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I was watching space shuttle launch recently, and at 2:53 you can see the space shuttle from an angle, even though it's high above the earth and you should see only the bottom.

Does the space shuttle fly straight up at its launch or does it have a curve in its path?

Launching space shuttle path

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A low earth orbit is not just "high up". It's "high up and moving really fast". –  Tim S. Mar 24 '14 at 13:49
Relevant what-if.xkcd, particularly this image. –  Emilio Pisanty Mar 24 '14 at 13:51
Strictly speaking, it no longer files anywhere. But the question and answers are applicable to any launch into Earth orbit. –  Keith Thompson Mar 24 '14 at 22:42

6 Answers 6

up vote 25 down vote accepted

Everything (not only space shuttles) that goes into the Earth orbit must curve its path on the way up. If a vehicle went straight up and did not achieve escape velocity, it would fall back to Earth after the fuel runs out.

The main objective of the rocket engine is not only to get the cargo above the atmosphere, but more importantly to accelerate it in horizontal direction to the orbital speed (7.5 km/s for the orbital altitudes of the Shuttle and International Space Station). That is why all rockets / shuttles curve their path gradually to horizontal direction and then burn a lot of fuel in horizontal direction. If the cargo did not reach enough horizontal speed, it would fall back to Earth.

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In the interest of pedantry, I will point out that you should have written "If a vehicle went straight up and did not achieve escape velocity, it would fall back to Earth after the fuel runs out. –  Malvolio Mar 24 '14 at 18:14
@Malvolio Your comment is very correct and since there is already 13 upvotes with it, I'm going to edit my answer accordingly :-) –  mpv Mar 26 '14 at 18:21

Any launch to orbit follows a common path. First they have to get above the atmosphere, to keep slowing from it down to a minimum. Then they need to re-orient to accelerate to maintain orbit. If you simply launched straight up, you'd fall straight down, it's the speed that makes the difference.

There's a pretty good wikipedia article about how this is managed called Gravity Turn. Basically, the rocket only goes straight up for a short period of time, then turns slightly off to allow Gravity to actually work for the rocket.

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It only goes straight up for a few seconds as it's clearing the pad. It turns to an angled path almost immediately after clearing the launch platform and begins traveling more horizontally than upwards very soon. If you listen to the audio from NASA TV during a shuttle launch, you'll hear them call out altitude, speed, and distance downrange at semi-regular intervals. By about 50 seconds, it is twice as far horizontally downrange as it is above the surface in altitude. You can listen to the audio in this shuttle launch from NASA TV. As others have said, this is because anything in low Earth orbit must accelerate to around 18,000 mph horizontally in order to stay in orbit. Basically, the speed tangent to the surface of the Earth must be such that the acceleration towards the Earth from gravity causes the object to fall around the Earth in a closed loop rather than falling into the atmosphere.

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at 2:53 you can see the space shuttle from an angle, even though it's high above the earth and you should see only the bottom.

Shuttle time-lapse image showing the curved path it takes after takeoff: National Geographic image of Shuttle launch

Actually, just after liftoff the Shuttle does a roll maneuver so it will fly with the top of the orbiter facing down. This is done to minimize the loads on the Shuttle components. Shuttle on its back

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I had some old optimal control software lying around. It's a really good match for this question, so I decided I would use that to attack the theoretical underpinnings of this question.

The central question I see is, does the shuttle want to roll to some non-vertical angle? Of course, by "the shuttle" I really mean the engineers who design it. And by the engineers, I really mean the underpinning mathematical reality. In short, will you burn less fuel if your trajectory is tilted somewhat? We would think the answer is "yes", but it's not easy to justify that. It's also not clear what the motivation is. If there were no atmosphere, we would prefer to launch horizontally, actually.

One way to answer this is to use numerical methods to brute-force the problem to a solution. Develop a figure-of-merit "score" function, and then vary the trajectory so that you get the best value. Numerically, you can probably accomplish this by guessing some initial trajectory and then using the calculated Jacobian and Hessian to approximately locate the nearest critical point, where the figure-of-merit is (hopefully) at a local minimum.

So to simulate, here are some of the things I take into account:

  • The thrust from the engines comes out to 3gs, and this is constant
  • Earth's gravity pulls it down
  • Atmospheric drag follows the drag equation, and I used Falcon9 mass and diameter parameters
  • The state vector gives the angles that the thrust is applied at, relative to vertical
  • A simple rk4 method is used to integrate this model given a certain control

At the end of the simulation, a figure of merit function tallies its score. These are chosen arbitrarily in order to get the outcome we want. In particular, I:

  • Penalize high eccentricities
  • Penalize a low semi-major axis
  • Penalize long burn times

The idea is that an optimizer program will find you the best combination of all these three for the lowest price. This is what I obtained in terms of the angle at which the thrust is applied. This is relative to the vertical, in the sense that 0 degrees would be applying thrust straight up. Note that thrust and velocity aren't always in the same direction. This represents the "best" way to angle the rocket to get the best performance.

Thrust angles

This type of method does often have problems with it. A lot of the "harry" behavior there is likely due to model errors as problems with the derivative calculations and other types of artifacts. The simulation runs over 500 intervals... and there are 17 independent variables for which the second derivative determines the system. Since it's near the critical point by definition, the figure-of-merit is highly insensitive to these variables and that can cause problems.

Nevertheless, it shows what I wanted it to show - the optimal trajectory clearly follows a non-vertical path near the surface. FYI, the initial angle here is about 27 degrees relative to the vertical. I would expect that real-world scenarios would be using numbers that are somewhat close to this.

Mathematically, that is a fairly rigorous justification for the angle at which rockets fly once they're clear of the pad. It would, indeed, be more optimal to tilt the launchpad itself. Logistically, that sounds like a really bad idea. So the engineers compromise by turning to the optimal angle after the launch pad has been cleared.

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"The thrust from the engines comes out to 3gs and this is constant" is badly wrong. As fuel is consumed, the mass of the vehicle decreases and the acceleration increases. Then the solids burn out and are jettisoned. The acceleration decreases a lot at that point. –  Ross Millikan Aug 24 '14 at 4:21

The Earth moves, and if the shuttle is travelling in a straight path it wouldn't make sense, because the Earth is moving, and we would perceive for it to be moving in a curved way. Not sure, but I think this is the coriolis effect.

Found a cool video on what happens if we dig a hole to the middle of the Earth. Somewhere in this video, it suggests that people wouldn't be able to dig a massive hole into the Earth and jump into it because of the coriolis effect.

Cheers, hope this helps.

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Could you please also link to the video that you mention? I find that unlikely, because by the time the Earth would rotate enough to throw you against the walls, you'd reach terminal velocity and the air you'd displace would form a cushion of increased air pressure around you and between you and the walls. You couldn't reach the center of the Earth because of the air pressure, its sheer pressure would liquefy it well before the depth of some 6,370 km. If you could somehow insulate it from the heat of the Earth's core, of course. If not, then it would be like jumping into a pressure cooker. –  TildalWave Mar 25 '14 at 9:58
Here ya go, sorry for the late reply youtube.com/watch?v=6TZVCxCiMHE –  Neil D'souza Apr 10 '14 at 12:46
OK, thanks, as I thought, that source completely neglects the fact that the Earth has an atmosphere, and doesn't even consider fluid dynamics in all of it. It would be more correct for an atmosphere-less body, say a large asteroid, but I'm afraid that it's oversimplifying for that as well. For example, it doesn't even mention the Coriolis effect. Or gravity gradient and a plethora of other forces. –  TildalWave Apr 10 '14 at 13:01

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