I Why do rockets turn horizontally so soon after launch?

[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.

 

[DaveC426913]

Because
1. the target altitude is only 200 miles. That's chump change compared to
2. the down range velocity required is Mach 25, which is the opposite of chump change, and besides,
3. the atmo thins out very rapidly with altitude, so atmo friction is of little concern.

 

[jbriggs444]

I have often thought this and have not found an explanation

Then you should read the rest of this thread. Because the answer was already here four years before you asked. If the answers are unsatisfactory, a more focused post looking for clarification (perhaps in a new thread) might yield more helpful conversation.

You were indeed correct to point out that a 1000 mph rotation of the atmosphere would not magically carry the rocket on an apparently horizontal trajectory relative to an earth that is also rotating at 1000 miles per hour. That explanation does not pass the sniff test.

 

[sophiecentaur]

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

Nice to pick up on an ancient argument. The tangential velocity at take off will correspond to some free energy, whatever. The choice of path must be affected by friction loss in the lower atmosphere and shortens the distance in the high drag phase. Thereafter, they choose a trajectory which makes best use of the fuel and gives the correct KE and GPE for the chosen low Earth orbit.

There are an infinite number of trajectories which could produce the wanted orbit but most of them would involve much more fuel. For instance, you could take a radial (vertical) path and then tip over to reach the right tangential speed (at the same time, providing enough radial thrust to keep you up there until the tangential speed caught up). You could also go up slower (= even more fuel). etc.etc

If there were no atmosphere, the optimum trajectory would be different and less fuel would be needed.

 

[Janus]

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.

Contrary, to what you might think, the most efficient way to reach orbit would be to lay the the rocket on its side at launch, accelerate it up to the speed needed to put in an elliptical orbit with the perigee at the surface and the apogee at the desired final orbital distance, Then upon arrival at apogee accelerating again to raise the perigee to the present altitude.

Of course that would require building a rail system long enough to support the rocket until it reached a minimum of 7.9 km/s, and building a rocket capable of being able to withstand traveling at that speed through sea level pressure atmosphere. ( such a "rail launch" system may be practical from an airless body like the Moon.)

The least efficient would be to fire the rocket straight up until it reaches orbital altitude, then doing a right angle turn and accelerating up to orbital speed.

Rocket launches from the Earth compromise. They start straight up, allowing them to pick up some speed, and then begin to tilt over, constantly accelerating as they do so. The angle changes as they gain speed and altitude.

The image you gave is looking "down range" so perspective make it looks like the "leveling off" is more extreme than it actually is.

 

[hutchphd]

Rocket launches from the Earth compromise.

This (and what preceeds) seems as complete and clear an answer to the initial question as one could devise. Thank you.

 

[Drakkith]

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.

This is incorrect. Rockets generally don't level off until they are done with their burn and have reached their orbital velocity. A near-ideal trajectory for a rocket* is to slightly pitch over soon after liftoff and then let gravity slowly 'pull the nose down' until the rocket reaches orbital height and velocity, at which time the pitch should be zero or near-zero. This is known as a gravity turn and saves fuel because you aren't spending fuel altering the spacecraft's direction, only its velocity. Gravity is doing the work changing the direction. This is why a rocket like the Japanese Lambda 4S, which has NO STEERING OR GUIDANCE, can be used to get payloads into orbit.

*More specifically, for a rocket doing a constant burn to enter a near-circular orbit. Other types of orbits, especially those that are highly elliptical or that are very high in altitude (like geosync/geostationary) require different launch trajectories.

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.

First, the rocket isn't horizontal until it reaches orbit or is close to orbit, so there is usually at least some vertical component to the thrust. For the first part of the launch this vertical component is enough to keep accelerating the rocket upwards against gravity even as it pitches over. It's only the latter part of the launch where the vertical velocity starts to fall as the rocket pitches over enough for gravity to overcome the vertical component of its thrust.

Second, if you're just eyeballing the altitude readouts then you really can't claim that the altitude increases at the same rate. You'd need to actually plot the readouts vs time, not just guess.

 

[sophiecentaur]

Contrary, to what you might think, the most efficient way to reach orbit would be to lay the the rocket on its side at launch, accelerate it up to the speed needed to put in an elliptical orbit with the perigee at the surface and the apogee at the desired final orbital distance,

Did you see the Spin Launch website? That project plans to inject a rocket into a high altitude using a centrifugal accelerator. But high speeds at low altitude involve a lot of drag!

