If we launch a Saturn V rocket from space (far away from Earth so that Earth's gravity has no effect on it) then what would be the maximum velocity achievable? How close could it get us to the speed of light?
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$\begingroup$ Could be easily calculated from the mass and thrust of each stage which you can easily look up on wikipedia, but I am too lazy to do so right now ;). But I can say with certainty that it wouldn't even bring it close to relativistic speed. $\endgroup$– PhilippMar 13, 2014 at 14:57
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$\begingroup$ By the way, being close to a gravity-well could actually help to reach higher speeds. $\endgroup$– PhilippMar 13, 2014 at 14:58
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$\begingroup$ @phillip if you could take the mass of the S-V, and replace the hydrogen/oxygen/RP1 with say Xenon and use a constant thrust Ion style thruster, how fast could you go? That could get much faster than LOX/LH or LOX/RP1. $\endgroup$– geoffcMar 13, 2014 at 14:59
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1$\begingroup$ @geoffc The NSTAR Xenon-ion thruster has an Isp of 3120 seconds, and the NEXT 4190, roughly ten times that of Saturn V's chemical engines, and Δv increases proportionally to Isp, so you'd be in the ballpark of 170km/s or 0.05% c. Remember that you'll have to get your Xenon-Saturn out of atmosphere and into orbit by some other means; you won't have nearly enough thrust to overcome gravity and atmospheric drag. en.wikipedia.org/wiki/NEXT_(ion_thruster) $\endgroup$– Russell BorogoveMar 13, 2014 at 20:41
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6$\begingroup$ @MarkAdler - It's too bad you deleted your answer. It was a very good WAG. I'm not using that term demeaningly. The other way around: It's a complement. One problem that keeps this site as a beta site is that too many questions only get one answer. There's nothing wrong with having a selected answer and a non-selected one. Apparently, it's a good thing in the eyes of the stackexchange powers that be that questions have multiple answers. $\endgroup$– David HammenMar 13, 2014 at 21:15
1 Answer
The amount by which a spacecraft is able to change its velocity is called it's Δv (delta-velocity) budget. You can calculate the Δv-budget of each stage of a rocket using the Tsiolkovsky rocket equation which reads: $$ \Delta v = I_{sp} * 9.81 * \ln \frac {Mass_{full}} {Mass_{dry}} $$
where Isp is the specific impulse ("fuel-efficiency") of the engine. The factor of 9.81 (the gravity of earth) in this context is the factor used to convert specific impulse to exhaust velocity. You could also express the formula above using exhaust velocity, but I decided to use the specific impulse, because the numbers I found for the efficiency of the Saturn V were in Isp.
The masses and Isp of each stage can be looked up on wikipedia. I made an overview as a table and calculated the delta-v using this handy online calculator:
| Individual stage | Total vessel |
Stage | Full mass | Dry mass| Full mass | Dry mass | Isp | Δv
------+-----------+---------+-----------+----------+-------+--------------
I | 2,300,000 | 131,000 | 2,900,000 | 731,000 | 263s | 3554.2 m/s
II | 480,000 | 36,000 | 600,000 | 156,000 | 421s | 5561.5 m/s
III | 120,800 | 10,000 | 120,800 | 10,000 | 421s | 8796.2 m/s
===========
17911.9 m/s
* masses are in kg
That means a Saturn V started in space could reach a speed of about 18 km per second or 64,500 km/h relative to its initial frame of reference. The speed of light is about 300,000 km per second, so this is still just about 0.006% of the speed of light.
This speed is just the rocket without any payload. When you add a payload, that mass needs to be added to the dry- and full-mass of the total vessel in each stage and the delta-v of each stage decreases accordingly. I didn't do the math, but when you would add the Apollo 11 payload, the 12 km/s stated in the answer by Mark Adler appears to be plausible (but note that the Apollo 11 payload would add three additional stages with additional Δv budget - the service module, the landing module descent stage and the landing module ascent stage).
Note that the first two stages of the Saturn V were optimized for operating in the atmosphere of Earth, not for the vacuum of space. When you would build a rocket only for operating in vacuum, you could expect each stage to be as efficient as the 3rd of the Saturn V. Without the acceleration requirement during launch, you wouldn't need such large and heavy engines like those on the first stages of the Saturn V. You would reach a much higher Δv by having only a single small engine and dropping only the empty fuel tanks themself. But that still would not nearly be enough to get close to the speed of light.
Another interesting trivia fact: Incidentally, 18 km per second is quite close to the current speed of Voyager I. Note that Voyager I was launched with a rocket much smaller than a Saturn V (a Titan IIIE). How could it still reach such a high speed? By performing multiple gravity slingshots around the various planets it flew by. What conclusion can we take from that? When you want speed, don't just add more power. Use the power you have in a smarter way.
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$\begingroup$ Is the constant 9.81 in the equation stands for Earth's g? If so how can we apply it in this scenario? $\endgroup$– jorelMar 13, 2014 at 15:52
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3$\begingroup$ @jorel - g is just a scale factor used to convert specific impulse expressed as a velocity (effective exhaust velocity) to specific impulse expressed in seconds. In other words, Philipp could have used specific impulse as a velocity and that scale factor g would have disappeared. People who use English units tend to express Isp in seconds. People who use metric tend to express Isp in km/s. $\endgroup$ Mar 13, 2014 at 16:05
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$\begingroup$ Phillip & @David Hammen - Thanks for the explanations. $\endgroup$– jorelMar 13, 2014 at 16:11
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2$\begingroup$ A little more history. Isp was and still is traditionally expressed as pounds (force) of thrust over pounds (mass) per second of propellant flow rate. That comes out to just seconds when you cancel the "pounds". However a pound of force is the force due to gravity of a pound of mass at the surface of the Earth. Hence the g. $\endgroup$ Mar 13, 2014 at 16:33
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$\begingroup$ @MarkAdler I think I fixed it. I am comming from a locale where the meaning of commas and periods in numbers is the other way around, so I keep mixing up the two when writing english texts. When you find some other place where I mixed it up, please suggest an edit. $\endgroup$– PhilippMar 13, 2014 at 16:45