I am very new to the rockets and this can be a very dumb question, just that I am not sure if my understanding here is right?

All of the rockets engines are at bottom which help it take off and after stage separation.

Why does it have to flip 180 degrees, this aligns the engines in direction of travel?

How does this help slowing down the rocket, it should only increase the speed of reentry when its fired?

From the videos the slowing down happens with the engines firing at bottom before landing. Does it do an additional 180 flip some where in between?

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    $\begingroup$ Why would firing engines with exhaust toward the direction of travel speed it up? $\endgroup$ Aug 23, 2016 at 12:13
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    $\begingroup$ Probably the OP is thinking of a rocket going straight up, then falling back. Pointing the engines up into space would then make things rather worse. An answer explaining orbits and the fact that the rocket is mostly going horizontal would probably be good, I'll write one when I have time if no-one beats me to it. $\endgroup$
    – Jack B
    Aug 23, 2016 at 12:55
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    $\begingroup$ I highly recommend playing around with Kerbal Space Program (game on Steam) to get a better understanding of orbital mechanics. $\endgroup$
    – Rozwel
    Aug 23, 2016 at 20:21
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    $\begingroup$ @Rozwel. Yeah, the only other alternative would be studying astrophysics. $\endgroup$
    – Zaibis
    Aug 24, 2016 at 5:44
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    $\begingroup$ @DewiMorgan well, at least somewhere between 90 & 180, anyway $\endgroup$ Aug 24, 2016 at 20:22

3 Answers 3


Hobbes has already showed you a diagram of the Falcon 9 launch profile, so I won't repeat that.

Note: This answer is not intended to be a complete, scientific treatment of the subject. I knowingly and deliberately simplify, gloss over and ignore things in several places, in order to explain this in a way that hopefully makes sense to the OP while still being reasonably correct.

There is a common misconception that space is simply "high up". It is not. You can get to orbital altitude quite easily, and if you do nothing else, once you turn off the engine, you will simply fall back down to the ground. This is because, to take one example, the gravitational attraction of Earth at the altitude of the International Space Station is approximately 90% of that on the surface.

Rather, space (or more accurately in this case, orbit) is when you are going forward very fast. In fact, you have to go forward so fast that your forward momentum matches the downward momentum imparted by gravity. In order to establish a stable, low (a few hundred kilometers up) orbit around the Earth, you need to move sideways at somewhere in the range 7,000 to 8,000 meters per second. This is called orbital velocity, and the value of it becomes lower as you move farther away from the central body (because the gravitational influence of the central body falls off with the square of the distance). If you are having trouble visualizing how fast this is, consider that the ISS, which is in low Earth orbit (about 400 km altitude) circles the Earth once every 90 minutes. The Earth's orbit around the Sun works in the same way, only that the Earth is moving at about 30,000 m/s to make a whole revolution around the Sun in about 365 days.

So a rocket spends most of its energy to gain that sideways velocity. It obviously spends some energy getting to the correct altitude, because altitude and velocity are intricately linked in the field of study we call orbital mechanics, but the vast majority is spent gaining forward speed, not altitude.

An orbit has a value known as its eccentricity. The eccentricity of an orbit is, very basically, a value indicating how circular it is. A perfectly circular orbit has an eccentricity = 0, and an elliptic orbit has an eccentricity in the range $0 < e < 1$. It is also valid to have $e \ge 1$, but we don't need to consider those cases here (those are escape trajectories because they are leaving the gravitational field of the central body). Perfectly circular orbits are a largely theoretical construct which mostly doesn't exist in real life; all practical orbits have at least a small, non-zero degree of eccentricity.

For orbits with very low eccentricity, and where the mass of the object in orbit is far smaller than the mass of the body it is orbiting (such as a spacecraft in circular orbit around the Earth), we can estimate the orbital period $T$ based on $$ v_o \approx \frac{2 \pi a}{T} \Rightarrow v_o T \approx 2 \pi a \Rightarrow T \approx \frac{2 \pi a}{v_o} $$ where $v_o$ is the orbital velocity and $a$ is what is known as the semi-major axis (half of the linear distance between the two points farthest apart in the orbit, which for circular orbits becomes the orbit radius). As you can see, there is a close relationship between $T$, $a$ and $v_o$.

