Why do rockets turn horizontally so soon after launch?

[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)

|

 

[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..

 

[DaveC426913]

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

Up-goer 5: