Why do rockets turn horizontally so soon after launch?

[jbriggs444]

Why does the order of applying velocities affect the final velocity? Is this a rocket / reaction engine thing to do with efficiency?

The comparison is between applying a delta v parallel to velocity versus applying a delta v perpendicular to velocity. Nothing specific to rockets. Nothing to do with sequencing. Purely a geometrical effect.

If I am coasting at 3 meters per second eastward and apply a 4 meters per second delta v northward, I get a new speed of 5 meters per second.
If I am coasting at 3 meters per second eastward and apply a 4 meters per second delta v eastward, I get a new speed of 7 meters per second.
If I am coasting at 3 meters per second eastward and apply a 4 meters per second delta v westward, I get a new speed of 1 meter per second.

Direction matters.

 

[sophiecentaur]

Of course the azimuth setting should be Eastwards. My issue is that you already have your Original horizontal velocity at take off. That is the same when you are in orbit and the engines provide the rest. If the Earth were a bit less massive and not spinning at all, would the answer to the OP be the same? Would they still go horizontal 'soon after lunch'? (Spelling in purpose)

 

[jbriggs444]

Of course the azimuth setting should be Eastwards. My issue is that you already have your Original horizontal velocity at take off. That is the same when you are in orbit and the engines provide the rest. If the Earth were a bit less massive and not spinning at all, would the answer to the OP be the same? Would they still go horizontal 'soon after lunch'? (Spelling in purpose)

Velocity is a vector. You use vector addition rules. The direction that you apply the delta v matters. This is not a complicated idea.

 

[CWatters]

Nothing to do with sequencing.

but sequencing was at the heart of the OP... why do they turn horizontal almost immediately and not after they clear the atmosphere?

 

[jbriggs444]

but sequencing was at the heart of the OP... why do they turn horizontal almost immediately and not after they clear the atmosphere?

Your response seemed to imply that they should never turn horizontal at all, that the eastward velocity of the launch pad was an advantage that would apply regardless. Unfortunately, that is incorrect. The advantage of the eastward velocity of the launch pad is squandered if the [much larger] delta v is applied vertically rather than parallel to the existing velocity.

Yes, one needs to get clear of the atmosphere. But one also wants to avoid squandering that free delta v from the eastward motion of the pad.

 

[CWatters]

Your response seemed to imply that they should never turn horizontal at all, that the eastward velocity of the launch pad was an advantage that would apply regardless. Unfortunately, that is incorrect. The advantage of the eastward velocity of the launch pad is squandered if the [much larger] delta v is applied vertically rather than parallel to the existing velocity.

No I can see your point about vector addition. The issue I have is only with the ordering. For example..

3 East + 4 North + 7 East
vs
3 East + 7 East + 4 North

give same result overall.

Except in the second case the velocity after the first addition is higher (7 vs 5). So the effect of turning horizontal earlier appears to make the velocity in the atmosphere higher without affecting the final velocity.

 

[sophiecentaur]

Your response seemed to imply that they should never turn horizontal at all, that the eastward velocity of the launch pad was an advantage that would apply regardless.

How could you have come to that conclusion? They, of course, need to turn eastwards to bring their tangential velocity to what's required. The eastwards component from the Earth is there all the time. Apart from the fact that it always gives you an advantage, you have given no reason why it affects the best time at which the rocket drops its nose.

The direction that you apply the delta v matters. This is not a complicated idea.

It actually is a complicated idea, in detail, which is why we are trying to think out the precise reason for using the available tangential deltaV so early. ~The reason must be to do with efficiency and the notion of Gravity Drag is at the bottom of it. Near the start of a vertical trajectory, the engines are doing very little work - only providing some GPE. Allowing the rocket to accelerate (in any direction) is good value and improved efficiency by having the engines working at low speeds for as short a time as possible.

 

[jbriggs444]

How could you have come to that conclusion? They, of course, need to turn eastwards to bring their tangential velocity to what's required. The eastwards component from the Earth is there all the time. Apart from the fact that it always gives you an advantage, you have given no reason why it affects the best time at which the rocket drops its nose.

It actually is a complicated idea, in detail, which is why we are trying to think out the precise reason for using the available tangential deltaV so early. ~The reason must be to do with efficiency and the notion of Gravity Drag is at the bottom of it. Near the start of a vertical trajectory, the engines are doing very little work - only providing some GPE. Allowing the rocket to accelerate (in any direction) is good value and improved efficiency by having the engines working at low speeds for as short a time as possible.

