This may seem a naive question, but as I understand it one of the reasons for the enormous energy expended by rockets is the extremely high speed (around 28,000kph) you need to achieve a stable orbit and avoid falling back to Earth.

The Saturn V mission profile was to reach Earth orbit, and then head toward the Moon.

Why not just go straight there and miss out the orbit? I realise that the earth's rotation would mean you wouldn't want to be pointing straight at the moon at lift off, but still it seems like a more efficient way. Just point it in the direction you want to go!

(Since this is my first question here I should add that I'm not anticipating having outsmarted everyone at NASA.)

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    $\begingroup$ related but different Do any launches bypass LEO? $\endgroup$
    – uhoh
    Commented Jul 25, 2019 at 14:37
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    $\begingroup$ You still need to escape earth orbit to go to the moon, so you have to go that fast regardless. "Stopping" in LEO doesn't really cost anything significant in terms of energy budget and it provides the opportunity to take a breath, check things over, make sure everything is working correctly, get ready for the TLI burn, etc. $\endgroup$
    – J...
    Commented Jul 26, 2019 at 13:22
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    $\begingroup$ Related : Orbiting Earth before heading to Moon $\endgroup$
    – J...
    Commented Jul 26, 2019 at 13:28
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    $\begingroup$ Because Stanley Kubrick thought it would be more dramatic! $\endgroup$
    – RonJohn
    Commented Jul 26, 2019 at 13:34

7 Answers 7


I think you may have a misunderstanding that isn't addressed by any of the other answers.

It is true that most of a rocket's work in entering orbit is building up enough speed to reach orbital velocity. But you have to build up even more speed to make it to the moon. In fact, while they were on their way to the moon they were still in orbit around the earth, just a very long skinny orbit with one end near the earth and the other end out where the moon actually orbits. If the moon had not been there, they would have kept looping around the earth, speeding up as they got close and slowing down as they got far away.

So no work was wasted. They first sped up enough to maintain a circular orbit around the earth, and spent enough time to make sure everything was OK. Then they added enough additional speed to make it to the moon. Doing it all in one shot would not have saved any fuel.

I recommend reading the other answers because they explain the positive benefits of having an initial parking orbit - this answer just explains why it's not inefficient. (Edited later to add this.)

  • $\begingroup$ There are some excellent answers here, but this one gets to the heart of the matter most succinctly. I learnt a lot from particularly the @Uwe and Steve Linton's answers $\endgroup$
    – Party Ark
    Commented Aug 3, 2019 at 11:53
  • $\begingroup$ This doesn't make sense to me. Some ineficiency must be present. You get the apoapsis for free, but you have to spend some extra delta-v to increase your periapsis in order to orbit the earth. This is delta-v you do not have to spend when going to the moon directly. $\endgroup$
    – Fax
    Commented Apr 16, 2022 at 2:20
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    $\begingroup$ There was no circularizing burn. TLI was done after one orbit, so at or near periapsis, resulting in a cigar-shaped orbit with apoapsis at lunar distance and periapsis close enough to reenter (with appropriate course correction), which was just what they wanted. See this question: space.stackexchange.com/questions/4797/… $\endgroup$ Commented Apr 17, 2022 at 11:13

Going directly to the Moon would require a very small launch window.

The Earth orbit before enabled a launch window of about 3 to 4 hours, see this question. Abort from an Earth orbit was possible when the second ignition of the third stage of the Saturn V failed using the Service Module engine to initiate a reentry.

Time in orbit was used to complete the long TLI checklist. If a severe malfunction would have been found, abort of the TLI (trans lunar injection maneuver) was possible.

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    $\begingroup$ The launch window and the failure modes are indeed the key. There is no energy difference between direct ascent and ascent via a parking orbit. $\endgroup$ Commented Jul 25, 2019 at 15:26
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    $\begingroup$ @MartinKochanski - Not quite true. A direct ascent is more energy efficient (but not a whole lot more efficient) than is using a parking orbit. $\endgroup$ Commented Jul 25, 2019 at 19:42
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    $\begingroup$ Does orbiting also help you slow down, and prevent crashing into the moon? $\endgroup$ Commented Jul 26, 2019 at 0:23
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    $\begingroup$ @axsvl77 Not on the launch side; the injection velocity is still the same. Using the capture at the destination site was used, of course, but even then, you could go straight from capture to landing, without estabilishing orbit first - it's just a lot of extra risks and difficulties for very little benefit (or no benefit, if you don't forget the command module had to stay in orbit anyway :)). $\endgroup$
    – Luaan
    Commented Jul 26, 2019 at 7:21
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    $\begingroup$ @DavidHammen I think you're overstating - the efficiency difference is so small as to be effectively negligible. $\endgroup$
    – J...
    Commented Jul 26, 2019 at 13:30

There is very little to gain by going straight to the Moon, and as @Uwe has said, it makes the timing of the launch extremely demanding. Let me have my go at explaining why there is very little to gain.

