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Reusability is a key strategy for SpaceX to lower the cost of orbital and interplanetary launches. With the Falcon 9 they are using the powered landing approach to recover the first stage for reuse.

Landing the Falcon 9 requires a suicide burn approach using just the the central Merlin engine. They have to time the burn just right because the Merlin can’t be throttled back enough to use a more measured, controlled landing.

With the Raptor engine being 2-3 times more powerful are they going to be able to land anything equipped with Raptor engines at all?

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  • $\begingroup$ You can make your own judgment as to the practicality of these claims, but they definitely plan to land their Rapor-powered BFSs and BFRs. $\endgroup$
    – Organic Marble
    Commented Mar 31, 2018 at 21:30
  • $\begingroup$ "...because the Merlin can’t be throttled back enough to use a more measured, controlled landing." Rapid deceleration at the last moment also consumes less propellant than slowing down gradually. These landings are generally incredibly measured and controlled and usually quite successful. $\endgroup$
    – uhoh
    Commented Apr 1, 2018 at 1:18
  • $\begingroup$ Thanks guys. I realize suicide burns are the most efficient way to land the boosters, it’s just mind blowing to me that they can hit almost 0ms within a couple of meters of the ground from orbit without easing it in $\endgroup$
    – Shane Taylor
    Commented Apr 1, 2018 at 21:13

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The premise of your question is incorrect.

The non-hovering suicide-burn approach isn't used because they can't throttle lower; it's used because it's the most fuel-efficient way to land. Every second in the air is a second that gravity is accelerating the first stage downward to the tune of 10 meters a second. The sharper the deceleration, the later you can start the burn, the shorter time it takes, the less fuel you have to spend on landing, the more fuel you can spend boosting payload.

In fact, they'd land hotter and faster if they could; they've been exploring the limits of the technology over the last couple of years, experimenting with three-engine final approaches to save a little bit of propellant.

That said: the Raptors may be more powerful, but the rockets they're going to be propelling will be larger than Falcon in the same proportion; the landing trajectories probably won't be vastly different from the ones they're using now.

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  • $\begingroup$ The BFR has many more engines, so I believe it may be capable of getting down to a 1:1 TWR or lower with subsets of the engines. All your efficiency points are valid though, and the real reason to continue suicide burns. $\endgroup$
    – Saiboogu
    Commented Mar 31, 2018 at 22:21
  • $\begingroup$ Perhaps coincidentally, the Falcon 9 booster can only barely throttle down to a hover. The dry mass of a Falcon 9 booster is about 25 tons, and the thrust of a single sea level Merlin 1D is 620kN. It seems like throttling was originally to 60% but "later refinements of the Merlin 1D can throttle down to 40%" (wikipedia), which would just be low enough for a hover. $\endgroup$
    – Steve Linton
    Commented Apr 1, 2018 at 12:27
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Based on available numbers we can get some ideas. Let's start with the BFS (the upper stage). Wikipedia gives a dry mass of 85 tons while a Raptor has a thrust at sea level of 1.7 MN. So to hover, an empty BFS would need one engine throttled to roughly 50%. I found one rather elderly report that the Raptor could throttle to 20% and another site from the same period that agrees but might come from the same source. If so, there is no problem (apart from fuel supply). This is probably good if they want to land humans this way.

The Booster has a dry mass around 155 tons according to one unsourced comment (harder data on this is hard to find) so could hover on one engine close to full thrust, although since it would be unmanned, I would expect them to save the fuel and come in hard on 3 or even 5.

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  • $\begingroup$ So they may be able to bring BFR boosters in with a less brutal, more controlled and accurate burn. Thanks $\endgroup$
    – Shane Taylor
    Commented Apr 1, 2018 at 21:22

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