Why always start in orbit?

[ErasmoGnome]

So I've been playing for a while, got a rover or two to Eve, Duna, Minmus, and the Mun, and am currently working on Dres and Jool's satellites. All the "rocket science" stuff fascinates me and I love googling little real-world tidbits as I learn it's KSP analogue.

One thing I've always wondered though, is that in every tutorial and video I've seen, from going interplanetary or just an excursion to the Mun, everyone always establishes an LKO before burning for their final destination. In all of my successful attempts so far, I've been skipping any attempt at a Kerbin orbit and just burning to where I want to go immediately. For instance, if I wanted to go to a planet with an orbit larger than Kerbin's I'll launch at sunrise, and just burn my apoapsis all the way to escape velocity to establish an independent solar orbit without every going into a low Kerbin orbit. That way, I'm basically already point where I want to go, and once I'm out of Kerbin's SoI I can fiddle with my solar orbit to get an intersect.

So, is there a reason that everyone always orbits Kerbin before blasting off to their other destinations? It seems to me that it's not that inefficient, as you can just burn to escape to convert the energy you used to get into the orbit into whatever direction you want, but it always seemed to be kind of unnecessary to me. Is there a reason?

Thanks.

[Norpo]

Mainly because of the convenience, as far as I know.

It'd be hard to get the aiming right, and besides, in orbit you can actually use maneuver nodes, and you have time to think about how you'll execute things, also, preperation by docking in LKO, etc. VERY important, especially when you're playing career/no quicksaves.

It's just easier.

[SRV Ron]

So, is there a reason that everyone always orbits Kerbin before blasting off to their other destinations? It seems to me that it's not that inefficient, as you can just burn to escape to convert the energy you used to get into the orbit into whatever direction you want, but it always seemed to be kind of unnecessary to me. Is there a reason?

Thanks.

Do some testing to find out with one design example.

1. Launch straight up to a Duna orbit intercept. Note how much fuel is left. Note what time of the day uses the least fuel to reach the objective.

2. Launch to orbit, then to escape, then to a Duna orbit intercept. Note how much fuel is left

3. Launch to a 100k orbit, then to a direct Duna orbit intercept. Note how much fuel is left.

You will quickly find out which method becomes the most fuel efficient and why the rest of us are using that method of flight.

[wibou]

Do some testing to find out with one design example.

1. Launch straight up to a Duna orbit intercept. Note how much fuel is left. Note what time of the day uses the least fuel to reach the objective.

2. Launch to orbit, then to escape, then to a Duna orbit intercept. Note how much fuel is left

3. Launch to a 100k orbit, then to a direct Duna orbit intercept. Note how much fuel is left.

You will quickly find out which method becomes the most fuel efficient and why the rest of us are using that method of flight.

For the few of us that don't actually know all that, could you details? I'm very interested by that.

[Empiro]

There were a number of threads that discussed this a while back. There is a maneuver planning reason:

You get to use maneuver nodes and fine-tune your maneuver, including adding normal / radial components so that you get an encounter straight away. I don't know how much you're burning once in a solar orbit, but if it's more than 200 m/s (and you're not going to Eeloo / Moho), then you're spending more Delta-V than needed.

There is also the efficiency / cost reason:

By burning straight up, you're either wasting Delta-V due to gravity drag, or you're having to use lots of engines to get a high TWR to minimize gravity drag. You can probably launch a craft that is lighter and cheaper that will do the same thing.

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For the few of us that don't actually know all that, could you details? I'm very interested by that.

The exact numbers are going to depend on your rocket design, but the basic gist of it:

1. You have lots of loss due to gravity drag. How much depends on your rocket. If you find yourself with more fuel than 3), then your rocket has way too many engines, and you'd be able to launch a rocket that's cheaper / lighter and still perform the same mission.