You are defending a claim that is different from the one that was made: that the launch pad's eastward speed is a free bonus that applies regardless. It is not.

 

[sophiecentaur]

You are defending a claim that is different from the one that was made: that the launch pad's eastward speed is a free bonus that applies regardless. It is not.

Why not?

 

[jbriggs444]

Why not?

Because it can be squandered. E.g. if the launch is purely vertical.

 

[DaveC426913]

Because it can be squandered. E.g. if the launch is purely vertical.

It seems to me, there is an issue of defining "vertical". To-wit: is that relative to Earth FoR or a remote FoR?

If the rocket maintains a vertical ascent wrt Earth's FoR (i.e. it remains over the same geographic location), then it carries with it the advantage of the Space Centre's rotation of 14 miles per minute East.

If the rocket maintains its vertical ascent wrt to a non-rotating FoR (say, the Earth as a whole, or just its centre), then it is true that the Earth's rotational boost is squandered.

(Of course, for the former to be true, it would have to start vectoring eastward just so as not to fall behind.)

 

[russ_watters]

Because it can be squandered. E.g. if the launch is purely vertical.

I guess I'm not following either. Perhaps the cases are getting mixed-up. I'll try to be more descriptive in defining them...

The OP was asking about a slow ascent to orbital altitude, then turning downrange and increasing speed to orbit. Let's assume for that case the vertical speed when you reach orbital altitude would be zero. (Case 1) Other cases would be:
2. Turning almost horizontal, at the right inclination almost immediately after the launch (implied but not necessarily intended by the OP to be "actual").
3. Rotating to the right inclination and pitching over gradually after launch (actual path).

To me it seems like the vector addition of increasing the speed and changing the direction *downrange* should be exactly the same in all three cases. And it should look like this:

Starting: 900 mph at 90 degrees (east)
Needed: 17500 mph at 135 degrees (angle guessed)
Assist: 636 mph
Wasted: 900-636=264mph
Delta-V: 16,863 @ 137 degrees

How is this math different for one case vs the others? The only difference I see between the three cases is in the gravity and aerodynamic drags, not the delta-V/vector addition.

 

[russ_watters]

It seems to me, there is an issue of defining "vertical". To-wit: is that relative to Earth FoR or a remote FoR?

I guess it is possible, but I would hope not, since we kind of already discussed this. If the rocket thrusts perpendicular to the Earth's surface, the tangential velocity is preserved forever and you eventually reach an altitude where the tangential velocity is above orbital velocity (albeit not at the right inclination). As far as I can tell, the portion of the tangential velocity that helps you is never lost regardless of how you get to orbit (unless you purposely angle your thrust against it).

A trajectory where you remain above the same point (again, inclination notwithstanding) requires pitching-over, so it *is* the real trajectory rockets. It's geosynchronous, only differing in altitude of the orbit/magnitude of the pitchover from LEO we've been discussing to this point.

 

[sophiecentaur]

Because it can be squandered. E.g. if the launch is purely vertical.

"Squandered" means it doesn't follow Newton's First Law then? We all know that throwing a ball vertical will result in it landing back in the hand. It's horizontal momentum has not been "squandered". Amazingly, it has been conserved, as will the momentum of the rocket.

 

[jbriggs444]

If the rocket thrusts perpendicular to the Earth's surface, the tangential velocity is preserved forever and you eventually reach an altitude where the tangential velocity is above orbital velocity (albeit not at the right inclination). As far as I can tell, the portion of the tangential velocity that helps you is never lost regardless of how you get to orbit (unless you purposely angle your thrust against it).

Yes, for a purely vertical thrust, the tangential velocity would be there, certainly. But as an increment to the resulting speed it would be small.

 

[jbriggs444]

"Squandered" means it doesn't follow Newton's First Law then? We all know that throwing a ball vertical will result in it landing back in the hand. It's horizontal momentum has not been "squandered". Amazingly, it has been conserved, as will the momentum of the rocket.

The relevant metric is not momentum. It is energy.

 

[sophiecentaur]

The relevant metric is not momentum. It is energy.

We must be talking at cross purposes. I am sure you know your Newtonian Mechanics. Are you actually saying that the horizontal momentum it started with is not there, once it reaches an orbital position? (The Earth's curvature could be a second order issue) The only thing that could change it would be the difference in the g vector angle.

 

[russ_watters]

Yes, for a purely vertical thrust, the tangential velocity would be there, certainly. But as an increment to the resulting speed it would be small.

What resulting speed? Could you please be more specific because it feels to me like this lack of specificity is the entire problem here.

