# Why can I achieve higher altitudes when flying in Orbit than straight up?

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Hey guys!

So today I was playing some KSP and I built a rocket which is meant to orbit Kerbin at a low altitude first and then increase the Apoapsis up to 1 000 000 km (So I can get some sweet science from low and high Orbit). After having achieved the 1 mil km I did a stupid mistake and I had to revert to start.This time, I decided to go for a more lazy approach. I decided not to achieve Orbit and just fly straight up. Now something happened which doesnt make any sense in my current understanding of the world. I was only able to achieve about 400 000 km - less than half the amount I was able to reach when flying in orbit.

I dont know how this can make any sense: In my first flight I was spending way more time in atmosphere, why I should have lost more speed than in my second attempt. Also I just dont understand how flying vertically results in much lower altitudes than flying horizontally "only".

I would be very happy if someone could explain the mathematics behind that. TY very much in advance!

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Two words: gravity losses. When you fly straight up, you're always burning against the force of gravity, which means a good part of the force generated by your engine gets cancelled out by gravity. Think of it like trying to swim upstream in a fast-moving river - you're going to have to exert a lot of force just to not get swept downstream. When you're flying to orbit, however, gravity losses are far less significant, since you spend most of your time burning perpendicular or near-perpendicular to the force of gravity. The end result of this is that, even though it seems like you need a lot more delta-V to fly to orbit and then raise your apoapsis, it ends up effectively cheaper due to all the gravity losses you avoid in the process.

The difference is exacerbated, of course, by KSP's toy solar system, since it's really easy to get into orbit in the first place. But that's a discussion for another time.

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28 minutes ago, Karol van Kermin said:

Hey guys!

So today I was playing some KSP and I built a rocket which is meant to orbit Kerbin at a low altitude first and then increase the Apoapsis up to 1 000 000 km (So I can get some sweet science from low and high Orbit). After having achieved the 1 mil km I did a stupid mistake and I had to revert to start.This time, I decided to go for a more lazy approach. I decided not to achieve Orbit and just fly straight up. Now something happened which doesnt make any sense in my current understanding of the world. I was only able to achieve about 400 000 km - less than half the amount I was able to reach when flying in orbit.

I dont know how this can make any sense: In my first flight I was spending way more time in atmosphere, why I should have lost more speed than in my second attempt. Also I just dont understand how flying vertically results in much lower altitudes than flying horizontally "only".

I would be very happy if someone could explain the mathematics behind that. TY very much in advance!

AWESOME question.

Here's an experiment. Go up to the roof of a tall building with at tennis ball and throw it straight up as hard as you can. Watch how far it goes. Then, try flinging it horizontally off the roof. Watch how far it goes.

Assuming your building was sufficiently tall, the ball will go much farther (in terms of total distance traveled) when you throw it off the side of the building. Why? Well, obviously, it has farther to fall.

What does this have to do with space? Well, if a rocket burns straight up into space, it will expend its fuel and then fall back down to the ground. However, if a rocket turns and burns horizontally, it will "fall" quite some distance from where it launched. If this distance is a substantial fraction of Earth's circumference, then the "surface" relative to the rocket's highest point will be substantially lower (from its point of view) than where it launched from.

If you burn horizontally fast enough, then you will miss the Earth completely. Earth's gravity turns as you fly, keeping you pulled toward the center of the Earth, which means you can continue to "fall" forever without hitting the ground, your velocity balanced against the centripetal pull of gravity. That's an orbit.

The first time you flew the mission, you turned horizontally and reached orbit pretty soon. Since you were no longer fighting gravity, you were able to use all your remaining fuel to add energy to your orbit, which raised your apoapse considerably. The second time, you burned all your fuel fighting gravity and so you didn't get nearly as high.

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Suppose your space vehicle has a TWR of 1.5. It can accelerate straight up at 0.5g as 1g goes into just hovering.

Now burn at an angle, with 1g up to cancel gravity and the rest on a horizontal vector sideways. By Pythagoras you accelerate at 1.1g sideways. (1.1g horizontal, 1g vertical, 1.5g hypotenuse).

