# Centrifugal force and rockets

1. May 17, 2012

### Gavroy

hi

i was wondering why rockets always start upright, as if they would start in a position parallel to the surface of the earth(or at least almost parallel to the earth, but at the equator this should not be such a problem, if there are no mountains or anything else close to the point where the rocket is launched) one could use centrifugal force to escape from the gravitational field of the earth.

why is this not a good idea, as afaik centrifugal force seems to be not used at all?

2. May 17, 2012

### DaveC426913

1] The most efficient use of fuel is to get the rocket to thinner atmo as soon as possible. Plowing through the thick atmo at surface altitudes is a huge waste of fuel. Furthermore, if it were still in the atmo for any subsequent part of its launch, it would be at risk of burning up due to frictional heating at the speeds it needs to reach.

2] They do take advantage of the rotation of the Earth. Sitting on the launchpad, they are already heading East at well over 750mph. The fact that the rocket is going straight up (from the point of the view of the Earth's surface) does not detract from this. In any inertial frame of reference, the rocket is indeed already moving at 750+mph.

Think of throwing a ball on a train. The ball is already moving at 60mph. Throwing the ball straight up (from the point of view aboard the train) does not decrease this at all. And throwing it forward doesn't increase the boost that the train gives it either. The train still contributes 60mph.

Last edited: May 17, 2012
3. May 17, 2012

### Gavroy

sorry I think I do not get this right. if there is no additional boost by starting with an angle that is closer to the surface of the earth, then why are they heading east?

let me point out my idea this way: if a rocket goes straight up then you still have a contribution by the rotation of the earth, but it is perpendicular to the motion of the rocket, so it would not contribute to the motion at all. therefore one should start heading east, but:

apparently, there must be a "best angle" to start your rocket, but how do you determine this angle?

4. May 17, 2012

### DaveC426913

There is an intrinsic boost Eastward, no matter what the rocket does. That 750mph gets it closer to its necessary 17,000mph.

Yes.

It doesn't contribute to its upward motion, but it does contribute to its Eastward motion, which is ultimately what it needs.

It will, yes, once it clears the thickest part of the atmo.

The best angle is 90 degrees.

Well, the best practical angle. Again, there's little point in applying any horizontal thrust (except that initial 750mph - it's "free", since the atmo is also moving Eastward at 750 mph) until it clears the thickest part of the atmo. So it goes up, then veers horizontal. It would be awkward to build launchpads at some rakish angle for the little you'd gain from it.

Last edited: May 17, 2012
5. May 18, 2012

### D H

Staff Emeritus
Assuming a rocket can be launched at any angle, the "best" angle for launch is not 90 degrees, where the "best" angle is that which minimizes the quantity of fuel needed to get the rocket to low Earth orbit.

Unfortunately, that assumption is unrealistic. Rockets cannot be launched from any angle. Tipping a rocket even slightly from vertical would require a much more massive launch infrastructure and would require a much more massive rocket. Rockets are launched from a vertical orientation because the only feasible orientation is horizontal, and that orientation is far from optimal.

6. May 18, 2012

### K^2

You can rotate the rocket almost immediately after the launch. Yet, the typical trajectory usually involves a good deal of vertical ascent. So the optimal angle does seem to be 90°.

7. May 19, 2012

### Gavroy

and what would be the theoretically best orientation? as far as i see, this angle could not differ that much from 90 degrees, as centrifugal force is compared to the velocity you need to escape from the earth relatively weak.

8. May 19, 2012

### DaveC426913

How does centrifugal force come into play?

9. May 19, 2012

### D H

Staff Emeritus
What centrifugal force? It is a fictitious force. You do not need to invoke centrifugal force; it is in fact easier to look at things from the perspective of a non-rotating frame in which there is no centrifugal force.

As far as "optimal" is concerned, what are you trying to optimize? There are multiple conflicting variables in play here, so there is no such thing as a trajectory that optimizes everything. The best one can do is some sort of compromise.

Start out easy by looking at launching from an airless, non-rotating planet with no concern for structural issues and with a rocket capable of extremely high (effectively infinite) thrust. Here the "best" thing to do is to launch horizontally with an impulsive burn that places the apofocus at the desired altitude. The rocket will perform a second horizontal impulsive burn to place the rocket in a circular orbit upon reaching apofocus. This is essentially a Hohmann transfer from the ground to on-orbit. Making all of the burns horizontal means there are no gravity losses.

Making the planet rotate just means the first burn needs to be to the east (in the direction of rotation) to minimize fuel consumption. It's still a Hohmann transfer with horizontal burns.

Adding an atmosphere makes things a lot more complicated. Air drag increases with the velocity squared while density decreases more or less exponentially with increased altitude. This means gives some advantage in gaining altitude at a relatively low velocity. However, this increases gravity losses. So a tradeoff.

Adding concerns about structural and payload integrity makes things even more complicated. The only orientations that make sense structurally are vertical and horizontal. Anything in between would require adding lot more structure to both the rocket and to the launch system, and even a horizontal orientation would require more structure in the rocket than would a vertical orientation. Another downside of a horizontal orientation is that this makes it hard to balance increase in velocity versus decrease in pressure. In technical terms, it would make for a very large max Q (maximum dynamic pressure).

As for the optimal trajectory, it's called a "gravity turn" (google that phrase).