My interpretation is that if you thrust vertically [perpendicular to Earth's surface], your tangential speed will start at and forever be about 900 mph and is totally decoupled from your vertical speed. Do you agree/disagree?

 

[jbriggs444]

We must be talking at cross purposes. I am sure you know your Newtonian Mechanics. Are you actually saying that the horizontal momentum it started with is not there, once it reaches an orbital position? (The Earth's curvature could be a second order issue) The only thing that could change it would be the difference in the g vector angle.

I agree that we must be talking at cross purposes.

The horizontal momentum is still there, certainly. It does not go away. But if you add a horizontal momentum of 3 units to a vertical momentum of 4 units, you get a diagonal momentum of only 5 units, not 7 units. The magnitudes of momenta do not add.

 

[jbriggs444]

My interpretation is that if you thrust vertically [perpendicular to Earth's surface], your tangential speed will start at and forever be about 900 mph and is totally decoupled from your vertical speed. Do you agree/disagree?

I agree.

 

[cjl]

Because it can be squandered. E.g. if the launch is purely vertical.

That's a pretty pointless hypothetical though, since basically all launches are to go to an orbit, and therefore will be gaining the great majority of their velocity parallel to the Earth's surface. Any eastward orbit will gain this benefit, regardless of when during the ascent the turn happens, so it doesn't really answer the question. In addition, polar and retrograde launches also pitch over soon after liftoff, so the fundamental reason here is clearly different.

 

[russ_watters]

Yes, for a purely vertical thrust, the tangential velocity would be there, certainly. But as an increment to the resulting speed it would be small.

Since vertical speed and tangential speed are totally decoupled, that last sentence cannot be claimed. The relationship can be literally anything. You are apparently assuming a large vertical speed, but that specifically contradicts the OP's scenario (my case 1) which includes a final vertical speed of zero.

 

[russ_watters]

That's a pretty pointless hypothetical though, since basically all launches are to go to an orbit, and therefore will be gaining the great majority of their velocity parallel to the Earth's surface.

What about geosynchronous orbits? Could you not thrust vertically for the entire flight to achieve it (not including the inclination issue - say the launch is from the equator).

I think it is more complete to say this initial velocity is never squandered for any eastern orbit direction.

[edit]
It does indeed appear that for geosynchronous orbit, the launch would be vertical if not for the need to re-locate to the equator and whatever point they want to be over, and then zero-out the east-west motion. Or to put it another way; if launched from the equator from the point it wanted to be above it would just launch vertically.

http://www.planetary.org/blogs/jason-davis/20140116-how-to-get-a-satellite-to-gto.html

You can see in the video that the satellite ends up having traveled about 3/4 of the way around the spinning globe in about 20hrs. That's commensurate with the starting speed providing all of the "downrange" motion. The one caveat being that because it starts off elliptical it has to start of fast and then slow down!

 

[jbriggs444]

Since vertical speed and tangential speed are totally decoupled, that last sentence cannot be claimed.

Sure it can. Vertical speed and tangential speed add to total speed according to the pythagorean theorem.

Or to put it another way; if launched from the equator from the point it wanted to be above it would just launch vertically.

Sure, you could use a pure vertical boost to attain geosynchronous altitude followed by a pure horizontal boost to attain geosynchronous speed. But that would be silly. A better approach would involve a vertical boost to clear much of the atmosphere followed by a horizontal boost to attain a circular low Earth orbit followed by a continued horizontal boost into a Hohmann transfer to geosynchronous orbit followed by a horizontal boost to attain proper geosynchronous velocity.

One way makes better use of the starting platform velocity than the other.

 

[russ_watters]

Sure it can. Vertical speed and tangential speed add to total speed according to the pythagorean theorem.

No, you didnt follow your own context, which I was explaining in the rest of the post. Your claimed relation is apparently based on knowing the final speed, which you don't. You said: "as an increment to the resulting speed it would be small." You don't know that because you don't know the final orbital speed over the rotating Earth. it could be near zero, making the relationship: 'as an increment of the resulting speed it would be large.'

Sure, you could use a pure vertical boost to attain geosynchronous altitude followed by a pure horizontal boost to attain geosynchronous speed.

No: You only need the vertical boost. You already have the speed! (for a hypothetical equatorial launch)

But that would be silly. A better approach would...

I'm not arguing what is the best approach. I'm simply explaining that the alternate approach works and making sure you understand what it is, because you appear to be misinterpreting the intent and as a result saying overly constrained things about it.

One way makes better use of the starting platform velocity than the other.