This is *much* better than trying to burn directly upwards!

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My understanding of this is that by flying at an angle we use gravity (slingshot) to burn less fuel.

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Another useful analogy:

Let's say you're on a bicycle and you're stopped on the side of a gigantic, smoothly sloping hill. Suppose you try to pedal as hard as you can for 10 seconds. Which direction will get you moving faster -- pedaling straight up the hill, or pedaling sideways around the circumference of the hill?

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19 hours ago, Cassel said:

My understanding of this is that by flying at an angle we use gravity (slingshot) to burn less fuel.

No, this isn't what a gravity slingshot is. A gravity slingshot is when you use a gravity well flyby to borrow velocity from the body at the centre of the gravity well. Relative to the planet you haven't gained or lost any speed, but in the context of the global reference frame you've added or subtracted some velocity due to the motion of the celestial body, accompanied by a course change.

------------------------------------------

What we're discussing is minimising gravity losses on take off. If you have a constant TWR of 5/3 then you can burn upwards but you lose 1g to gravity and so get an acceleration of only 2/3g.

If instead you burn sideways you hover at 1g upwards with 5/3g on the hypotenuse and a 4/3g horizontal component.

You accelerate twice as fast horizontally as you could vertically and because the earth is a ball the surface will eventually fall away from you so that that horizontal velocity becomes vertical anyway. So for the the same fuel, engine and throttle settings, you'll minimise gravity losses by burning horizontally.

In KSP you'll often see players turn horizontally immediately on airless bodies. This is why. As long as they sustain just enough vertical thrust to prevent them from crashing back to the surface the most efficient use of thrust is to burn sideways. Often there are other practical considerations that mean you want to fly a more lofted trajectory - terrain avoidance and atmospheric drag.

Edited by RCgothic

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The Oberth effect may come into play to some degree here too. But the Oberth effect might as well be black magic to me, so I'm not sure.

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7 minutes ago, Mad Rocket Scientist said:

The Oberth effect may come into play to some degree here too. But the Oberth effect might as well be black magic to me, so I'm not sure.

It does not. Oberth doesn't ask where you are, but how fast you're going. Oberth helps less during launch than it does in orbit.

No the whole argument for going sideways at launch is to avoid having to counter gravity's pull, and that only works because planets are spheres so if you go sideways fast enough, you can set it up so you never fall back to the ground.

Edited by 5thHorseman

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1 hour ago, 5thHorseman said:

It does not. Oberth doesn't ask where you are, but how fast you're going. Oberth helps less during launch than it does in orbit.

No the whole argument for going sideways at launch is to avoid having to counter gravity's pull, and that only works because planets are spheres so if you go sideways fast enough, you can set it up so you never fall back to the ground.

Well, fighting gravity less means you get more actual delta-v, which increases efficiency of the propellant's energy.

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15 hours ago, Mad Rocket Scientist said:

But the Oberth effect might as well be black magic to me

I'm going to commit a travesty of physics by oversimplifying this explanation:

Kinetic Energy = K = 0.5*mass*velocity squared = 0.5*m*v^2

Lets say we have an orbit where the speed is 2200 m/s. This is roughly a circular Low Kerbin Orbit.

v1 = 2200 m/s, so K1 = 0.5*m*2200^2 = m * 2,420,000

Now we have another orbit where the speed is 2000 m/s. This is a slightly higher, but still fairly-Low Kerbin Orbit.

v2 = 2000m/s, so K2 = 0.5*m*2000^2 = m * 2,000,000

What happens if we do a 100m/s burn at either of these two points? (We'll play with infinite fuel on, and say that m doesn't decrease; it would decrease by the same amount in both examples anyway.)

v1' = 2200 + 100 = 2300, K1' = m * 2,645,000

v2' = 2000 + 100 = 2100, K2' = m * 2,205,000

K1' - K1 = delta K1 =  m * 2,645,000 - m * 2,420,000 = m * 225,000

K2' - K2 = delta K2 = m * 2,205,000 - m * 2,000,000 = m * 205,000

delta K1 / delta K2 = 1.098

Same magnitude of burn, same delta-v expended, but you get almost 10% more energy from the burn. Is it free? No, because you're changing the energy of the fuel you're blowing out the back by more as well.

