"Gravity drag" and conservation of energy.

[The Pink Ranger]

Assuming you are lifting off from a body with no atmosphere why is a steep ascent path less efficient than a shallow one? The amount of kinetic energy required to achieve a certain orbit should be exactly the same in either case shouldn't it?

Edited August 8, 2014 by The Pink Ranger

[GoSlash27]

Negative. An orbit is just "falling around" a body, so the most efficient orbit is just off the body's surface unless it has an atmosphere to slow you down. Any orbit above that requires more energy to attain.

In this case, the shortest distance (and hence most efficient path) between you and orbit is to go that-a-way, not up there and hang a left.

Best,

-Slashy

[Superfluous J]

No it shouldn't.

At a given altitude, a minimum given speed (in ANY direction) will give you a valid orbit, so long as your periapsis is above the surface. It doesn't matter how you achieve that speed, you just have to get there.

If you burn straight up, every second you burn you're "wasting" 9.8 (or less, as you rise) m/s of thrust. If you burn sideways, you are NOT wasting that.

Simply try it. Get a rocket with about 7-8km/s of dV and launch straight up until your apoapsis is at 80km. Then coast up there and then burn sideways until you're in orbit. Assuming your rocket can do that (it's actually quite hard) note how much dV you have left. Then launch the same rocket the "normal" way where you go straight up to 10km, then start your gravity turn. You'll be surprised at how much dV you save.

[Master Tao]

I'll try to help you visualize it. For the most part, the difference is that the kinetic energy needs to be added in the right direction. Here are a couple of ways of visualizing this:

1. You can picture a ship in a circular orbit as having a constant horizontal velocity vector. Gravity just pulls the orbit circular. When you burn up, your goal is not to establish orbit, but to clear the atmosphere and terrain so you don't crash while burning horizontally.

2. Imagine a smooth, frictionless world. Picture two burns using a ship where the TWR is less than 1:

• Burn straight up. All that energy is cancelled by gravitational pull and the ship goes nowhere.
  
• Burn horizontally. Initially, gravity will keep you pinned to the surface, but as you get fast enough, you lift off and establish an AP on the far side of the world. You'll only need to make a small circularization burn to finish the orbit.
  

[Eric S]

Another way to look at it is this. Imagine you have a craft with an acceleration of 2 m/s, and local gravity is 1 m/s. If you thrust straight up, you lose half of your thrust to gravity, so only half of your thrust goes to increasing the energy of the craft's orbit. Now, instead of thrusting straight up, thrust 30 degrees above the horizon. Your 2 m/s of acceleration now results in 1 m/s of vertical acceleration and just over 1.7 m/s horizontal acceleration. Gravity will kill the 1 m/s of vertical acceleration, but the horizontal acceleration, which is greater than the remaining acceleration of the vertical launch, is untouched, resulting in 70% more acceleration for the same amount of fuel. Furthermore, as you increase your horizontal velocity, gravity starts having less effect on the trajectory (as a function of distance, not time), so you don't even need to maintain 1 m/s vertical acceleration in order to continue in a circle.

As you might already see, as the TWR of the craft increases, this factor becomes narrower. However, then you start getting into how much energy is required to circularize, assuming you're going for an orbit and not just escape velocity, and the greater horizontal velocity of the launch profile with the more aggressive turn means that it needs less delta-v to circularize.

EDIT: Ugh, corrected the percentage, 0.7 m/s is 70% of 1. Yes, it's also 35% of 2 which is what I was thinking for some stupid reason, but that's not the point.

,

Edited August 8, 2014 by Eric S

[The Pink Ranger]

I'll try to help you visualize it. For the most part, the difference is that the kinetic energy needs to be added in the right direction. Here are a couple of ways of visualizing this:

1. You can picture a ship in a circular orbit as having a constant horizontal velocity vector. Gravity just pulls the orbit circular. When you burn up, your goal is not to establish orbit, but to clear the atmosphere and terrain so you don't crash while burning horizontally.

2. Imagine a smooth, frictionless world. Picture two burns using a ship where the TWR is less than 1:

• Burn straight up. All that energy is cancelled by gravitational pull and the ship goes nowhere.
  
• Burn horizontally. Initially, gravity will keep you pinned to the surface, but as you get fast enough, you lift off and establish an AP on the far side of the world. You'll only need to make a small circularization burn to finish the orbit.
  

