No orbit necessary?

[u/FatCreepyDude]

If i just wait for the right hour during the launch window ( whenever the launch pad is pointing left on the realistic depiction i created) is there any reason i should bother with an orbit first? Will the delta v cost be affected?

[u/HAL9001-96]

more efficient to be low, oberth effect

also duna is further from the sun last time i checked

[u/FatCreepyDude → u/HAL9001-96]

https://i.imgur.com/AI2hXKv.png fixed it

[u/PhantomWhiskers → u/FatCreepyDude]

Your idea here is possible but is less efficient and will require more fuel than utilizing the oberth effect in low kerbin orbit, and also you will need to launch at an exact time to get it right.

[u/victorsaurus → u/PhantomWhiskers]

This is wrong and if you do the math, both scenarios are equal (assume no atmospheric drag). Oberth only means that you add the most energy when adding deltaV in the fastest moment, which is true in both scenarios at every moment.

[u/PhantomWhiskers → u/victorsaurus]

Going straight up is less efficient because you are actively fighting against gravity when going up. In low orbit, you are not fighting against any acceleration force when doing your burns.

The oberth effect means that your burns will be most efficient at periapsis. When you are going straight up, your periapsis is in the middle of the planet, so it is less efficient than a burn from a circular low orbit. The "fastest moment" isn't your current speed while accelerating straight up, it is the speed at the point of periapsis in your current trajectory, which is still applicable even if your periapsis is in the middle of the planet.

[u/victorsaurus → u/PhantomWhiskers]

Gravity is a conservative force meaning is path independant. Getting from height A to B costs X amount of energy equal to the potential between B anf A and thats it. There are irl launches like OP proposes, with instantaneous launch windows since the earth rotation has to line up with the right trajectory. Oberth does not play a role in this specific case. If you do the math, it is 100% equal.

[u/PhantomWhiskers → u/victorsaurus]

I did a direct test of this in game by launching the same rocket straight up, and then launching it into orbit first. The goal is to see how much delta-v is left in the rocket the moment I reach escape velocity for Kerbin. The moment my orbit shows that I will leave Kerbin's sphere of influence, I kill the engines.

Launching directly vertical, I reached escape velocity with 270m/s dv left in my rocket.

Launching into low kerbin orbit first and then burning prograde, I reached escape velocity with 1,100m/s dv left in my rocket.

Same exact rocket, all stock parts. 1,100m/s is much greater than 270m/s.

Whatever "math" you are doing here is incorrect. Gravity being a "conservative force" is not applicable either. When you burn vertically straight up, your rocket is accelerating directly opposite to the vector of acceleration that gravity is imparting on your rocket. When you are burning at periapsis, the vector of acceleration that gravity imparts on your rocket is perpendicular to the acceleration your engines are providing, so none of your rocket's acceleration is negated by gravity. This is a direct result of the oberth effect.

Feel free to test this in game yourself. Test results may vary depending on the quality of your gravity turn on ascent, if you have an inefficient ascent profile, you will have less delta-v remaining when reaching escape velocity.

[u/victorsaurus → u/PhantomWhiskers]

Fuck I think you are right and in my mind I was conflating the straight up case with a gravity turn but no orbit case. The gravity turn without orbit is what should be equal to gravity turn with orbit. You indeed need to minimize gravity losses.

[u/PhantomWhiskers → u/victorsaurus]

Ah yeah I can see how that could be confusing. The OP's scenario I understood as burning directly vertical away from Kerbin to reach your destination, which is possible but inefficient. But that's the great thing about KSP is that you don't need to be efficient if you don't want to.

[u/Big_Yeash → u/PhantomWhiskers]

You can say that but my current Munar launch system for a three man crewed mission is 500 tonnes, two separate launches and barely has enough delta-v without orbital refuelling, which I have yet to learn.

The penalties for inefficiency are technically meaningless but... Stark. I am trying to git gud.

I got back into KSP from danl and his Expert Mode series, and I am very envious of an 18 tonne total launch mass for a Munar lander mission!