Since orbits are really defined by how much energy they have, you can go "farther" for the same amount of fuel, by burning lower.

Hopefully that takes some of the black magic out of it.

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17 hours ago, Mad Rocket Scientist said:

The Oberth effect may come into play to some degree here too. But the Oberth effect might as well be black magic to me, so I'm not sure.

Good example by @FleshJeb above.

Another, less mathy way of looking at it:

Suppose you do a burn out in empty heliocentric space. You're pushing fuel out the back of your rocket, which slows the fuel down (relative to the sun) and speeds you up. You're throwing something in one direction and it pushes you in the other direction. It's an exchange of momentum. Makes sense so far, right?

Now, suppose you do a burn deep in a planet's gravity well. This time, you're still doing an exchange of momentum (pushing the fuel in one direction which pushes you in the other direction), but this time, the fuel you're pushing away has much more momentum to begin with because it's been sped up by the gravity well. When you push it away, you're pushing against something that is, effectively, "heavier"; as a result, it gives you more of a kick, relative to the original heliocentric orbit.

This doesn't violate conservation of momentum, because the momentum left behind is (essentially) added to the gravity of the planet.

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1 hour ago, FleshJeb said:

I'm going to commit a travesty of physics by oversimplifying this explanation:

Kinetic Energy = K = 0.5*mass*velocity squared = 0.5*m*v^2

Lets say we have an orbit where the speed is 2200 m/s. This is roughly a circular Low Kerbin Orbit.

v1 = 2200 m/s, so K1 = 0.5*m*2200^2 = m * 2,420,000

Now we have another orbit where the speed is 2000 m/s. This is a slightly higher, but still fairly-Low Kerbin Orbit.﻿﻿﻿

v2 = 2000m/s, so K2 = 0.5*m*2000^2 = m * 2,000,000

What happens if we do a 100m/s burn at either of these two points? (We'll play with infinite fuel on, and say that m doesn﻿'t dec﻿rea﻿s﻿e; it wo﻿uld decrease by the sa﻿m﻿e amount in both examples anyway.)

v1' = 2200 + 100 = 2300, K1' = m * 2,645,000

v2' = 2000 + 100 = 2100, K2' = m * 2,205,000

K1' - K1 = delta K1 =  m * 2,645,000 - m * 2,420,000 = m * 225,000

K2' - K2 = delta K2 = m * 2,205,000 - m * 2,000,000 = m * 205,000

delta K1 / delta K2 = 1.098

Same magnitude of burn, same delta-v expended, but you get almost 10% more energy from the burn. Is it free? No, because you're changing the energy of the fuel you're blowing out the back by more as well.

Since orbits are really defined by how much energy they have, you can go "farther" for the same amount of fuel, by burning lower.﻿﻿

Hopefully that takes some of the black magic out of it.

32 minutes ago, sevenperforce said:

Good example by @FleshJeb above.

Another, less mathy way of looking at it:

Suppose you do a burn out in empty heliocentric space. You're pushing fuel out the back of your rocket, which slows the fuel down (relative to the sun) and speeds you up. You're throwing something in one direction and it pushes you in the other direction. It's an exchange of momentum. Makes sense so far, right?

Now, suppose you do a burn deep in a planet's gravity well. This time, you're still doing an exchange of momentum (pushing the fuel in one direction which pushes you in the other direction), but this time, the fuel you're pushing away has much more momentum to begin with because it's been sped up by the gravity well. When you push it away, you're pushing against something th﻿at is, effectively, "heavier"; as a result, it gives you more of a kick, relative to the original heliocentric orbit.