On Earth, if you want to slow down, you can simply stop applying forward thrust. Friction will then eventually (generally quite soon) cause your vehicle to lose forward velocity, and it will eventually come to a stop in one manner or another. This is independent on the construction of your vehicle; it could be a car, a bicycle, an airplane, a helicopter or a race horse, and that doesn't matter.

In space, there is no (or rather, for all intents and purposes and over short periods of time, negligible) friction. Hence, if a spacecraft turns off its engine, the spacecraft will simply continue in the orbit caused by its current velocity and current position within the local gravitational field. Unlike an airplane, the spacecraft can be pointing in literally any direction, and orbital mechanics doesn't care; unlike an aircraft, a spacecraft is not operating within any fluid medium. To turn around, change its direction of travel, or stop, in space, a spacecraft needs to apply additional thrust in some direction other than the direction of travel. (There are other ways, such as gravitational slingshots, but those are basically a special case of this, and those don't apply to Earth-orbit launches anyway.)

Such thrust is commonly, but far from always, applied retrograde. Retrograde is one of those funny words in rocketry and orbital mechanics; here, it means "opposite to the direction of travel". In a car, it would be called "in reverse and with wheels pointing straight ahead".

Applying thrust partially or fully retrograde has the effect of slowing down the spacecraft's forward speed. Because this means that it can no longer keep up its current orbit, it will start falling in the direction of the prevalent gravitational field, which in the case of a low Earth orbit and thrusting for reentry eventually causes the spacecraft to reenter the Earth's atmosphere as it falls toward the Earth. With less thrust, or by thrusting in a direction other than fully retrograde, it will cause the spacecraft to settle into a different orbit, which may or may not intersect Earth ground or atmosphere. The change of velocity resulting from this thrust is what we call delta-v, from the Greek letter $\Delta$ which in mathematics is used to indicate difference, and $v$ which by convention is the symbol associated with velocity. The change of velocity needed to settle into a different orbit, even an Earth-intersecting one, is commonly far lower than what was required to establish the original orbit; for example, the Space Shuttle was commonly in an orbit similar to that of the ISS, but only needed approximately 90 m/s of delta-v under power to commit to landing. Once committed to landing, the space shuttle did not have enough engine power to get back into a stable orbit around the Earth.

As the engines are fixed in place at one end of the spacecraft and the spacecraft is generally pointing prograde (in the direction of travel) after the engines have been shut down, the spacecraft has to turn around before it can apply any significant thrust retrograde. Thus, it has to flip over. Because there is so extremely little friction in space, it can do so leisurely, without really affecting its direction of travel much at all; in an atmosphere, the vehicle would probably start tumbling and possibly break up because of aerodynamic forces, and aircraft need special considerations to be able to perform anything resembling similar maneuvers.

The rest is merely an elaborate dance to get the spacecraft to settle down slowly and safely in an appropriate spot on the Earth's surface. Phew. Putting it that way makes it sound downright simple.

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    $\begingroup$ As Michael said (obligatory XKCD) : what-if.xkcd.com/58 $\endgroup$
    – Steve Ives
    Aug 23, 2016 at 14:34
  • $\begingroup$ @SteveIves Indeed, I very specifically had that one in mind when I wrote especially the first part of this answer and even considered incorporating some of the images in it, but decided that the answer was already plenty long enough even without those. $\endgroup$
    – user
    Aug 23, 2016 at 15:18
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    $\begingroup$ Or, to paraphrase Douglas Adams, "[orbit] is throwing yourself at the [Earth] and missing." $\endgroup$
    – JS.
    Aug 23, 2016 at 23:41
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    $\begingroup$ @JS. True, but you have to miss it just right, or your orbit has $e \ge 1$ and you have a bad day. $\endgroup$
    – user
    Aug 24, 2016 at 6:39
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    $\begingroup$ @JS. I was thinking more along the lines of you will stay in space today and every day. Strictly speaking, $e\gt 1$ implies a hyperbolic trajectory, and it doesn't take $e$ much greater than 1 to be on a hyperbolic escape trajectory. Wikipedia has some nice examples, citing comet C/2006 P1 on a hyperbolic closed Sun-centered orbit at $e=1.000019$ with an orbital period of $10^5$ years (speculation: because of the planets), and C/1980 E1 on an escape trajectory with $e=1.057$. $\endgroup$
    – user
    Aug 24, 2016 at 16:34