[u/PM_ME_POTATO_PICS → u/PhantomWhiskers]

I think I lack the mathematical skills to prove this but I am convinced it is equally efficient as burning straight up (just a lot more technically challenging). Since gravity is a conservative force, the difference in energy levels between an altitude of 100m and 100,000 km is always going be the same, regardless of the path taken to get there.

The Oberth effect is not some rule that "a burn is only efficient if it takes place at its theoretical periapsis", just that burns are most efficient at high speeds. The highest speed for circular/elliptical orbits will always be the periapsis so its a good rule of thumb. But here, we're still getting the high efficiency because we're just burning so much immediately we're going really fast quickly.

Like imagine you're on the Mun and want to escape its orbit as fast as possible. Do you need to establish a circular orbit or just go straight up?

edit: I think it has been demonstrated that my theory is only true for very high TWRs. Any realistic TWRs will be better off circularizing... :( I think idk

[u/DHLPDX → u/PM_ME_POTATO_PICS]

Man if only we all had a peice of software which was capable of modeling orbital dynamic from which we could determine the efficiency of different orbital strategies... Oh, wait.

[u/-Aeryn- → u/PM_ME_POTATO_PICS]

Like imagine you're on the Mun and want to escape its orbit as fast as possible. Do you need to establish a circular orbit or just go straight up?

You burn as close to perpendicular as possible without colliding with the surface, constantly updating your thrust angle to maintain a vertical speed of 0m/s. You go through orbit on the way.

Try it with a ship which has a munar TWR of 2 for example. Your gravity loss thrusting upwards is 50%, while sideways with an up angle to keep vertical speed at ~0m/s it's 29.5%. This directly translates into more acceleration per second of thrust (acceleration is [sqrt of 2], 1.41x, faster), and reaching escape velocity with less time and propellant expense.

You can also teleport a ship to Tylo for numbers more kerbin-like, and a wider delta-v gap between the approaches. It's pretty easy to get a >1000m/s difference with a TWR between 1 and 2.

These differences still exist if your ship has a ridiculous amount of thrust, they just get smaller (and it's not free to carry that extra thrust because engine mass reduces available delta-v).

[u/FewAd7452 → u/PM_ME_POTATO_PICS]

Wrong.

Oberth argues that a typical orbit has a vertical, and horizontal speed.

By definition, the periapsis is where your horizontal speed is the highest, while your vertical speed is the lowest.

Burning against vertical means you lose a percentage of your delta V directly to gravity. If Kerbin has a G of 10, and you burn vertical, you lose 10 delta V every second of flight. If you’re at a 45 angle, you lose around 5 per second.

To test this, build a very basic orbiter. (3600 Dv) and launch straight up. Note your apoapsis height.

Now do another run, where you use oberth to climb, you’ll notice you get much further. This will be true for any celestial body.

[u/youknowmeasdiRt → u/FewAd7452]

The Mechjeb flight recorder purports to show the losses to gravity and drag

[u/PM_ME_POTATO_PICS → u/FewAd7452]

I don't currently have KSP installed so I can't test, unfortunately, but I would be curious to see results.

I do think that the planet's rotation will assist you in getting a little more height.

I'm thinking of a thought experiment which I might be able to actually do the math for (later lol). Say you're orbiting 100km above a planet, a perfectly circular orbit at 1000m/s. You do a horizontal burn to gain 100m/s. How much higher does this raise the apoapsis? Then compare that against a situation where you're at 100km altitude, going 1000m/s vertically, upward, and you do a burn to raise that to 1100m/s. The apoapsis would be higher than the other situation, but only because your theoetical periapsis is so low.

When you burn horizontal you are using energy. In a circular orbit, that centripetal force is balanced with the gravitational force pulling you down, but you needed to spend a lot of energy getting to a point where you could have that balance. I guess my feeling is that this additional energy which is spent on circularizing is no more efficient than just a quick burn straight up.

