13
$\begingroup$

SpaceX's attempt at a reusable first stage rocket sounds amazing, but how they're going to get it to land back on the launchpad is beyond me. Assuming that the first stage will separate with a speed of about 3 km/s at a downrange distance of around 150 km, they'll essentially need to provide more than 3 km/s retrograde delta-v and somehow cover the 150 km back to the pad. Or am I being stupid and missing something obvious? I would have thought that a mobile water-based pad would be ideal, as it could save a ton of the required delta-v, but I haven't heard anything about such plans.

So, does anyone know how they're going to do it? Does the Falcon 9 v1.1 really have enough fuel to provide enough delta-v to get back to the pad and slow down sufficiently before doing a controlled landing?

EDIT: I'm trying to create a rough simulation using some differential equations, but I cannot find a reliable enough source for the downrange distance, altitude and velocity (i.e. x and y-components or at least speed and flight path angle above horizon) of the first stage at MECO and separation. I know these values will be different depending on the mission, but are there any stats for, say, a resupply mission to the ISS?

Improve this question
$\endgroup$
2

5 Answers 5

Reset to default
13
$\begingroup$

They have said that they will reserve around 15% of the fuel capacity of a first stage for reusability operations. At the point they need to impart the Delta-V to return to base, they will thus be 85% empty.

Thus the need for only 3 engines instead of the 9 main engines for the retro propulsion burn.

I won't try to do the math, but they claim they have, and think it will work.

There was some discussion I did not entirely follow about whether 2 or 3 burns are required to return. They need to cancel the velocity in the forward direction, and start heading back. Then need to control entry back into the atmosphere in a survivable fashion.

Remember they are lofted upwards and forwards when MECO-1 and stage separation occur. They can ride the upward component and focus on the forward vector to cancel it out.

For Falcon Heavy it is trickier, since the middle core stage burns till much later and is quite a bit higher and faster at it's MECO.

Some have speculated that launching out of Texas, and recovering in Florida might buy some margin on reusability operations.

For the CRS-5 flight (Dec 2014 scheduled date) Musk has said they will use a barge being built in Louisiana (50mX70m) to try to land on at first, until they can prove to the FAA they can control the landing accurately enough. Musk predicts only 50% chance of success on first attempt.

By the CRS-8 flight, Apr 8, 2016, they successfully landed a first stage on the ASDS barge "Of Course I Still Love You". They made several attempts in which they literally hit the barge, demonstrating the ability to get back to the barge, and finally the ability to land. Elon Musk in the press conference after the launch suggested that fully 1/3 of their future flights would use the barge for landing. Usually heavy GEO missions that needed the extra performance.

Improve this answer
$\endgroup$
11
  • 1
    $\begingroup$ 15% of the fuel mass is 65.7 T or so, the stage 1 mass is about 28 T empty, full cluster thrust is 598.8 T or so... a single thruster is capable of more than 1G; the stock burn time is about 170 sec for 9 engines; 15% of the fuel on a single engine of the cluster of 9 is about 230 sec and still producing positive thrust... throttled back, 3 engines should easily be able to do so. $\endgroup$
    – aramis
    May 7, 2014 at 1:21
  • 4
    $\begingroup$ The planned SpaceX space port on Texan Boca Chica Beach, Brownsville on the border to Mexico, 1500+ km west of Florida, certainly suggests the concept of Texas-launch and Florida-landing. But I imagine that soon rockets will fly from the factory to the launch pad, to near LEO, back to ground somewhere and then to another launch pad where another satellite is waiting. They are rockets. They are built to move. Why intermittently turn them into cargo on inferior transports such as ships or trains? $\endgroup$
    – LocalFluff
    May 7, 2014 at 15:05
  • 1
    $\begingroup$ @LocalFluff Almost entirely will depend of lifespan of the engines. If you only get 15 firings of a Merlin, then delivery from factory to test facility is 1. Test run 2. Fly to launch site 3. Launch 4. Land and return to launch site 5. That ate up a good possible segment of reuse of the engine. Now that number of 15 is totally arbitrary. If a truck can cheaply/quickly move it, saving a full engine firing may be worth it. $\endgroup$
    – geoffc
    May 7, 2014 at 15:14
  • 4
    $\begingroup$ @LocalFluff Those "inferior transports" have far greater reliability, safety, fuel efficiency, cost per kilogram of payload, less specialized "launch" and "landing" facilities, less specialized "launch" crew, and don't put extra stress on a very, very, very expensive cluster of rocket engines. SpaceX ain't gonna change that in 10 years. $\endgroup$
    – Schwern
    Nov 12, 2014 at 18:50
  • 1
    $\begingroup$ @Schwern "SpaceX ain't gonna change that in 10 years" --Correct, it looks like they changed it in 2 years. $\endgroup$
    – Mike Vonn
    Apr 12, 2016 at 20:31
7
$\begingroup$

Based upon the statement of 15% of fuel being reserved, and three engines used for the landing phase... and doing a simple "back of the envelope" kind of calculation...

