# Launch Advantages in the East Direction?

1. May 3, 2014

### student34

I have read numerous explanations as to why it is beneficial to launch a rocket in the eastern direction. I know that it's true because NASA's website http://spaceplace.nasa.gov/launch-windows/en/ even claims this. But it makes absolutely no sense to me.

I understand that the Earth is spinning at about 410 m/s towards the east. I also understand that the 410 m/s is an initial tangential speed that lessons the mechanical energy needed to reach an orbit.

So then what advantage does an easterly launch bring? The conservation of angular momentum conserves the tangential speed of 410 m/s, so that will be there whether it is a vertical launch or an easterly launch. Therefore, it seems that any easterly motion is just going to add to the 410 m/s that the rocket starts with. And the vertical launch costs the same amount of potential energy no matter what angle it gets launched at.

Can someone please make sense of this.

I left out the orbital help of the Earth around the Sun to keep this simple for me.

2. May 3, 2014

### Staff: Mentor

Vertical launch? Launching vertically doesn't get you into orbit. The choices are east and west. If you launch to the east, you need to gain 820 m/s less speed than if you launch to the west. That's it.

3. May 3, 2014

### Drakkith

Staff Emeritus
The energy expended to reach a certain vertical altitude is the same no matter which way you launch. The key lies in the fact that to establish an orbit a ship needs to reach a certain velocity around the Earth. If you launch towards the east, you have an advantage in that the ship is already moving. Launching to the west means that not only do you not have this added velocity to help you, but you actually have to spend more energy to get yourself going in the other direction.

4. May 4, 2014

### A.T.

Not into a low Earth orbit, but eventually into some higher orbit.

5. May 4, 2014

### jbriggs444

If you are moving 3 meters per second east and you want to end up moving at least 5 meters per second in some direction, how can you do so most cheaply?

You can thrust east and spend 2 meters per second of delta-v.
You can thrust north and spend 4 meters per second of delta-v.
You can thrust west and spend 8 meters per second of delta-v.

The choice is simple.

6. May 4, 2014

### dauto

That's also why it is preferable to launch from a base near the equator than from one at high latitudes. the closer to the equator, the faster is your motion due to earths spin.

7. May 4, 2014

### Staff: Mentor

As Drak wrote, you need some minimum velocity to stay on the orbit (in Polish it is called 1st cosmic speed, no idea what is the English name). To leave the planet you need to reach so called escape velocity - around 11.2 km/s for Earth. Starting to the east you have already part of that speed.

8. May 4, 2014

### Drakkith

Staff Emeritus
9. May 4, 2014

### Staff: Mentor

No, that's something a little bit different. 1st cosmic is the lowest orbital speed required to keep an object on a circular orbit[STRIKE] (so the orbital speed can take any value between 1st cosmic and escape velocity). [/STRIKE]

Numerically it is escape velocity/square root of 2.

Last edited: May 4, 2014
10. May 4, 2014

### Drakkith

Staff Emeritus
Not sure I understand. Orbital speed for a circular orbit starts off very high when orbiting close to the body, and ends up very low when very far away, right? So the lowest orbital speed would be nearly zero when the object is extremely far away. Or have I misunderstood something?

11. May 4, 2014

### Staff: Mentor

Sorry, I am an idiot. I got caught by the way it is explained in most books here.

Let's say I am on the surface of the Earth and I want to throw an object horizontally, so that it will stay on the orbit. The lowest speed required is around 7.9 km/s and that's the 1st cosmic speed.

The important part being "the lowest at the distance R, equal to planet radius".

12. May 4, 2014

### dauto

I think that's called the LEO speed where LEO stands for Low Earth Orbit

13. May 4, 2014

### voko

The first cosmic speed is a term rarely used in English literature. I believe the term, along with the terms second, third and fourth cosmic speed was introduced by Ary Sternfeld, a Jewish-Polish-Soviet rocket science pioneer, possibly in the course of his correspondence with Tsiolkovsky.

All of those speeds are functions of the distance from some point of interest. The first and the second speeds are usually given at the surface of a gravitating body.

14. May 4, 2014

### student34

Ah, of course, thanks everybody!

15. May 5, 2014

### QuantumPion

If you launch straight up you will never get into orbit, you will always fall back down to the earth, unless you have escape velocity, in which case you would be in orbit around the Sun.

16. May 5, 2014

### A.T.

Why not? If I launch vertically at the equator, I have a tangential speed of 465m/s, which is the circular orbital speed at some height.

Last edited: May 5, 2014
17. May 5, 2014

### Staff: Mentor

Are you considering the rotation of the earth and the initial transverse velocity it gives the rocket?

Include it and the angular momentum that it implies, and you'll end up in some elliptical orbit. I don't have time now to calculate whether that orbit intersects the surface of the earth, may try later if someone else doesn't get there first.

18. May 5, 2014

### jbriggs444

If it is a closed orbit (ellipitical) rather than open (hyperbolic, parabolic) then clearly it intersects the surface of the earth. Perigee is attained prior to launch and is subterranean. In a little less than one complete orbit the rocket will be back at perigee again, below the surface.

19. May 5, 2014

### Staff: Mentor

Right - thanks.

20. May 5, 2014

### A.T.

If you start vertically at the equator and fire your engines long enough, then you can reach a closed orbit that doesn't intersect the Earth.