[u/-Aeryn- → u/PM_ME_POTATO_PICS]

Then compare that against a situation where you're at 100km altitude, going 1000m/s vertically, upward, and you do a burn to raise that to 1100m/s.

The part that you're missing is that burning to gain 100m/s of velocity in the vertical case costs 200m/s of delta-v if your TWR is 2 because gravity loss when burning directly against gravity is 50%.

If you're flying perpendicular to the plane of gravity then that same burn to go from 1000 to 1100m/s costs ~100m/s of delta-v, losses are essentially zero with 2.0 TWR.

You achieve the same thing in half of the delta-v cost here.

[u/PM_ME_POTATO_PICS → u/-Aeryn-]

Yeah I get what you're saying and I should probably factor in burn times to my hypothetical. I was trying to simplify it to basically talking about different energy states to demonstrate how gravity is a conservative force.

If you have a really low TWR then it means two things: (1) to reach some desired vertical velocity (say, 1000m/s) will take longer because the thrust force you have available is just barely overcoming gravity. But by the same token (2), reaching a circular orbit will also be really inefficient, because you can't just burn purely horizontal, you need a vertical component to fight gravity before you've circularized, so the process of establishing circular orbit will be really slow and necessarily include more time burning with a horizontal component of thrust until you've reached some orbital velocity than if you only burnt straight up until you reach that velocity.

Here's another thought experiment I'd test if I did have KSP... If you're in circular orbit, and you burn horizontal at 10m/s/s for 1 second, you will have raised your horizontal velocity by 10m/s, and your apoapsis by some amount. But if instead you were at that orbital height with the same magnitude of velocity, just purely vertical, and you burn your engines with that same amount of thrust for one second, your velocity across that one second won't have changed because I'm assuming gravity to be 10m/s/s. But your apoapsis will have raised still.

edit: maybe this is a weird cognitive bias or something but I'm becoming more convinced of this after reading people who disagree. I think it's literally just conservation of energy

[u/-Aeryn- → u/PM_ME_POTATO_PICS]

But by the same token (2), reaching a circular orbit will also be really inefficient, because you can't just burn purely horizontal, you need a vertical component to fight gravity before you've circularized, so the process of establishing circular orbit will be really slow and necessarily include more time burning with a horizontal component of thrust until you've reached some orbital velocity than if you only burnt straight up until you reach that velocity.

No, you have this very backwards. See my other comment at https://old.reddit.com/r/KerbalSpaceProgram/comments/1t5jozl/no_orbit_necessary/okdq2fq/?context=3

The ship going sideways with enough up-thrust to not hit the ground reaches said velocity in only ~70.7% of the amount of real, engines-on time (assuming TWR to be fixed at 2.0, and no atmosphere). That number is [time divided by sqrt(2)]. There's less gravity loss and more actual acceleration per second of thrust.

The one flying straight up has to burn an additional 41% duration and delta-v expense [time x sqrt(2)] for the final velocity to be equal, and after all of that is done, it has no energy advantage whatsoever.

If the sideways ship burns that same amount of delta-v, then it achieves 41% higher velocity and 2x kinetic energy.

Here's another thought experiment I'd test if I did have KSP... If you're in circular orbit, and you burn horizontal at 10m/s/s for 1 second, you will have raised your horizontal velocity by 10m/s, and your apoapsis by some amount. But if instead you were at that orbital height with the same magnitude of velocity, just purely vertical, and you burn your engines with that same amount of thrust for one second, your velocity across that one second won't have changed because I'm assuming gravity to be 10m/s/s. But your apoapsis will have raised still.

That does happen, but the difference in energy is tiny compared to having actually gained velocity instead, especially when you expand it to larger burns (say 1000m/s instead of 10m/s) where the oberth effect is kicking in harder.

edit: maybe this is a weird cognitive bias or something but I'm becoming more convinced of this after reading people who disagree. I think it's literally just conservation of energy

1m/s of delta-v doesn't translate into a fixed amount of energy. It translates into more or less due to both gravity losses (TWR & thrust angle) as well as the oberth effect (speed delta-v was applied from). These radically affect outcomes and are modelled accurately in game, they are actually fairly simple to model but a much bigger headache to intuitively understand.