2. You're not maximizing the Oberth effect. You'll find that you're burning 950 m/s to escape, and then about another ~500 m/s to get into a Duna intercept

3. Takes advantage of the Oberth effect. You'll only need to burn about 1080 m/s from orbit to get the Duna intercept. This difference gets bigger the further out you go, and makes a huge difference when travelling to Jool or Eeloo.

[Beachernaut]

From orbit, I can set up an intercept and aerobrake at Duna, Jool, or Eve in one burn. Not so much when launching strait from the launchpad.

[problemecium]

I second the statements about maneuvering and stuff, but the most efficient way to go from Kerbin to another planet is actually neither going to orbit first nor going straight up - it's launching using a gravity turn as if going to orbit, but instead of circularizing, continuing to burn horizontally until the intercept is reached. The most efficient time to do this is actually about 45 minutes after sunrise or sunset due to some physics related to the ejection angle. Long story short, you want your escape path to be due east or west at the point where you leave Kerbin's SOI, and since your trajectory will curve, you should burn at a moderate angle towards or away from the Sun.

For evidence of this effect, I recommend trying out the "Impending Impact" scenario and observing the fuel savings when the Mun escape maneuver is done with the ejection angle in mind.

[SAI Peregrinus]

KSP Trajectory Optimization Tool can be used to help plot the optimum transfer. As parameciumkid said, that will be a gravity turn to a burn at the right ejection angle. That said, it's a lot easier to do if you plot leaving from orbit, due to maneuver nodes.

[Scarecrow88]

Starting from orbit rather than burning straight out also allows you to build more interesting ships to explore those further planets, the sort of ships that wouldn't get off the launch pad as a single design.

[pandaman]

With my current multi ship mission to Duna I am pretty much doing what parameciumkid mentioned above with some success. I wait or timewarp until KSC is close to the optimum ejection angle for the 'normal' transfer burn from orbit then launch, level out and keep burning, rather than circularise first and leave from orbit. I found I don't need to fiddle much if at all to get intersection.

[KerikBalm]

A proper gravity turn to escaep velocity and straight into an intercept, would be the most efficient way... good luck doing it. And it will save you barely any dV

Gravity turn-> escape velocity -> heliocentric orbit (or even just somewhere along the escape trajectory) -> burn for intercept is inefficient due to the oberth effect.

You want to burn as much as possible when you are going as fast as possible, this means burning as low as possible (when there is no atmosphere to consider), so that you don't slow down some more before your second burn.

A proper gravity turn to a circular orbit prevents you from slowing down at all, and you can complete the intercept burn at your leisure in a very efficient manner

[Sirrobert]

It's far easier to plot an accurate trajectory with a manouver node from Orbit, than it is to try and calculate the exact time to launch, at the exact direction, all from the ground (without manouver node)

[OhioBob]

One thing I've always wondered though, is that in every tutorial and video I've seen, from going interplanetary or just an excursion to the Mun, everyone always establishes an LKO before burning for their final destination. In all of my successful attempts so far, I've been skipping any attempt at a Kerbin orbit and just burning to where I want to go immediately. For instance, if I wanted to go to a planet with an orbit larger than Kerbin's I'll launch at sunrise, and just burn my apoapsis all the way to escape velocity to establish an independent solar orbit without every going into a low Kerbin orbit. That way, I'm basically already point where I want to go, and once I'm out of Kerbin's SoI I can fiddle with my solar orbit to get an intersect.

So, is there a reason that everyone always orbits Kerbin before blasting off to their other destinations? It seems to me that it's not that inefficient, as you can just burn to escape to convert the energy you used to get into the orbit into whatever direction you want, but it always seemed to be kind of unnecessary to me. Is there a reason?

In either method we are creating a parking orbit from which the spacecraft is injected into its transfer orbit. In your method you are establishing a parking orbit around the sun (Kerbol), while the method used by others is to establish a parking orbit around Kerbin. The latter is far more efficient.