The most fuel efficient way for a rocket to get from Earth to the Moon is basically to accelerate as close to the Earth as possible until it is moving at the necessary speed (which is about 40% faster than orbital velocity) away from Earth and towards the Moon. It then coasts, being slowed gradually by Earth's gravity until it just reaches the point where the Moon's gravity is strong enough to pull it the rest of the way. By missing the Moon slightly, and using rockets to slow itself a little, it can enter an orbit around the Moon. This is due to something called the Oberth effect which is discussed often on this site.

Now, it might seem obvious to pick a moment when the Moon is more or less straight overhead and accelerate upwards, but, as @Machavity says, this ignore the useful bit of extra velocity you can get from Earth's rotation, which means it's actually better to do most of the acceleration horizontally West to East (also good to start as close to the equator as you can). Once you're doing that, there will come a moment when you are moving at orbital velocity, so if you just turn your rocket off at that point you will be in Low Earth Orbit. Then you wait until your orbital velocity is pointing more or less at the Moon (actually at where it's going to be in a couple of days), turn the engine back on and finish your acceleration.

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    $\begingroup$ You could gain a bit of extra delta-V by accelerating while still deep in Earth's gravity well, but that would really only work if there was no atmosphere. I'm not going to do the math, but I expect the savings would be offset by air resistance until you're really close to a decent parking orbit anyway. Doing the same on the return trip might be worthwhile, but that would also mean you'd have to land the whole (return) craft on the Moon, which will probably be worse fuel economy anyway. $\endgroup$
    – Luaan
    Commented Jul 26, 2019 at 7:24
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    $\begingroup$ @Luaan crucial here is that low orbits are essentially “ground-level orbits”, approximated as well as the atmosphere allows. Even with Earth's dense atmosphere, this is a pretty decent approximation, on the Moon even better. On Titan, this isn't really possible. $\endgroup$ Commented Jul 26, 2019 at 10:27
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    $\begingroup$ @Luaan Low Earth Orbit is deep in Earth's gravity well. $\endgroup$ Commented Jul 26, 2019 at 10:35

This question seems to hinge on a fundamental misunderstanding about space, that is, to be fair, extremely common among the general public. It's the idea that space has no gravity, so things in space are weightless.

"But wait!" you say. "I've seen videos of astronauts in space, and they sure seem weightless to me." And you'd be right, they do seem weightless... but they're not. They're in a state known as "freefall".

Now, "freefall" as a technical term means that the object doesn't have any forces other than gravity acting on it. So, if you shot a gun straight up, then it would be in freefall from the moment it left the barrel to the moment it hit the ground (ignoring air resistance), even though it was only "falling" in the colloquial sense for half of that time. Things in freefall appear weightless to other things that are also in freefall, so that's how astronauts can float around a spaceship. In fact, it's been proven that you can't tell the difference between actual weightlessness and freefall in a uniform* gravitational field.

So, if you try a direct ascent from the Earth, you'll be fighting gravity the entire time. If you tilt over and thrust horizontally, however, you can build up your speed without having to fight gravity, which makes intermediate parking orbits much more efficient than direct ascents**, even before you take into account things like the Oberth Effect.

You mentioned that they would have had to achieve about 28,000 km/h to enter a stable orbit, implying that a direct ascent would be slower. But that's not true. Yes they had to go that fast to orbit, but then they had to accelerate even more (by about another 6,000 km/h) to go fast enough to avoid falling back to Earth, and even then, they were only going in the neighborhood of 3,000 - 5,000 kph when they reached the point where the Moon's gravity became greater than the Earth's and they started accelerating again. That's the same speed that a direct ascent would have to achieve, albeit without the aforementioned efficiency boost.

