to all be coordinated over 2 day orbits.  Yikes.

If you have three full vehicles (two refuelers + mission ship) in LEO, it's wasteful to take all three to your highest orbit.

Instead, you want a lower orbit (requiring less delta-v), such that they arrive two-thirds full. Then you dump one tanker into the other two vehicles, and the nearly empty tanker returns to Earth. Then your two (now full) vehicles repeat as before.


Don't worry about the measly two extra refueling events. If you had three full tankers in LEO, you already did 18 refuelings just for that. So in reality, all this does is is go from 19 refueling events to 21 refueling events (~10% more).


Yes, it's a lot of math. You have to reverse-calculate the height of each elliptical orbit, based on the tanker parameters, the mission vehicle parameters, AND the individual payload mass. That's why I made my big tanker ladder spreadsheet: it does all the math for you.


P.S.:  Let's not go from arbitrarily assuming a 2 week period HEEO to now arbitrarily assuming a 48 hour period HEEO.    That would be missing the forest for the trees in my post here.

My point is that we should choose the elliptical orbit based on the mission constraints themselves. What we should not do is choose an elliptical orbit by throwing darts at a dartboard ("2 weeks", "48 hours", etc), then only afterwards calculate the relevant mission constraints on that orbit, and finally throw up our hands in defeat because we found exactly one orbit that doesn't work. 

[Twark_Main]

You'd have to boost two refuelers to the HEO and that'd have to all be coordinated over 2 day orbits.  Yikes.

If you have three full vehicles (two refuelers + mission ship) in LEO, it's wasteful to take all three to your highest orbit.

Instead, you want a lower orbit (requiring less delta-v), such that they arrive two-thirds full. Then you dump one tanker into the other two vehicles, and the nearly empty tanker returns to Earth. Then your two (now full) vehicles repeat as before.

...

There's an interesting middle ground here as well. You can have your tankers do a single tandem refilling, then have the remaining full tanker rendezvous with the mission vehicle in the highest possible orbit.

The advantage is that your mission vehicle only has one refilling event in HEEO, and it limits the number of passes through the Van Allen belt.

This exact refilling ladder can't be calculated using my refilling spreadsheet, but the math is pretty easy. The trick is to remember that for optimal rocket staging each stage provides half the total delta-v. Since tanker ladders are essentially identical (mathematically) to parallel staging, this means that the delta-v from LEO to the intermediate HEEO is exactly half that of the delta-v to the final HEEO.

This allows us to construct an equation:

v2 = 2 * v1

Where v1 is the excess velocity (from LEO) of the intermediate HEEO, and v2 is the excess velocity of the second HEEO.

Solve for v1. 

We know we need the tanks to be full at the end, which allows us to construct more equations. Let's call the mass of our final propellant transfer X, and use the rocket equation.

Mission vehicle burn directly to HEEO2:
v2 = Isp * ln(m1/m0) = Isp * ln[ (mission vehicle wet mass ) / (mission vehicle wet mass - X) ]

Tanker burn from LEO --> HEEO1 and HEEO1 --> HEEO2:
v1 = Isp * ln(m1/m0) = Isp * ln[ (tanker wet mass) / (tanker wet mass - (mission vehicle prop capacity - X)) ]

Substitute these into the first equation and simplify:

ln[ (mission vehicle wet mass) / (mission vehicle wet mass - X) ] = 2 * ln[ (tanker wet mass) / (tanker wet mass - (mission vehicle prop capacity - X)) ]

Insert numerical values:

ln[(120 + 150 + 1200) / (120 + 150 + 1200 - X) ] = 2 ln[ (100 + 1200) / (100 + 1200 - (1200 - X)) ]

Get lazy and just use Wolfram|Alpha  :

x = 786.47 tonnes

Substitute back into the original equation to find v2 and v1:

v2 = Isp * ln[ (mission vehicle wet mass) / (mission vehicle wet mass - X) ]
v2 = 375 s * 9.80665 m/s^2 * ln[ (120 + 150 + 1200) / (120 + 150 + 1200 - 786.47) ]
v2 = 3677.49375 m/s * ln[ 1470 / 683.53 ]

v2 = 2816 m/s
v1 = 1408 m/s


Sorry for the long post, but hopefully this "worked example" is helpful.

[Barley]

There's an interesting middle ground here as well. You can have your tankers do a single tandem refilling, then have the remaining full tanker rendezvous with the mission vehicle in the highest possible orbit.

The advantage is that your mission vehicle only has one refilling event in HEEO, and it limits the number of passes through the Van Allen belt.

This exact refilling ladder can't be calculated using my refilling spreadsheet, but the math is pretty easy. The trick is to remember that for optimal rocket staging each stage provides half the total delta-v. Since tanker ladders are essentially identical (mathematically) to parallel staging, this means that the delta-v from LEO to the intermediate HEEO is exactly half that of the delta-v to the final HEEO.

Optimal is a widely misused term.  You should always ask optimal with respect to what?
In this case optimal means sizing the two stages.  Since we have existing Star Ships and tankers we are solving a different problem.  If we want to force the use of that "optimal" to size the stages were are constrained by the availability of small integers, and fractions composed of small integers.