So, this is the part I first asked for clarification of. I think what you are saying here is wrong (and is your basis for use of the word "squandered"). Could you please explain in more detail why you think this is true...perhaps by responding to post #82 or providing some actual vector math.

Edit: We can simplify if you want: if we launch from the equator, the boost from eatrh's rotation is 1,037mph and delta-v 16,463mph to LEO regardless of trajectory taken. If you disagree, please explain why, preferably with math.

 

[jbriggs444]

No, you didnt follow your own context, which I was explaining in the rest of the post. Your claimed relation is apparently based on knowing the final speed, which you don't. You said: "as an increment to the resulting speed it would be small." You don't know that because you don't know the final orbital speed over the rotating Earth. it could be near zero, making the relationship: 'as an increment of the resulting speed it would be large.'

Nobody wastes more delta v getting to a given orbit than they have to. In almost all (*) cases, eastward horizontal delta v is more efficient in attaining a given orbital energy than vertical delta v (barring atmospheric interference). Part of that is making efficient use of the initial velocity.

(*) Not all. Polar or retrograde orbits might have different requirements.

 

[cjl]

What about geosynchronous orbits? Could you not thrust vertically for the entire flight to achieve it (not including the inclination issue - say the launch is from the equator).

Nope. For one, if you launch vertical, your horizontal speed is what is maintained, not angular speed, so you wouldn't have nearly the horizontal speed needed for geosynchronous orbit once you get there. Secondly, you nearly always launch into a low Earth orbit followed by a transfer orbit, since it is much more favorable energetically than trying to launch straight to GEO.

I think it is more complete to say this initial velocity is never squandered for any eastern orbit direction.

[edit]
It does indeed appear that for geosynchronous orbit, the launch would be vertical if not for the need to re-locate to the equator and whatever point they want to be over, and then zero-out the east-west motion. Or to put it another way; if launched from the equator from the point it wanted to be above it would just launch vertically.

http://www.planetary.org/blogs/jason-davis/20140116-how-to-get-a-satellite-to-gto.html

You can see in the video that the satellite ends up having traveled about 3/4 of the way around the spinning globe in about 20hrs. That's commensurate with the starting speed providing all of the "downrange" motion. The one caveat being that because it starts off elliptical it has to start of fast and then slow down!

Again, even if launching from the equator, you'd want to launch into low Earth orbit first and transfer. The overall delta V is significantly less that way. I can go into the math later if you're interested.

 

[jbriggs444]

I think what you are saying here is wrong (and is your basis for use of the word "squandered"). Could you please explain in more detail why you think this is true...perhaps by responding to post #82 or providing some actual vector math.

Post #82 mentions three launch scenarios but then only shows a launch profile for one of them. That launch profile is a failure -- the result would be a crash and burn. Here is that profile:

Starting: 900 mph at 90 degrees (east)
Needed: 17500 mph at 135 degrees (angle guessed)
Assist: 636 mph
Wasted: 900-636=264mph
Delta-V: 16,863 @ 137 degrees

As I read it, that's an impulsive burn with a delta v of 16,863 mph on a ramp angled 47 degrees above the horizontal. Plus the Earth's rotation gives you a launch angle of 45 degrees true. That gives you an elliptical orbit that intersects with the surface of the Earth.

You have enough energy to get into a circular low Earth orbit. But you do not have the right angle to do so. With this launch profile, you are going have $\frac{\sqrt{2}}{2}v$ too much vertical velocity and $(1-\frac{\sqrt{2}}{2})v$ too little horizontal velocity. You would need a circularizing burn somewhere to make up the difference. More delta v to spend.

By contrast, if you had simply thrust horizontally, an initial delta v of only 17500-900 = 16600 mph would have done the job and put you into a circular low Earth orbit. Less delta v. And no need to spend more to avoid crashing.

To some extent this is an apples and oranges comparison. Your launch profile plus circularizing burn puts the craft into a high Earth orbit. The horizontal launch profile puts the craft into a low Earth orbit. The two are not the same. [To say nothing of the pesky atmosphere]

But here is the thing. Mechanical energy is conserved. It does not matter what direction the craft is moving at a particular speed. Its energy at a given altitude is a function of speed alone. An impulsive horizontal burn of 16,863 mph added to a starting velocity of 900 mph gives you an orbital speed of 17,763 mph. An impulsive burn of 16,863 mph at a 47 degree angle skyward added to a starting velocity of 900 mph horizontal only gets you to 17,500 mph. You've squandered energy and you're not getting it back.