 

[hutchphd]

But high speeds at low altitude involve a lot of drag!

I took @Janus to mean "ignoring air drag" because

( such a "rail launch" system may be practical from an airless body like the Moon.)

This entire argument requires only two words: Hohmann transfer.
In practice the effects of finite available acceleration and low altitude drag sometimes demands a trajectory that dips below "horizontal"

 

[jbriggs444]

In practice the effects of finite available acceleration and low altitude drag sometimes demands a trajectory that dips below "horizontal"

If we are going to ignore air resistance and contemplate subterranian trajectories then we may be able to do better than a horizontal eastward launch. We should thrust west instead. Or move the launch site to one of the poles.

Start the trajectory by killing all horizontal velocity. Let the vehicle free fall to the center of the earth. The craft will be have about 42 megajoules/kg at this point if I've eyeballed the PREM gravity profile correctly. This is about 9000 meters per second. Now you do an impulsive burn and let the vehicle free fall back up and out the far side of the earth.

Some engineering work may be needed to create and evacuate the tunnel through which the vehicle will travel.

If you spend 1600 kph (450 m/s) delta V on the initial de-orbit burn and another 450 m/s delta V on the impulsive burn at the center of the Earth, that will buy you about 4 megajoules/kg bringing you to 46 megajoules/kg at the center of the earth. The result is that the vehicle emerges on the far side of the earth with that +4 megajoules/kg of kinetic energy intact. That is enough for 3000 m/s (10,000 kph). You spent 3200 kph delta V and received 10000 kph delta V.

This is the Oberth effect.

 

[DaveC426913]

contemplate subterranian trajectories

Er. I don't think anyone was doing that were they? I assumed a below-horizontal trajectory would be accompanied by a positive altitude!

 

[hutchphd]

Er. I don't think anyone was doing that were they?

Well not on purpose at least. There was a russian "proton" launch .....(they installed the gyro upside down)

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[sophiecentaur]

Let the vehicle free fall to the center of the earth. The craft will be have about 42 megajoules/kg at this point if I've eyeballed the

Hangonabit. All that 'free' Kinetic Energy would be lost by the time you reached to far surface. KE + PE will be the same throughout the trajectory. Motion underground would approximate to SHM for uniform density.

Also, you mention killing horizontal velocity but the horizontal velocity of the walls of the tube is not the same over the whole distance so would you have a spiral path (or wheels on the side)?b

Unless you can take the vehicle through a low vacuum, drag will always be an issue

 

[hutchphd]

Start the trajectory by killing all horizontal velocity

While your invocation of the Oberth effect is on point I believe the implementation proposed is far from optimal. The method of "slamming on the brakes"makes no sense to me.
If we bestow upon the spacecraft the ability to travel through planetary matter unaffected by hull drag then any "burst" firing of the engine should be delayed until the distance of closest approach to earth center. This will provide the corrct generalization of the Hohmann transfer.

 

[sophiecentaur]

While your invocation of the Oberth effect is on point I believe the implementation proposed is far from optimal. The method of "slamming on the brakes"makes no sense to me.
If we bestow upon the spacecraft the ability to travel through planetary matter unaffected by hull drag then any "burst" firing of the engine should be delayed until the distance of closest approach to earth center. This will provide the corrct generalization of the Hohmann transfer.

I'm even more confused now. I need someone to take me through this in 'stages' (pun intended). Whilst in the underground phase, the normal orbital mechanics wouldn't apply, would they? Force Proportional to distance rather than Inverse square law., for a start and the path would be 'constrained' in a bore.

I'm almost looking for April 1st here.

 

[hutchphd]

I think Groundhog Day a more fitting holiday!

I will try to start. Because the real Earth is nonuniform in radial density, the actual path for a satellite will be more complicated, so closed form solution not usually tractable.. However the force will still be central and angular momentum will be conserved.
The Oberth effect criterion, as I understand it, is to expend your fuel in a way that maximally uses its kinetic energy (relative to the Center of Mass system of the orbit). This means to get close to periapsis before firing. I presume this is calculable if we assume the earth to be uniform, but this is not really required.
It is not necessary "slam on the brakes" in order gain the Oberth bonus. Just shape your tunnel to accommodate the coriolis forces and the sudden acceleration at perigee. I think @jbriggs444 was concocting a worst case to effectively prove his point..