Lets start with three ships, two tankers with dry mass 100 tonne, a mission vehicle with dry mass 120+150 tonne, all vehicles hold 1200 tonne of fuel, Isp is 375s so exaust velocity is 3677 m/s

Burn two tankers in parallel using half the fuel.
m₀ = 100 + 1200 + 100 + 1200 = 2600
m₁ = 100 + 100 + 1200 = 1400

m₀/m₁ = 1.857
Δv₁ = 3677 * ln(1.857 ) = 2276 m/s

Meanwhile we burn the mission ship to the same delta_v  Since the exhaust velocity is the same the mass ratio will be the same so
m₁^mission = (1200 + 120 + 150)/1.857 = 792

As a sanity check this is between the fully loaded and fully empty masses of the mission ship.

Now we "stage" by moving the remaining 600 tonne of fuel from one tanker into the other and "discarding" the dry weight of the empty tanker.  We will burn the remaining tanker and mission ship in parallel until 1200 tonne of fuel remain to fill the mission ship.

m'₀ = 1200+100+792 =2092
m'₁ = 1200+100+120+150 = 1570
Δv₂ = 3677 * ln( 2092/1570) = 1055

So the intermediate HEEO is at LEO + 2276
And the final HEEO is at LEO + 3331

I have not shown these are optimal (although within the constraints they are) but they are clearly possible, better than your "optimal" and don't need Wolfram alpha, just a log table.

[Keldor]

There's an interesting middle ground here as well. You can have your tankers do a single tandem refilling, then have the remaining full tanker rendezvous with the mission vehicle in the highest possible orbit.

The advantage is that your mission vehicle only has one refilling event in HEEO, and it limits the number of passes through the Van Allen belt.

This exact refilling ladder can't be calculated using my refilling spreadsheet, but the math is pretty easy. The trick is to remember that for optimal rocket staging each stage provides half the total delta-v. Since tanker ladders are essentially identical (mathematically) to parallel staging, this means that the delta-v from LEO to the intermediate HEEO is exactly half that of the delta-v to the final HEEO.

Optimal is a widely misused term.  You should always ask optimal with respect to what?
In this case optimal means sizing the two stages.  Since we have existing Star Ships and tankers we are solving a different problem.  If we want to force the use of that "optimal" to size the stages were are constrained by the availability of small integers, and fractions composed of small integers.

Lets start with three ships, two tankers with dry mass 100 tonne, a mission vehicle with dry mass 120+150 tonne, all vehicles hold 1200 tonne of fuel, Isp is 375s so exaust velocity is 3677 m/s

Burn two tankers in parallel using half the fuel.
m₀ = 100 + 1200 + 100 + 1200 = 2600
m₁ = 100 + 100 + 1200 = 1400

m₀/m₁ = 1.857
Δv₁ = 3677 * ln(1.857 ) = 2276 m/s

Meanwhile we burn the mission ship to the same delta_v  Since the exhaust velocity is the same the mass ratio will be the same so
m₁^mission = (1200 + 120 + 150)/1.857 = 792

As a sanity check this is between the fully loaded and fully empty masses of the mission ship.

Now we "stage" by moving the remaining 600 tonne of fuel from one tanker into the other and "discarding" the dry weight of the empty tanker.  We will burn the remaining tanker and mission ship in parallel until 1200 tonne of fuel remain to fill the mission ship.

m'₀ = 1200+100+792 =2092
m'₁ = 1200+100+120+150 = 1570
Δv₂ = 3677 * ln( 2092/1570) = 1055

So the intermediate HEEO is at LEO + 2276
And the final HEEO is at LEO + 3331

I have not shown these are optimal (although within the constraints they are) but they are clearly possible, better than your "optimal" and don't need Wolfram alpha, just a log table.

The optimal refueling altitude is always going to be as low as possible.  This is because otherwise we're just moving extra tank mass around when we don't have to.

For a high energy mission, we'd want to start by topping off the payload vehicle as well as another tanker in low orbit, then both will burn into a high elliptical orbit, where the second tanker will top off the payload vehicle one final time before it leaves Earth orbit for good.  What we don't want to do is try to do all our refueling in high orbit, where we pay to drag dozens of nearly empty tankers all the way up.

[Greg Hullender]

Yes, it's a lot of math. You have to reverse-calculate the height of each elliptical orbit, based on the tanker parameters, the mission vehicle parameters, AND the individual payload mass. That's why I made my big tanker ladder spreadsheet: it does all the math for you.

The way I like to think of the basic problem is that you start with one fully fueled Starship and one depot in circular LEO with at least some fuel in it, and you want to end up with a fully fueled Starship in HEEO and an empty depot. This means you'll want to accelerate them in tandem until the point where, between the two of them, they have exactly enough fuel for a single Starship. Then you transfer all the remaining fuel from the depot to the Starship. Since the velocities have to match, so do the mass ratios, and that lets you solve for everything.

