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I Why do rockets turn horizontally so soon after launch?

  1. Apr 15, 2018 at 1:30 PM #1

    DTM

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    My 12 year old asked me this question. I have a MS in Mechanical Engineering, so I can usually answer his physics questions, but this one stumped me. When lifting off, why do most rockets turn close to horizontal almost immediately? Of course we know they need mostly horizontal speed to achieve orbit, but they also need altitude. If the rocket gained all it's altitude first, vertically, and then made a 90deg turn to the horizontal and then sped up horizontally, the majority of it's high speed flight would be outside of the earths atmosphere, greatly decreasing air resistance and wasted energy. Wouldn't this be much more energy efficient?
     
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  3. Apr 15, 2018 at 1:45 PM #2

    russ_watters

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    No, because quite a bit of the energy would be wasted just in holding the rocket up against gravity. I read somewhere that at liftoff the shuttle astronauts experienced 3g's of acceleration, which means that fully 1/3 of the energy expenditure is just to hold it up.
     
  4. Apr 15, 2018 at 1:48 PM #3

    A.T.

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    Watch a successful launch, not just the crashes.
     
  5. Apr 15, 2018 at 1:58 PM #4
    Rocket do travel mostly vertically in the parts of the atmosphere that are thick to minimize atmospheric drag. They then turn gradually sideways (usually in some variant of a gravity turn to avoid a large angle of attack that would tear the rocket apart) to minimize the effects of gravity drag/loss. The earlier and faster you go sideways the less you have to fight against gravity. It's a bit un-intuitive but the same effect is why after reaching orbital velocity gravity is the only thing needed to maintain orbit.

    Your method is actually very inefficient, the entire energy used to achieve orbital height is wasted.
     
    Last edited: Apr 15, 2018 at 2:06 PM
  6. Apr 15, 2018 at 2:04 PM #5

    DTM

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    I'm not sure I understand that. 3g's of acceleration is their rate of change of speed. About 96 ft/s^2. The larger their acceleration the quicker they can get into orbit. Not sure how that equates to 1/3 of the energy expenditure is "just to hold it up"? I would suggest a rockets energy can be divided into 3 parts. 1. The energy needed to accelerate it up to orbital speed (~17,000 mph) (KE or Kinetic energy). 2. The potential energy it gains by getting to orbital altitude (about 150 miles for LEO). and 3. the energy it wastes by pushing the air out of the way on it's way up (Air resistance). I'm still not sure why they don't gain the potential energy first, which would be done at a slower speed and therefor much less air resistance which is proportional to speed^2 or even speed^3. Then turn horizontal to gain the KE. Yes this would have to be done at an angle of attached pointing away from the earth to counteract gravity, but it still seems like a more efficient trajectory. Although I'm sure I must be wrong, or the experts would do it this way. Just trying to understand why I'm wrong.
     
  7. Apr 15, 2018 at 2:09 PM #6

    DTM

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    [QUOTE="Your method is actually very inefficient, the entire energy used to achieve orbital height is wasted.[/QUOTE]

    Wasted? You have to achieve the potential energy of your orbital altitude (say 150miles). You have to gain that PE sometime, why is it wasted if you do it right off rather than gradually?
     
  8. Apr 15, 2018 at 2:16 PM #7

    russ_watters

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    Acceleration felt includes the 1g you feel just by standing on Earth's surface.
    At a force required for 1 g of acceleration, per f=ma, provided by the engines, the rocket would simply hover just above the launch pad. 3x more force provides provides 2g of speed increases in addition to that 1g of "just holding it up".

    Thinking about it more, my answer was only half an answer though. That explains why fast acceleration is better than slow acceleration, but not necessarily why pitching-over sooner is better than later. The other half is that the atmosphere gets thin really fast. Even with the fast pitch-over, the shuttle would reach its maximum aerodynamic pressure in less than a minute, at 35,000 feet, and would experience almost no aerodynamic pressure after 2 minutes:
    https://www.nasa.gov/pdf/466711main_AP_ST_ShuttleAscent.pdf

    Total time to orbit is about 9 minutes; the rest is spent just accelerating.