Trajectory of the Falcon 9 first stage:
F9 trajectory

Graphic courtesy ZLSA Design (zlsa.github.io)

As you can see, before the boostback burn, the stage flips so the engines point in the direction of travel. When the engines fire, this slows down the stage.

This trajectory is used when the stage returns to the launch site (and for some early experiments where the ASDS landing ship was placed close to the launch site). For landings downrange to the ASDS, the boostback burn is skipped. The booster still has to flip over, so the engines are pointed in the direction of flight for the reentry burn.

  • $\begingroup$ That diagram doesn't look right. A boostback burn is used when returning to (near) the launch site to land on "Landing Zone 1" (LZ1). In the case of landing on the drone ship, there is no boostback burn, so the trajectory looks more like an arc. $\endgroup$ Aug 23, 2016 at 22:44
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    $\begingroup$ @CraigMcQueen: Incorrect, boostback burns are also used on some ASDS landings (the ones where the barge is more like 200km out to sea, instead of 450km or so). For GTO launches, they don't have enough fuel for a boostback burn, which gives the arc trajectory you mention, but also means an extremely high-speed re-entry. For LEO launches, even if not returning to LZ1 or similar, a pre-re-entry boostback burn means the re-entry velocity is a lot lower, which is safer for the rocket. $\endgroup$
    – CBHacking
    Aug 24, 2016 at 0:30
  • $\begingroup$ @CBHacking: Do you have a reference for a drone ship landing that includes a boostback burn? Is it possible that SpaceX did it in the past before they had permission for LZ1 landing, but they would not plan to do so in the future now that LZ1 is permitted? $\endgroup$ Aug 24, 2016 at 0:49
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    $\begingroup$ The diagram was correct for the first ASDS experiments where the ASDS was close to shore. IDK if this flight profile is still used. I've amended my answer. $\endgroup$
    – Hobbes
    Aug 24, 2016 at 8:12
  • $\begingroup$ Cool diagram. +1. But really, nice answer. $\endgroup$
    – J Sargent
    Aug 25, 2016 at 2:43

A rocket launches, initially straight up to get out of the dense part of the atmosphere, but very soon turns to accelerate horizontally.

Orbiting is all about going fast enough around the planet (or gravity source) that the balance of centripetal force and gravity are the same.

Orbital speed is very high, on the order of 18,000 MPH. So the payload, in order to orbit needs to attain that velocity.

When the first stage separates, it is about 60 kilometers above the ground, going about 3000 KPH. (Different missions profiles differ a bit with SpaceX since a GTO mission requires more performance from the first stage than a light LEO mission).

In order to land, the stage had to reenter, which is down, but also slow down, and lose that 3000 KPH velocity.

Being empty, (most of the fuel and oxidizer is used) the mass of the stage drops from the 1.5 million lb range down to the 150-200 KLb range. Thus the engine thrust is more effective at tranferring momentum, working on a smaller mass. Thus they only use three engines to slow it down.

When there is sufficient margin, (payload was light enough, or orbit was easy enough) they not only cancel their forward velocity (Thus the 180 degree flip) but also to fly back the distance they covered already, to land back on land in Florida (Or potentially Vandenberg, or even Boca Chica in the future).

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    $\begingroup$ Pounds and miles in space travel? Hasn't it cost enough?! $\endgroup$
    – yo'
    Aug 24, 2016 at 8:48
  • $\begingroup$ Actually, the trajectory of most launches follows a path known as a Gravity Turn: en.wikipedia.org/wiki/Gravity_turn $\endgroup$ Aug 24, 2016 at 21:06

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