Another way to look at it is this. Imagine you have a craft with an acceleration of 2 m/s, and local gravity is 1 m/s. If you thrust straight up, you lose half of your thrust to gravity, so only half of your thrust goes to increasing the energy of the craft's orbit. Now, instead of thrusting straight up, thrust 30 degrees above the horizon. Your 2 m/s of acceleration now results in 1 m/s of vertical acceleration and just over 1.7 m/s horizontal acceleration. Gravity will kill the 1 m/s of vertical acceleration, but the horizontal acceleration, which is greater than the remaining acceleration of the vertical launch, is untouched, resulting in 35% more acceleration for the same amount of fuel. Furthermore, as you increase your horizontal velocity, gravity starts having less effect on the trajectory (as a function of distance, not time), so you don't even need to maintain 1 m/s vertical acceleration in order to continue in a circle.

As you might already see, as the TWR of the craft increases, this factor becomes narrower. However, then you start getting into how much energy is required to circularize, assuming you're going for an orbit and not just escape velocity, and the greater horizontal velocity of the launch profile with the more aggressive turn means that it needs less delta-v to circularize.

Yea I thought about it for a bit and realized you can burn off all your potential on the pad without actually moving anywhere. Both of these responses really helped me solidify the idea though so thanks. I think I have an idea of what mechjeb means by "cosine losses" now.

Edit: Repped both of you.

Edited August 8, 2014 by The Pink Ranger

[QuesoExplosivo]

No, Mechjeb's cosine losses are when engines are angled off to the side (so they end up fighting each other somewhat).

[The Pink Ranger]

Ah gotcha.

[UmbralRaptor]

Assuming you are lifting off from a body with no atmosphere why is a steep ascent path less efficient than a shallow one? The amount of kinetic energy required to achieve a certain orbit should be exactly the same in either case shouldn't it?

There's more than one thing going on here, even on airless bodies. I actually have a fairly relevant blog post

1) Rockets work by changing their momentum (I suppose one could call ÃŽâ€V a measure of momentum change with the mass abstracted out), but orbital parameters work by changing energy. Hence the Oberth effect, etc. Vertical burns tend to keep your speed down, hence the greater ÃŽâ€V requirement.

2) With non-zero burn times, gravity drag comes in. Eric_S gives a good overview.

[Kasuha]

Assuming you are lifting off from a body with no atmosphere why is a steep ascent path less efficient than a shallow one? The amount of kinetic energy required to achieve a certain orbit should be exactly the same in either case shouldn't it?

Remember, the best way to change a part of an orbit is from the opposite side of that orbit. Apoapsis is best changed from periapsis and vice versa.

Now, as soon as you liftoff, you're in orbit. A very eccentric, elliptical one, with its periapsis deep under the surface. In map view, if you zoom out sufficiently, you can often even see that ellipse. And you are near its apoapsis. The best thing you can do at that spot is raise your periapsis, that means getting into circular orbit with Pe/Ap at these few meters above ground where you are just at the moment. And only after that to raise the apoapsis - on the other side of the planet.

Of course you need to make sure you don't crash back onto the surface or into a nearby hill. But these are the inefficient parts of your maneuver. The efficient part is getting circular as soon as possible.

[EdFred]

On non-atmo'd bodies, I immediately pitch to 45, and start shallowing out my ascent from there.

[The Pink Ranger]

Yea I've definitely got the concept down. I remember someone mentioning it was more efficient to burn below the horizon once you've cleared most of the atmosphere around 30km and try to keep your orbit as circular as possible as you're pushing it out, now I know why. Anyway how do I set the topic title to "answered"? I don't see the option anywhere when I try to edit the post.

[Superfluous J]

Yea I've definitely got the concept down. I remember someone mentioning it was more efficient to burn below the horizon once you've cleared most of the atmosphere around 30km and try to keep your orbit as circular as possible as you're pushing it out, now I know why. Anyway how do I set the topic title to "answered"? I don't see the option anywhere when I try to edit the post.

You have to edit the first post, then go into advanced mode, and it's the very top of the page.

Not at all obvious but there you go

[geb]

I like to think of it this way. If you have a rocket with thrust-to-weight of exactly one, it needs to burn a lot of fuel to remain in a static hover. In an ideal situation you could hold altitude forever altitude on zero energy, since there's no change in potential, but the rocket can't do that.