    As a simple estimate, you could calculate the angle that gets you to 150 miles altitude and 17,500 mph in 9 minutes assuming uniform acceleration and a linear trajectory pretty easily...
     
  9. Apr 15, 2018 at 2:17 PM #8
    In order to reduce the effects of atmospheric drag, rockets guide their thrust to get above the densest portion of the atmosphere near the Earth surface is quickly as possible, (say around 80000 feet). They direct the thrust near vertical. At the same time, they need to achieve orbital velocity (which is quite high) so if they usually direct their thrust closer to horizontal they can use the big thrust form the first stages, rather than relying on the smaller thrust from the later stages. All stages will be necessary, but the more velocity gained by the rocket in the first stage, the less velocity will need to be gained from the smaller engines in the second and subsequent stages.

    It turns out, the orbital velocity for the rocket at the Earth surface is not that much greater than the orbital velocity at a typical low earth orbit about 150 miles altitude. If you (both) want you can verify this by taking a square root. Orbital velocity is square root of GM/R (see wikipedia). In one calculation,use R is the radius of the Earth. Then compare it to a similar calculation with R + 150 miles substituted for R. (Square roots were not taught in school for 12 year olds in my day, these days with calculators, I would not be surprised if kids understand this.)

    The rocket climbs near vertical, not to be able to attain a smaller orbital velocity, but to get above the atmosphere to reduce drag and structural loads on the rocket. The rocket pitches over relatively quickly to use the big engines, and save the smaller ones till later.

    It might also be hard to direct the thrust vertical until a height, and all of a sudden turn near 90 degrees all at once,
     
  10. Apr 15, 2018 at 2:26 PM #9

    sophiecentaur

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    It’s worth pointing out that a circular orbit has equal amounts of Kinetic and Potential Energy so the vehicle has to be given both. Simply lifting it up by 150 miles and letting go will have it crashing into the ground.
     
  11. Apr 15, 2018 at 2:34 PM #10

    russ_watters

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    That's true only for one orbit, as potential energy increases but kinetic energy decreases with altitude.
     
  12. Apr 15, 2018 at 2:52 PM #11

    A.T.

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    The vertical thrust component has to offset gravity, and only the remainder adds energy to the rocket, while the entire horizontal thrust component adds energy to the rocket. This has to be balanced against the losses due to drag, and the optimum is some gradual transition from vertical horizontal.
     
  13. Apr 15, 2018 at 2:55 PM #12
    Concerning the last two posts. My Resnick / Halliday freshman/sophomore (college) physics text gives the gravitational potential energy at r to be - GMm/r
    For a circular orbit, you can verify that (with the equation for orbital velocity in a circular orbit that I mentioned earlier) KE = 0.5* m v squared = GMm /(2r).

    The kinetic energy for a circular orbit is (only) half as large as the potential energy. Moreover, they are not even the same sign. Potential energy is less than zero, and kinetic energy is greater than zero.

    At the end of my posting, I (jocularly) was about to suggest the original poster, compute the orbital velocity (perhaps with his 12-year-old), for the two different altitudes over a bowl of Cherrios, next morning. Getting all these concepts straight is going to take several bowls of Cherrios.
     
  14. Apr 15, 2018 at 3:02 PM #13

    russ_watters

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    A couple of things to note:
    1. That equation is potential energy at a radius relative to a point mass. A rocket launched from the surface of the Earth.

    2. I think your note about the sign convention adds unnecessary confusion: The rocket is doing positive work because it is applying force in the direction of motion.
     
  15. Apr 16, 2018 at 11:12 AM #14

    russ_watters

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    I can at least give quantitative answers to the first 2:

    Per kg, a common low earth orbit (150 miles/240km) requires:
    2.4 MJ of potential gravitational energy
    30.7 MJ of kinetic energy

    So the kinetic energy is greater by nearly a factor of 13.

    In terms of the angle, assuming a flat earth and linear trajectory, the angle with respect to the ground would be 6.6 degrees. That's pretty low, but since earth is curved, it would be even lower.

    So going up first is really going in the wrong direction, so aerodynamic drag would need to be pretty significant to make up for the otherwise bad trajectory.

    wikipedia has some good content here:
    https://en.m.wikipedia.org/wiki/Gravity_turn

    Basically, you pitch over a little and then let it fall the rest of the way.
     