If I were doing it right, I'd save that couple of hundred mph of delta v and go straight into a Hohmann transfer from a circular low Earth orbit. My Hohmann transfer is more efficient than yours due to the Oberth effect. I get to start my burn moving low and fast. You start your burn moving high and slow.

 

[OmCheeto]

I made a graph of "pitch angle" vs "time" for a shuttle launch.

The initial pitch number of 78° is from a NASA web page; "By about 20 seconds after lift-off, the vehicle is at 180 degrees roll and 78 degrees pitch." [ref]

Graphed against time, the change in pitch angle looks pretty smooth to me, and doesn't go horizontal until the craft crosses the Kármán line.

 

[JohnM]

Don't forget there is a constraint on the flight path from the (crash) landing of the first & second stages as well. They have to land in the sea within the range boundary or in the case of Russian rockets in uninhabited land. The (crash) landing boundary is normally to the East of the launch point to make use of the Earth's rotational valocity. The need for the first and second stages to land somewhere prevents the UK being used as a launch site as the discarded stages would crash into Europe.

You also want to launch from near the equator to gain the maximum velocity from the Earth's rotational speed hence the ESA space port in South America.

In the days of ICBMs being in the UK the first stage landed in the North Sea and the second stage in the north of Norway.

 

[olaney]

Even if the rocket traveled "straight up", the surface of the Earth is turning, which carries the observer away from the origin of the trajectory. In this respect, it is the observer who moves horizontally with respect to the rocket rather than the other way around. To this visual effect, add whatever course correction the rocket guidance system actually makes. At the equator, the velocity of Earth's rotation is around 1000 mph, which makes the apparent horizontal component large. Most launches are planned to take advantage of the rotational velocity at the launch latitude. This is why most launch centers are located at low latitudes.

 

[gary350]

I am a HOBBY rocket person. The reason rockets turn depends on how far away the target is. Nothing scientific about that. If the rocket is going to hit a target 100 miles away it does not need to travel higher than the radius of 100 miles half circle = 50 miles. If the rocket travels out of the atmosphere then it may burn up when it returns if it does not have a heat shield so there it a limit how high up it can go. If the rocket is going to travel 1/2 way around the world it can fly about 60 miles up all the way around to the other side of the world. After the rocket is launched at the target it turns to 45 degrees up to 60 miles then turns again to fly parallel to Earth surface. If the rocket is going to the moon or another planet it still may need to turn to go in the direction of the target. Military rockets turn and fly at low elevation to reach the target fast. Some military rockets are launched at a low angle anything from 3 to 15 degrees targets might be 1/4 mile to 5 miles away.

 

[Anno]

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 Earth's atmosphere, greatly decreasing air resistance and wasted energy. Wouldn't this be much more energy efficient?

I have often thought this and have not found an explanation that satisfies me!!! Most rockets level off very soon after liftoff when they are no more than 10 to 20 miles high, meaning they have an awful long way to go to reach their final altitude. However, even after flying horizontal they still seem to reach orbit in the same time they would if they continued vertically which of course is impossible. Why not fly to 50/60 or even 100 miles before starting to level off?? When you watch Space X launches the altitude displayed on screen also increases at the same rate even after the rocket changes it's pitch, which makes no sense. My intuition tells me something is off and until someone gives an explanation I am happy with it will continue to do so.

 

[Anno]

Even if the rocket traveled "straight up", the surface of the Earth is turning, which carries the observer away from the origin of the trajectory. In this respect, it is the observer who moves horizontally with respect to the rocket rather than the other way around. To this visual effect, add whatever course correction the rocket guidance system actually makes. At the equator, the velocity of Earth's rotation is around 1000 mph, which makes the apparent horizontal component large. Most launches are planned to take advantage of the rotational velocity at the launch latitude. This is why most launch centers are located at low latitudes.

The atmosphere obviously rotates at the same speed as the Earth so until the rocket leaves the atmosphere it's trajectory would remain in synch with the Earth's rotation if flown straight up. The observer would not be 'carried' anywhere.
Rockets enter an arc trajectory soon after takeoff. That's a fact, not a visual effect caused by the Earth's rotation.

 

[Anno]

The atmosphere obviously rotates at the same speed as the Earth so until the rocket leaves the atmosphere it's trajectory would remain in synch with the Earth's rotation if flown straight up. The observer would not be 'carried' anywhere.
Rockets enter an arc trajectory soon after takeoff. That's a fact, not a visual effect caused by the Earth's rotation.

When viewed from an angle rockets never actually travel straight up and immediately enter an arc, only adding further to the mystery.