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I guess my question is related to somehow assembling and fueling a falcon 9 in LEO then launching it. I would say if all stages were fully fulled and the Merlins were all vacuum optimized.

Situation 1: How fast could it get up to with no destination in mind and you burned all the fuel.

Situation 2: Lets say its heading towards Mars and you wanted to leave enough fuel to get into mars orbit. Then how fast could you get there?

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    $\begingroup$ The vacuum optimization for Merlin involves adding a very large engine nozzle extension; there wouldn't be room for 9 of them at the base of the first stage. $\endgroup$ Commented Jan 27, 2020 at 22:07
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    $\begingroup$ Several issues: first, due to the Oberth effect, it matters a great deal when, where, and in what direction you perform your burn. Second, just computing the delta-v available relies on details of the Falcon 9's mass ratios and stage dry masses that can only be estimated from public information; as Russell Borogove mentioned, the engines don't fit; and the Falcon 9 upper stage can only last a few hours due to LOX boiling off, RP-1 getting cold and gelling, and batteries running down. You're better off fully defining a Falcon 9-like vehicle. $\endgroup$ Commented Jan 27, 2020 at 23:02
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    $\begingroup$ The fact that the rocket is launching in a vacuum would be insignificant compared to the fact that the rocket would already be in orbit, meaning it would already be travelling at 8km/s. A launch with a reasonable ascent profile does not actually waste very much energy overcoming the atmosphere. $\endgroup$
    – Avi Cherry
    Commented Jan 27, 2020 at 23:17
  • $\begingroup$ Here's a figure of 1.3-1.8km/s total drag to LEO, with a reference $\endgroup$
    – Avi Cherry
    Commented Jan 27, 2020 at 23:24

2 Answers 2

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It depends on if you want any payload on it. Using the standard Merlin 1Ds on the first stage (Merlin Vacs won't fit), and using the specifications for the Falcon 9 FT given on Spaceflight101, I compute the following delta-V figures using the rocket equation:

  • 0 payload: 15645 m/s
  • 10 tons payload: 11497 m/s
  • 20 tons payload: 9776 m/s

(These are probably a little generous; I'm assuming all of the given propellant mass can be used, but normally engines are shut down prior to complete depletion for safety reasons.)

Situation 1: How fast could it get up to with no destination in mind and you burned all the fuel.

Adding the 7800 m/s you started with in low Earth orbit, even a 20-ton payload gets you up to 17500 m/s, comfortably higher than Earth escape velocity, comparable to New Horizons' departure speed.

Situation 2: Lets say its heading towards Mars and you wanted to leave enough fuel to get into mars orbit. Then how fast could you get there?

The upper stage of Falcon 9 has a very limited battery life, and can't keep its liquid oxygen from boiling off over that kind of journey, so it won't be able to get into Mars orbit by itself. If the payload includes a storable-fuel rocket that can do its own orbital insertion, then you can use all of the F9 to leave Earth orbit. Assuming the orbital insertion stage and payload is around 20 tons, and we're thus leaving Earth at New-Horizons-like speeds, it's about 80 days to Mars, in contrast to the usual 150 to 300 days it takes for fuel-efficient trajectories.

Note that since you’re arriving at Mars at such high speed, you’re going to need a lot of fuel to slow down into orbit, so not much of the 20 tons is going to be available for useful payload.

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  • $\begingroup$ I'm confident that your answer is the result of some reasonable calculation, but for future readers to judge for themselves is it possible to add a short explanation for how you compute these numbers and what assumptions were used? Thanks! $\endgroup$
    – uhoh
    Commented Jan 29, 2020 at 2:29
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    $\begingroup$ It's a simple question of weight, er, mass ratios, as usual, but I've added a link to the rocket equation. $\endgroup$ Commented Jan 29, 2020 at 4:02
  • $\begingroup$ This is more or less exactly what i was looking for! Also love all the Monty Python references. Would there be a large difference if they did 5 Vac Merlins (or however many would fit?) instead of the 9 1D? Also does this include the upper stage? could that be used for orbital insertion? Thanks a ton! $\endgroup$
    – Stickyz
    Commented Jan 29, 2020 at 13:50
  • $\begingroup$ This is using both the first and second stage for departure. The second stage can’t retain power and LOX for 80 days. Merlin vac on the first stage would give you something like another 500 m/s off the top of my head. $\endgroup$ Commented Jan 29, 2020 at 16:15
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Partial answer only.

This could turn out to be quite a complicated question to answer fully, but it turns out that they've done the experiment for the Falcon Heavy and we know the answer for that one at least.

The Tesla Roadster launched first to a high apoapsis (about 7000 km) elliptical LEO orbit and stayed there for 6 hours and then went off to deep space by burning every last bit of propellant.

That resulted in a heliocentric orbit with a perihelion of 0.99 AU and aphelion ~1.7 AU.

It is now catalogued as such.

You can find out where it is now and where it is going and how fast at https://www.whereisroadster.com/

So a "how fast could it go" question really depends on what orbit you put it in. If you aim for a highly elliptical heliocentric orbit then at perihelion (closest to the Sun) it will be going a lot faster than at aphelion one-half period later.

I think that ultimately you question could best be phrased as

What is the air speed velocity of an unladen swallow Falcon?

To which the answer is

African or European?

or even better:

What heliocentric $C_3$ can a Falcon 9 achieve with zero payload?

For more on $C_3$ see https://space.stackexchange.com/search?q=user%3A265+C3

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    $\begingroup$ for the air speed velocity of an unladend swallow see 1, 2 $\endgroup$
    – uhoh
    Commented Jan 28, 2020 at 4:26
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    $\begingroup$ Isn't the crux of this question related to having the first stage lit off in LEO? Which the roadster thing didn't do. $\endgroup$ Commented Jan 28, 2020 at 5:24
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    $\begingroup$ @OrganicMarble yes, that is why this is labeled "Partial answer only" up at the top. Several comments below the question address how to approximate drag losses, this is another part of the puzzle useful towards formulating a complete answer but too long to leave as a comment. $\endgroup$
    – uhoh
    Commented Jan 28, 2020 at 5:31
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    $\begingroup$ Total +1 for Holy Grail references, that in context, make a lot of sense. Well played sir, well played indeed! $\endgroup$
    – geoffc
    Commented Jan 28, 2020 at 15:38

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