This ends up being a function of the wet mass of the Starship (S), the wet mass of the Depot (D), the amount of fuel in the Depot (F), and the specific impulse of the Starship (k). Then

∆v = k ln((S+D)/(S+D-F))

Which is quite elegant; the argument of the logarithm is just the mass ratio of the two ships taken as a unit. (Imagine that they're bolted together.)

It doesn't matter how big the payload is (assuming we're counting that as part of the wet mass) nor how much fuel the Starship actually holds. However, if F is too big (about 3000 tons of fuel in my examples), then your fully fueled Starship ends up on a hyperbolic orbit, not a high Earth orbit, and that spoils the plan, which called for doing a big Oberth burn at the next perigee.

For the examples I played with, this required a 2.5 kps burn, putting the vehicles into a 13.6-hour orbit with an apogee of 51,000 km (from Earth's center). For the payloads I was playing with, this amounted to 3.5 kps extra hyperbolic excess velocity (on top of 10.5 kps I got without this trick), which is not too shabby.

By the way, I might have found an error in your spreadsheet. We get the same answers for delta-v but different answers for apogee. They're only different by about 6.5%, but that's a lot since the other numbers are within 0.1%. You're using a very complicated formula for apogee, and that may be the problem. I'm using r_a = (r_p*v_p)^2/(2*mu-r_p*v_p^2) where r_a is the apogee (measured from the center of the Earth), r_p is the perigee, v_p is the velocity at perigee, and mu is the standard gravitational parameter for the Earth.

I did not explore your idea of using two or three extra depots.

[Greg Hullender]

And the final HEEO is at LEO + 3331

Hmm. I compute escape velocity from a circular orbit with an altitude of 200 km to be 3226 m/s, so that's likely to be a problem. :-)

[Greg Hullender]

Okay, thinking about the multiple-depot case, I think I see the optimal (i.e. max ∆v) way to use a "wagon train" comprising a fully fueled Starship and n fully fueled depots. Let S be the mass of a fully fueled Starship (including payload). Assume all depos have the same wet mass (D) and the same mass of fuel (F). Assume that the specific impulse is the same for all vehicles.

In the case with a Starship and a single depo, we wanted to boost until F fuel was consumed between the two vehicles. At that point, fully fueling the Starship fully drains the depot. That this is optimal probably doesn't need elaborating. The boost that accomplishes this is ∆v = k ln((S+D)/(S+D-F)). (The trick to finding this is to see that the mass ratios for the two vehicles must be equal because both have the same ∆v.)

When you add another depo, the main thing you want to do is to discard excess mass at the first opportunity. Otherwise, you're wasting fuel. Therefore, just as in the single-depot case, you want to do a single burn that consumes F fuel. After that, you can redistribute what's left, and you will discard exactly one depot. Then repeat.

It shouldn't take much thought to see that ∆v_n = k ln((S+n*D)/(S+n*D-F)). Then sum these up to get the total ∆v--note that this does not include the ∆v from the Starship itself.

However, using what I think are reasonable numbers, if you start from a 200-km-altitude orbit, you can't use even just two depots; between the two of them, you'll go 800 m/s over escape velocity. So this trick is likely to be of interest only for a single depot.

[Greg Hullender]

The optimal refueling altitude is always going to be as low as possible.  This is because otherwise we're just moving extra tank mass around when we don't have to.

For a high energy mission, we'd want to start by topping off the payload vehicle as well as another tanker in low orbit, then both will burn into a high elliptical orbit, where the second tanker will top off the payload vehicle one final time before it leaves Earth orbit for good.  What we don't want to do is try to do all our refueling in high orbit, where we pay to drag dozens of nearly empty tankers all the way up.

I think all anyone is talking about is putting fuel depos in LEO--as low as practical--and then filling them up with a series of "tanker" flights from the ground. The depots are likely to be "stretched" Starships, holding perhaps 1500 tons of fuel each.

The simplest use case is for a Starship launched from the ground to rendezvous with a single depo (in LEO), refuel itself, and then do a big burn to higher orbit, to the moon, or elsewhere. If that works, that's what you ought to do--no question. (I don't think anyone is debating this point.)

But this doesn't seem to have enough ∆v for some missions.

So assume you've got a fully fueled Starship in LEO and a second, fully fueled depo in LEO. How do you leverage that second depo to add some ∆v to your Starship? This is the problem everyone is trying to solve.

Boosting them both into the same HEO and then transferring the remaining fuel from the depo to the Starship seems like the best way to do that. When the fully fueled Starship comes back to perigee, it does the burn for its final destination. This can buy you as much as 2.5 kps at perigee--3.5 kps of additional hyperbolic excess in the examples I played with.

Potentially, the (almost) empty depo does aerobraking to eventually get back to LEO.

Even then, though, as you say, you want to make the fuel transfer in the lowest possible orbit. Too low and the Starship won't have room for the fuel. Too high, and the depo won't have enough fuel left to fully fuel the Starship.

[Bob Shaw]

If your launch costs are unreasonably cheap then all these edge of case studies don't matter. You accept the losses, fuel up and fire up. If Starship works we're no longer in marginal economics mode - the old paradigms simply don't apply.

We must wait and see.