Please note that I'm not a physicist, so I'm probably not explaining this very well. I'm using "fight gravity" in a colloquial sense. Basically, gravity is always pulling you down, so any fuel you use to go up is opposed by gravity. As an extreme case, picture a rocket weighing a total of 100kg, with an engine that produces 981N of thrust. If pointed straight up, the thrust will be perfectly balanced by its weight (ignoring mass reduction due to fuel burn), so it will waste all its fuel hovering in one place going absolutely nowhere. Turn it on its side, however, and suddenly it's accelerating faster than a Ferrari, going from 0 to 100km/h (60mph) in 2.8 seconds.

The same thing happens in space. Any vertical thrust (e.g. radial out (away from the planet) or radial in (toward the planet)) will have to overcome both the spacecraft's inertia and gravity, while thrusting horizontally (prograde (forward), retrograde (backward), normal (left), or antinormal (right)) only has to contend with the spacecraft's inertia, and so is more efficient. You can see this during a launch: rockets have to thrust radially for a short time to get them above the thickest part of the atmosphere, but then they pitch over to a horizontal attitude as soon as they reasonably can in order to avoid burning more fuel then they have to. Low Earth orbits are on the order of about 7.5km/s, but spacecraft launched from the surface typically have a delta-v capability of 8 to 8.5 km/s, that extra delta-v being lost to gravity during the short vertical ascent phase.

During the planning stages of the Apollo program, direct ascent was considered as one possible launch strategy. It's benefit was that it would be a much simpler plan than worrying about an orbit, and they were in a major time crunch. One of the reasons it was shelved is that they didn't have a facility big enough to build the enormous rocket that such a plan would require.

*Note that Earth's gravitational field isn't precisely uniform. Thus, astronauts do feel a very tiny bit of gravity that changes in strength and direction depending on where they are in the ship. That's why official NASA literature refers to "microgravity" rather than "zero gravity".

** Okay, technically you don't have to actually orbit to get the efficiency boost. It's the horizontal thrust that's important here. So you can thrust horizontally from suborbital speed all the way up to escape velocity, without ever achieving an official "orbit". But stopping partway to check that everything's working and that you're properly aligned for the next engine burn is just good engineering, as @jamesqf mentioned.

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    $\begingroup$ What does "fight gravity" mean? If you're at some location above the surface of the Earth, the force of gravity acting on you at that point will be the same regardless of how fast or which direction you are moving. Every other answer here says that a direct ascent would be anywhere between very slightly more efficient to no more efficient than "stopping" in a circular orbit first. Yours is the only answer that says stopping first is "much more efficient." What do you know that the others don't? $\endgroup$ Commented Jul 26, 2019 at 17:34
  • $\begingroup$ @SolomonSlow My response is too long for a comment, so I edited it into my answer. $\endgroup$ Commented Jul 26, 2019 at 20:32
  • $\begingroup$ Imagine a hypothetical rocket that, at any given instant, makes thrust exactly equal to the rocket's own weight. You seem to suggest that this rocket could leave the ground if you "turn it sideways." How is that? If it's pointed straight up, as you said, it will balance on its tail and go nowhere. But if you point it in any other direction, then vertical component of the thrust will be less than the weight of the rocket. Turn it parallel to the ground, and its horizontal acceleration may be faster than a Ferrari, but it also will accelerate vertically, at $9.8 ms^2$ toward the ground. $\endgroup$ Commented Jul 26, 2019 at 22:58
  • $\begingroup$ I think when you say "horizontal," you are imagining a ship that is already in orbit. Once in orbit, it doesn't take any energy to stay there. I think that's your "doesn't fight gravity." But I think you are ignoring the enormous energy that was spent to achieve orbit. Let's say the goal is to reach altitude Y. A ship orbiting at altitude X where X<Y must spend some extra energy to raise its orbit to Y. But let's say we have a rocket that travels straight up to X, and we want to modify it to reach Y. Does that take any more extra energy? IDK, and I'm too lazy to do the math right now. $\endgroup$ Commented Jul 26, 2019 at 23:15
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    $\begingroup$ @SolomonSlow When I said "horizontally", I meant just that: parallel to the ground. A rocket that's horizontal, lying on the ground, but accelerating will eventually leave the ground, when the curve of its freefall trajectory becomes flatter than the curvature of the Earth. Or you can think of it as centrifugal force, if you like, it gets you to the same place. I'll reiterate what I said in my answer: the only reason that direct ascent was considered for the Apollo missions was that it's simpler for people who'd never gone to the moon and were under a major time crunch imposed by (cont) $\endgroup$ Commented Jul 27, 2019 at 1:49

In addition to the physics reasons, there's a practical engineering reason: you can use the time in Earth orbit to check your spacecraft systems to see if they've been damaged by the stresses of launch. If they have, you can quickly abort and return to Earth. Likewise with the lunar orbit: you can check the LM before committing to a descent burn, and re-dock with the CM if there's a problem.