In both cases there are going to be gravity and drag losses while ascending from Kerbin, but let’s ignore those and call it a wash. Let’s consider only the velocity needed to escape Kerbin gravity, which, from an altitude of 75 km altitude, is 3235 m/s. (I’m using 75 km because that’s my typical parking orbit altitude.) Note that it actually takes a little less to escape Kerbin because the game uses a simplified patched conic method. We don’t actually have to reach escape velocity, we just have to reach the sphere of influence.

In your method we need to reach about 3235 m/s to escape Kerbin and enter into a solar orbit. The resulting orbit would essentially be the same size as Kerbin’s orbit around Kerbol. We then need to perform a second burn to either decrease our periapsis (if traveling to an inner planet) or increase our apoapsis (if traveling to an outer planet) in order to establish an intercept with our target. Let’s say we want to go to Duna, this would require a ∆V of about 950 m/s (varies about 850-1050 m/s), resulting in an orbit with a periapsis near Kerbin and an apoapsis near Duna. The total ∆V, therefore, is 4185 m/s. (With gravity and drag losses this would be closer to 6450 m/s.)

When using the Kerbin parking orbit method we again need to reach a velocity of 3235 m/s to escape Kerbin, it’s just done in two steps. We first reach an orbital velocity of 2287 m/s (75 km) and then perform a second burn to escape. The big difference is that from Kerbin orbit we can take the time to create a maneuver node, establish a good intercept with our target planet, and inject directly into our transfer orbit.

Let’s say that from our Kerbin parking orbit we increased our velocity to exactly escape velocity. Once we’ve left Kerbin space we’d be traveling at zero velocity relative to Kerbin. In order to reach Duna we’d need to add another 950 m/s (same as your method). Let’s now say that as we leave our parking orbit we accelerate to little bit greater than escape velocity. In this instance when we reach the edge of Kerbin space we’d have some left over velocity. This residual velocity is called hyperbolic excess velocity. We can plan our burn so that our hyperbolic excess velocity is exactly the 950 m/s we need to reach Duna.

The equation for hyperbolic excess velocity, V_{∞}, is (that subscript is the 'infinity' symbol if it's too small to read)

V_{∞}² = V_bo² – V_esc²

where V_bo is the burnout velocity and V_esc is escape velocity.

In our Duna example we know that escape velocity is 3235 m/s (75 km parking orbit), and the required V_{∞} is approximately 950 m/s. We can therefore calculate the required burnout velocity.

V_bo = SQRT( 950² + 3235²) = 3372 m/s

Thus we see that by performing our transfer orbit injection from low Kerbin orbit, it takes only a small amount of ∆V to produce the V_{∞} needed to reach our target planet. In this case the ∆V needed to reach Duna is only 3372 m/s (or about 5635 m/s including gravity and drag losses). This is considerably less than it takes to perform the transfer injection from a solar orbit.

Of course it is theoretically possible to launch directly from the surface into an escape trajectory that has the required V_{∞}, but it’s nearly impossible to obtain the proper intercept with the target planet with the tools available to us in KSP. This is why the Kerbin parking orbit method is preferred. It allows the use of maneuver nodes to properly plan and execute the transfer burn while still being highly efficient in terms of ∆V.

Edited January 11, 2015 by OhioBob

[OhioBob]

A proper gravity turn to escaep velocity and straight into an intercept, would be the most efficient way... good luck doing it. And it will save you barely any dV

I agree that the ∆V savings aren't worth the trouble. In fact, if you don't get the intercept just right, you're likely to burn up whatever ∆V savings there were in having to make course corrections. Better to go into orbit first and take your time making a good clean transfer burn.

I suppose it might differ from person to person on how they build their rockets and how they perform their ascents, but in my case I estimate that the ∆V savings would amount to only 30 m/s.