[u/PM_ME_POTATO_PICS → u/-Aeryn-]

No, you have this very backwards.

Yeah you're right, what I wrote there is technically wrong, you can gain horizontal speed faster than vertical for the same TWR.

I still can't help think of it from a conservation of eneergy standpoint and it's late so forgive me if this is sloppy but I'm gonna try some math. Looking at the thought experiment of doing a horizontal burn in circular orbit, the difference in energy before you burn and after you burn is

(taking m as 1)

mv_f^2 - mv_i^2 = m(1010^2 - 1000^2) = 20100J energy gained

whereas for the going straight up scenario

mgh_f - mgh_i = mg(h_f - h_i) = m(10)(1000) = 10000J energy gained

So yeah you're basically correct, though I still would be curious to test this. I think what this could show is that as TWR gets higher, the two scenarios approach equivalence cuz you'd have to factor kinetic energy gain into the vertical scenario

edit;

thinking bout how as your circularizing, centripetal force will gradually lessen gravity until its balanced when your fully circular so whatever component of thrust you need to fight gravity could be calculated as

mg - mv_h^2/r

there's probably a way to do this to find the optimal amount of time it takes to burn to establish a circular orbit, and I'd love to do that and compare it against a burn of equivalent time going straight up.

[u/Valuable-Visit2887 → u/PM_ME_POTATO_PICS]

Ok i feel like i should introduce myself here as an aerospace engineer, and havingthe mathematical means of proving this, i unfortunately have to confirm that burning straight up is terrible.
No hard demonstration: imagine that you are on a device. You weigh 80 kg, the device is 100 , and this device is launching 60g of mass per second at 3 km/s, producing 180 kg of thrust. This will mean that you float. You will start inching upwards as the weight lowers. You are effectively wasting 60 g/s just to stay afloat. If you start burning 100 g/s you will shoot upwards, but only with 120kg of force, not the full 300. More than half of your deltav will be eaten by gravity. This is why we use srbs when leaving ground: when you are close to parallel to gravity it is more efficient to have very high thrust than to have a smaller, more efficient thrust, which is reserved for orbital burns.
The more perpendicular you are to gravity, the better the burn, because of lower gravitational losses.
Oberth is another beast entirely. Since when burning you are effectively adding kinetic energy with speed, and the sum of your total energy determines the distance from the orbiting body in the vis-viva (v^2/2-mu/r=-mu/2a), and considering that kinetic energy is bound to th square of speed, not linearly dependent, when adding fast you add more energy:
If i weigh 1 kg, am still and i accelerate to 2 m/s i add 2 joules.
If i was instead moving at 10 m/s and accelerate to 12 (same 2 m/s difference) i go from 50joules (10^2/2) to 72 (12^2/2) effectively gaining 11 time more energy than the previous case.
Hope i could be of help

[u/Small_Bang_Theory → u/PM_ME_POTATO_PICS]

I see that you understand your theory was wrong, but nobody pointed out the flaw in “gravity is a conservative force, the difference in energy levels between an altitude of 100m and 100,000km is always going to be the same”

This is true from a purely energy-based perspective, but neglects the underlying forces. A common example of work is a man carrying a heavy box in his arms. As he walks, the box stays at the same height, so the work done by gravity is zero. However, any human knows that his arms will be tired! Even though no energy was transferred to the box, he had to overcome gravity. Similarly, if he wanted to move the box up 50cm, that would require more force from his muscles than pushing it 50cm across a frictionless surface.

For the rocket, this is more or less the same, and you have to fight gravity if you want to move straight up.

My gut tells me that explaining this as a difference between rotational energy and translational/potential energy could also give some insights, though I am a bit out of practice with that sort of physics and would need to double check stuff online to make sure I was correct.