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    $\begingroup$ This "time in orbit" option was quite useful for Apollo 12, which got struck by lightning on the way up. $\endgroup$
    – Mark
    Commented Jul 26, 2019 at 22:14

The INS, the internal Navigation system, of the time were sensitive to vibration and the Saturn five first stage was likely to roll all the but the navys ICBM gyros in the first stage. The first two stages not much more than a collection of well tested 1959-1962 era ICBM parts.

It is well documented that TLA burn was figured and cross checked by both ground base IBM mainframes and the Apollo Command Module (CM) legendary computer. There was a calibration step of the acclerometers, alignment of the CMs reference plane that having high athmith angles between ground stations and RCS system that should be done in earths orbit because that would be the automated self contained way home in case the radio comms went away. There are 80 other reasons for a earth orbit check out including crew adjustment to weightlessness.

Could the make it to the moons orbit by with a toggle switch, a wrist watch, some grations etched on the window, and manual controls of the service modules gimbals and RCS? Even the orbital calculator, Buzz says NO...finding the CQ offset was just impossible. NASA packed 2 extra 300 lbs of computers and gyros and 1 way oversized LM rocket engines to avoid doing that.

If either alignment sextant was hazed over by some odd outgasing, there would be no leaving earths orbit, no way to allign the platform for the lunar orbit burns where tracking sites were all pointing in the same 3 degrees. If the CM leaked at the rate of a basketball that goes flat in 3 days, no point in leaving. If the LEM computer or comms were broke, little need to leave orbit. If the real time clocks could not be adjusted to agreement no lunar docking after a landing. If any of the fuel cells were broken the mission over.

In earths orbit, just about any radio called retrograde burn with astronauts seeing the earths horizon will get your crew on an aircraft carriers deck in 3-12 hours.


Some key points to consider:

  • The longer you are applying thrust against gravity, the more fuel you require.
  • An orbit is an orbit; low-Earth orbits, the Moon's path around Earth (or more correctly, the Earth-Moon barycenter), and even Earth-Moon or Moon-Earth transfers are all orbits, and arguably all Earth orbits. Earth-Moon transfer is a special case because the most expedient Earth-Moon transfer is an Earth escape, but still a solar orbit.
  • The higher your orbital altitude, the more orbital energy you have, even though both angular and tangential velocity are lower. Thus the Moon, despite its Month-long period (travels around Earth at about 2,300 mph) has more orbital energy per unit mass than a satellite zooming around Earth at 17,000 mph.

What this means is that in order to reach the Moon, a rocket must acquire more kinetic energy than it takes to achieve a low-Earth orbit. It also means that getting up to orbital velocity as quickly as you can minimizes the fuel needed to do so.

The flight profile to go from Earth to Moon is thus an optimization, trading off a number of factors.

  • Get above the atmosphere (or a good portion of it) before building up too much speed to avoid aerodynamic drag and heating. Go too fast in thicker air and you waste energy on drag, and risk overheating parts of the spacecraft.
  • Get up to orbital speed as quickly as possible, given limitations on structures and human tolerance for g-forces. More g's means you have to build stronger, which adds weight, and your human occupants will suffer.
  • Once you're in at least a low Earth orbit, you can take more time, and you don't need to pile on the g's to go further; a smaller thruster will do the job, it will just take longer.
  • To get to the Moon from low-Earth orbit means that at a minimum you need to accelerate until you achieve an elliptical orbit having an apogee pretty much at the Moon's orbital distance from Earth. It's not exactly that (I don't know the math/physics to work it out precisely) but the velocity required at LEO altitude will be something well in excess of a circular LEO velocity; in other words, you need to add several thousand mph to your LEO velocity.

To summarize:

Attaining LEO velocity is not a wasted effort; it is:

  • A necessary step on the way (you still have to go faster)
  • An optimization (minimizes the fuel used)
  • An opportunity to check that everything is OK before moving on to the next step.

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