**Stock Data**
  28.0 T   Falcon 9 v1.1 dry mass (est)
   5.0 T   this author's wing-it for the landing legs on the v1.2
 411.0 T   Fuel Mass (est)
 598.7 TT  Full thrust (at MSL) 9 engine cluster
  66.5 TT  Thrust per engine
 170.0 sec Burn Duration
1530.0 e•s engine-seconds burn duration

This gives us enough to work from.

  28.0 T   Falcon 9 v1.1 dry mass (est)
   5.0 T   this author's wing-it for the landing legs on the v1.2
  61.6 T   Fuel Mass (est) at separation
  94.6 T   est. mass at separation
 199.5 TT  Thrust on 3 engines at full throttle.
  66.5 TT  Thrust per engine
  46.5 TT  Thrust per engine minimum (70%)
  76.5 sec Burn Duration 3 engines full throttle
 229.5 e•s engine-seconds burn duration
 327.9 e•s burn duration engine seconds at 70% throttle (minimum)

It looks to me like they separate at about 144 seconds, fall for a bit, then begin powered flight on a single throttled back engine. Elon has said the goal is 40% throttle minimum, and so I would expect the Merlin 1E to go lower still - but the thing's still got ≥1G at landing. with the 1D @ 70%. At separation, it's got about 0.6G per engine. More than adequate to cancel forward velocity, and begin the landing sequence.

The problem is throttle-back, not adequate thrust!

Improve this answer
$\endgroup$
6
$\begingroup$

Horizontal component of first stage velocity will be under Mach 6 or under 2,000 meters/sec as Elon Musk stated previously (in a Popular Mechanics magazine article from 2012) for recovery of his 1st stage. Remember that SpaceX announced F9R wind tunnel tests at NASA-Marshall up through Mach 5. Remember that overcoming gravity losses is ~ 1,500 meters/sec of 1st stage delta V so total 1st stage burn out delta V is ~ 3,500 meters/sec = 2,000 meters/sec horizontal velocity + 1,500 meters/sec gravity losses.

That 15% number is a reduction of orbital payload by 15% (or 30%) and probably not a massive 15% fuel reserve on 1st stage of 60+ tons propellant. A 32 ton propellant reserve is more likely.

Air Force RLV booster studies and Kistler had estimated rocket-back maneuver to launch pad, including bleeding off Mach 4 to 7 in horizontal velocity, to require ~ 3,000 meters/sec delta V (according to the cited AF RLV booster return study that states 11,000 fps - 1,000 fps = 10,000 fps = 3,000 m/s as delta V needed for return to launch pad). This is closer to 32 tons propellant reserved for the 3,000 m/s delta V needed for the 1st stage to rocket back to the original launch pad.

Elon Musk has already said that F9R landing legs weight are less than Tesla Model S weight of 2.1 tons and that mass fraction of F9R 1st stage is under 4%. You should assume 18-tons including legs in empty weight and ~ 32-tons fuel to bring 1st stage back to pad. They need to be able to throttle Merlin 1D back to 30% thrust (i.e. 20-tons thrust) for a soft landing of an empty 18-ton booster.

Air friction alone can slow booster to subsonic velocities, but it can also tear booster apart. SpaceX has to control reentry and position vehicle using multiple smaller firings so it does not "belly flop" (Elon Musk's words) into the atmosphere.

Use these numbers as your estimates:

418-ton fueled 1st stage with 18-ton empty weight (i.e. 16-tons for 4% mass fraction and 2 tons for the new landing legs) -- http://www.spaceflight101.com/falcon-9-v11.html;

50-ton burn out weight of 1st stage with 32-tons fuel left for rocket-back maneuver;

2nd stage fueled weight of 75-tons, empty weight of 5 tons, and payload weight of 13-tons, for 88-tons total payload + 2nd stage weight carried by 1st stage; and

total F9R rocket weight of 506 tons to place 13-ton payload into LEO and to rocket back 1st stage to launch pad.

I also assume 300-sec average Isp on 1st stage and 340-sec Isp on 2nd stage.

The above estimates fit with the F9R v1.1 weight and performance estimates made by this web site -- http://www.spaceflight101.com/falcon-9-v11.html

The above estimates should work in your calculations for F9R taking a 13-ton payload to LEO and then the F9R 1st stage having the delta V of 3,000 meters/second (obtained from 32-tons reserved propellant) to be able to perform a rocket-back maneuver to the launch pad.

I estimate that it will be less than 3-tons of added weight penalty on the F9R 2nd stage for the propellant, heat shield, (4) Super Draco thrusters, and landing legs needed to recover and reuse the F9R 2nd stage similar to recovery and reuse of the Dragon-2 spacecraft.

This means that a fully reusable F9R that reuses both its 1st and 2nd stages could carry 10-tons to LEO and less than 1-ton to GTO.