[Liowen]

I would say one nice thing about being in orbit is if you mess up the burn you can go back to the last save, but if you launch from the pad and something goes wrong you have to do the launch all over again. Also you can do multiple launches for other missions and leave them in orbit (unless of course life support is being used) and eject them as the windows become available, this is great for probe missions in career mode to have them ready to go at a moments notice.

[ErasmoGnome]

That's a great explanation, OhioBob! I liked everyone else's talk about convenience which totally makes sense, but the math behind hyperbolic excess velocity is fascinating - thanks for sharing!

Edited January 11, 2015 by ErasmoGnome

[Starhawk]

I was wondering about this exact thing just two or three weeks ago. I was fairly convinced that it was better to start from Kerbin orbit, but even though I speak some orbital dynamics I wasn't sure of the exact reason. So I checked it out myself. Tried some transfers both ways with the same craft. My testing confirmed that starting from Kerbin orbit is much more efficient, and after the many excellent posts above, I now have a better understanding of why.

It may be easier in some sense to just jump out of Kerbin's SOI and then play with a maneuver node in your (solar? kerbolar?) orbit until you get an encounter, but it costs quite a bit of delta-v.

[OhioBob]

... the math behind hyperbolic excess velocity is fascinating - thanks for sharing!

If you really find it that interesting I can easily show you the derivation of the equation. A body moving only under the influence of gravity has constant energy (i.e. law of conservation of energy). Of course the energy of a body is the sum of its kinetic and potential energy. This is given by the following,

E = mv²/2 - GMm/r

where m is the mass of the body, v is its velocity, r is its radial distance from the center of the planet, and GM is the gravitational parameter of the planet (i.e. constant of gravitation G times mass M). The term mv²/2 is the kinetic energy and -GMm/r is the potential energy.

If r₁ and v₁ are the distance and velocity of the body when it is near the planet, and r₂ and v₂ are the distance and velocity of the body when it is far from the planet, then, since energy is constant, we have

mv₁²/2 - GMm/r₁ = mv₂²/2 - GMm/r₂

Dividing through by m we get

v₁²/2 - GM/r₁ = v₂²/2 - GM/r₂

To obtain the hyperbolic excess velocity we want r₂ to be some distance far from the planet. As r₂ approaches infinity, -GM/r₂ goes to zero, therefore we have

v₁²/2 - GM/r₁ = v₂²/2

Multiplying by 2 we get

v₁² - 2GM/r₁ = v₂²

The equation for escape velocity is V_esc = (2GM/r)^1/2, therefore we can substitute as follows

v₁² - V_esc² = v₂²

Since v₁ is the velocity near the planet we can call this the burnout velocity. And since v₂ is the velocity at some infinite distance, it represents the hyperbolic excess velocity. Thus the final equation is,

V_bo² - V_esc² = V_{∞}²

I hope I didn't bore you with all that, but I think it really helps the understanding to know the derivation of an equation.

[Zourin]

I often circularize to hit my LKO refuel depot before I travel on if I'm cutting it close (and I usually am to cut back on weight). I try to hit it on the way back to 'drop off' excess fuel that I don't need to return to Kerbin. The more fuel you can conserve in orbit, the less you have to burn to get more out there.

Sometimes a good answer doesn't involve a lot of math ^.^

Edited January 11, 2015 by Zourin

[OhioBob]

v₁² - 2GM/r₁ = v₂²

The equation for escape velocity is V_esc = (2GM/r)^1/2

One additional clarification on the part above. Note that escape velocity is the special case where the burnout velocity is exactly that needed for the body to reach infinity but with zero residual velocity. In other words, escape velocity is the value of v₁ where v₂ is equal to zero. Thus,

v₁² - 2GM/r₁ = 0

V_esc = (2GM/r)^1/2

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Sometimes a good answer doesn't involve a lot of math ^.^

True, but for those who don't mind the math it can be helpful. Some people aren't satisfied just to know that an equation works, they want to know why it works. Those who don't like math can simply skip over it. We can aim to please both the math geeks and those who are math adverse.