  16. Apr 16, 2018 at 2:43 PM #15

    ZapperZ

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    I do not understand this, and I question the validity of the observation.

    What does it mean that the "... rockets turn close to horizontal almost immediately... "? I've watched many rocket launches. They do NOT "turn close to horizontal almost immediately". In fact, they only change orientation after several stages of launch!

    So please define "immediately" here.

    Zz.
     
  17. Apr 16, 2018 at 3:06 PM #16

    jbriggs444

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    The Pythagorean theorem is a hint that making a 90 degree turn is not optimal. If you add a vertical delta-v to a horizontal delta-v the magnitude of the total is less than the sum of the magnitudes of the components.

    The Oberth effect is a better clue. Maximum energy efficiency is obtained when thrusting parallel to the current velocity vector and doing so when the craft is at the highest available speed already. Waiting for the craft to come to a stop some hundreds of miles in the air before thrusting horizontally fails to make use of this.

    Edit: Barring atmospheric effects and subterranean trajectories, I believe that the most efficient launch would be an impulsive horizontal burn into a Hohmann transfer orbit.
     
  18. Apr 16, 2018 at 4:31 PM #17
    I agree with your figures. The kinetic energy requirement is 13 times the potential energy requirement.

    KE = 1/2 (1 kg) vorbital squared = 30.1 Million Joules (per kilogram) at r = ( 6370 km + 240 km ): location of orbit.

    PE at orbit location is twice this and negative = - 60.2 Million Joules at r = ( 6370 km + 240 km ): location of orbit, and
    PE surface of the Earth = - GMm / (R) = - 62.6 Million Joules at r = ( 6370 km ) : location of Earth Surface

    At launch you have - 62.5 million Joules and you need - 60.2 Million Joules. You have to supply the difference: ( - 60.2 - - 62.6) = 2.4 million Joules.
    It is like you have to supply energy to get less in the hole. With this interpretation, the total (potential + kinetic) energy for a bound (ellipse or circular) is always negative.

    You never quoted it but the comment by Sophiecentaur that the gravitational potential energy is equal to the kinetic energy for a circular orbit is incorrect. (As the above calculation supports, the potential energy is twice as great and negative from the kinetic energy). Sophicentaur's point that without horizontal velocity, the rocket would eventually fall back is true.

    The Wikipedia article you supplied has good detail.

    Any comments by anyone on my note that the early stages with greater thrust contribute horizontal velocity imposing less "velocity to be gained" demand on the later stages with lesser thrust to achieve orbit
     
  19. Apr 16, 2018 at 4:41 PM #18

    russ_watters

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    The observation is wrong, but not that wrong: the "immediately" part is correct. The attitude adjustments start as soon as practicable after clearing the tower. Googling, I'm seeing T+15s for the shuttle (which I think is actually 11s after launch).

    The "close to horizontal" is the most wrong; it isn't anywhere close to horizontal for a long time. It's a much more gradual arc.

    [edit] Looking at some videos, the pitch-over starts immediately after a roll to get the orbiter to the right orbit inclination. It is noticeable, but very slow.

    Edit2:
    Here's a launch profile:
    https://spaceflightnow.com/shuttle/sts124/fdf/124ascentdata.html

    You can see from it that it travels vertically until about the time it finishes its roll (+16sec), then it starts pitching and traveling downrange. But it is quite slow. But after a minute, it is 6.8 miles up and 3.3 miles downrange; a 30 degree angle from vertical.
     
  20. Apr 16, 2018 at 5:18 PM #19

    ZapperZ

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    Please note that in the first post of this thread, the OP claim that the rocket "... turn close to horizontal almost immediately... ". "30 degree angle from vertical" is not what I consider to be "close to horizontal". This is what I am disputing.

    Zz.
     
  21. Apr 16, 2018 at 5:25 PM #20

    russ_watters

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    Fair enough; taken as one statement it is pretty wrong. I was responding to what appeared to you to be taking it as two statements. In particular I thought this was confusingly put:
    I may not be understanding what you mean by "several stages of launch", but I think it should be clear the orientation starts changing almost immediately after launch.
     
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