Improve this answer
$\endgroup$
7
  • 1
    $\begingroup$ I've often ran into problems with this general type of thinking for gravity losses. 1,500 m/s is a tempting figure to use, but its clearly flawed and far too high. If you do a Hohmann Transfer calculation you get something closer to 120 m/s. But which model is more accurate? For climb to orbit, its a mess of a combination of them. After first stage, I expect that air and gravity drag will be small, so I would endorse a ballpark calc of 10 or 9.5 - (7.9-2), for about 3.8 km/s contribution. That's what you wound up with anyway. $\endgroup$
    – AlanSE
    May 24, 2014 at 17:41
  • $\begingroup$ citation for the mass of the landing legs? $\endgroup$
    – aramis
    May 24, 2014 at 18:45
  • $\begingroup$ twitter.com/elonmusk/status/330054002148515841 $\endgroup$
    – Anom
    May 24, 2014 at 23:23
  • $\begingroup$ popularmechanics.com/science/space/rockets/… $\endgroup$
    – Anom
    May 24, 2014 at 23:28
  • $\begingroup$ enu.kz/repository/2010/AIAA-2010-8672.pdf $\endgroup$
    – Anom
    May 24, 2014 at 23:42
4
$\begingroup$

You're missing something obvious: Air friction.

The terminal velocity, that is the speed, the rocket would have when air friction and gravity are in balance, is a small fraction of orbital velocities in a few kilometers altitude. So just by reentering the atmosphere, they kill almost all of their momentum.

After reentry, they may also use aerodynamic forces to help get back to the launch pad, but I don't know if that will be a significant factor. In any case, while they need to speed the rocket up quickly during launch, they don't need (or want) a lot of speed to return to base.

Improve this answer
$\endgroup$
7
  • $\begingroup$ I'll do some calcs when I get home, but if they're going to rely on air friction instead of engine thrust for the bulk of the slow-down, then by the time the first stage re-enters the atmosphere for sufficient drag to slow it down, it'll probably be closer to 200km downrange and at an altitude of only about 50-40km. That seems like it'll be quite low, but if my calculations are correct, it'll still have about 3.65 km/s delta-v left to get back to the pad, so maybe this will do. $\endgroup$
    – InquisitiveInquirer
    May 7, 2014 at 14:54
  • 2
    $\begingroup$ After doing a quick test in the Orbiter sim it looks like using friction is a complete no-go if the F9 first stage has any hope of getting back to the launch pad. Friction substantial enough to produce any sort of slow-down only occurs at around 45km altitude, but even worse, by that time the first stage is more than 1000km from the launch pad! $\endgroup$
    – InquisitiveInquirer
    May 7, 2014 at 18:29
  • 1
    $\begingroup$ But if launched from "Boca Chica Beach" (Brownsville) in Texas and landed in Florida, about 1500 km to the east, then your calculation show an advantage for SpaceX F9r, doesn't it? $\endgroup$
    – LocalFluff
    May 8, 2014 at 11:00
  • 1
    $\begingroup$ Yes, that definitely sounds like a plausible solution, but then they'll have to contend with 1) Strenuous re-entry conditions and 2) Have the FAA (or some similar body) give them the go-ahead to allow the F9 to operate close to inhabited areas (you don't exactly want a rocket to land in your back yard if something goes awry during its ascent stage :P) $\endgroup$
    – InquisitiveInquirer
    May 8, 2014 at 11:36
  • 1
    $\begingroup$ Launch from Texas, fly over gulf, land in Florida, ship back seems like an elegant cycle. And if there's any issue, you have a fair chance of ditching in the gulf. Operating close to inhabited areas might not be the biggest problem in Florida, protected wetlands might be a bigger issue. $\endgroup$
    – MSalters
    May 8, 2014 at 19:51
3
$\begingroup$

The last water landing was about six miles down range from the launch site. About as close as they wanted to get. The goal is to come back to the same launch site, and have the mobile launch tower go and get it very soon after landing. This might be a few hundred feet from the assembly and launch site. I did see where the planned landing sight is at the California complex. I asked as was assured that SpaceX increased the Falcon 9 from the ten to 13 metric tons was so they could fulfill commitments but use the fly-back system and it might lower the payload buy up to that much because of fuel needs to do fly back to the launch sight. I would have designed it so that it could land a few hundred mile down range close to a significant airport that could handle heavy air transport and use that to bring the stage back to the launch sight. It would use the tired wheeled launch tower as in in California instead of the first one in Florida on rails. The erection and launch tower can drive over to where the booster has landed, grab it, bring it down, drive back to the horizontal assembly tower, or possibly drive into a transport boat, train, and or plane. At the assembly area they would check it out, attach new payloads, possibly sometimes change some of the engines, do any other maintenance needed, and reuse it as a new rocket.

The back and bottom of the booster does have a very thin heat shielding material. Early plans called for this supplying most of the breaking at reentry, but current plans are to use fuel to do most of that, which is one of the reasons a larger booster and more return fuel is needed. It might take up to thirty percent of the payload as fuel to return.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.