This doesn't violate conservation of momentum, because the momentum left behind is (essentially) added to the gravity of the planet.﻿﻿﻿﻿﻿

These are both great explanations, thanks. I was thinking it somehow increased the amount of Delta-V you got from the same mass of propellant.

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15 minutes ago, Mad Rocket Scientist said:

These are both great explanations, thanks. I was thinking it somehow increased the amount of Delta-V you got from the same mass of propellant.

That's actually almost correct. You DO get more work out of the same mass of fuel. The only reason it's not correct is that Delta-V is specifically designed as a "mass-agnostic" term that provides an versatile benchmark for comparing what you can do with various spacecraft. It still matters how and where you apply it. Much like ISP is a generic benchmark for comparing engines, but TWR is an additional factor to consider in design and piloting.

I don't have a perfect understanding of this myself, and I hope @sevenperforce will correct anything I've gotten wrong.

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1 hour ago, FleshJeb said:

That's actually almost correct. You DO get more work out of the same mass of fuel. The only reason it's not correct is that Delta-V is specifically designed as a "mass-agnostic" term that provides an versatile benchmark for comparing what you can do with various spacecraft. It still matters how and where you apply it. Much like ISP is a generic benchmark for comparing engines, but TWR is an additional factor to consider in design and piloting.

I don't have a perfect understanding of this myself, and I hope @sevenperforce will correct anything I've gotten wrong.

@Mad Rocket Scientist So here's the kicker.

Adding 100 m/s of dV adds 100 m/s of dV. That doesn't change. 100 m/s of dV is the same instantaneous change in velocity regardless of where it is applied, at that moment.

However, if you add 100 m/s of dV to your velocity while your fuel is moving very very fast, then you end up leaving with far more velocity than if you added that 100 m/s while your fuel is moving relatively slow, because fuel that is moving fast has a greater amount of inertia/momentum/impulse than fuel that is moving slow.

That's the way to think about it, conceptually. The math is...the math. The math works out on its own.

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10 hours ago, sevenperforce said:

Even in plane changes ?

I guess the answer is around and about gravity losses and near-Oberth effect, but I don't know.

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53 minutes ago, YNM said:

Even in plane changes ?

Yes, because velocity is a vector. delta-V isn't change in speed. It's the magnitude of difference in velocities.

11 hours ago, sevenperforce said:

However, if you add 100 m/s of dV to your velocity while your fuel is moving very very fast, then you end up leaving with far more velocity than if you added that 100 m/s while your fuel is moving relatively slow

I believe, you mean energy here. This can mean more velocity elsewhere in the orbit, but instantaneous change will be exactly the same dV, just as you explain above.

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17 hours ago, FleshJeb said:

No, because you're changing the energy of the fuel you're blowing out the back by more as well.

Although you've shown me the math behind this, I still cant wrap my head entirely around it.

Since the thrust is in both cases the same, the delta V between the fuel and the spacecraft should be the same. How does a faster speed of the fuel and the spacecraft relative to the spectator have more energy, even though the delta V stays the same.

I know that the math makes sense, but I just simply dont understand why it makes sense.

EDIT 1:

Also, after having thought about it, I came to the realization that you need more energy to accelerate an already moving object than a resting object, even in a perfect frictionless enviroment.

Edited by Karol van Kermin

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59 minutes ago, Karol van Kermin said:

Also, after having thought about it, I came to the realization that you need more energy to accelerate an already moving object than a resting object, even in a perfect frictionless enviroment.

You're basically on track for resolving your conceptualization problem. Same amount of fuel burned by the same rocket should give it the same change in velocity. Yet rocket gains more energy if it goes from 100m/s to 200m/s than if it goes from rest to 100m/s. Where does extra energy for this come from? Well, how does kinetic energy of fuel/exhaust change in both of these scenarios?

Edited by K^2

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3 hours ago, K^2 said:

Yes, because velocity is a vector.

... but not "speed", energy or (magnitude of) angular momentum.

Edited by YNM

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Work is force times distance. If you're going faster you apply the force over a greater distance and so do more work.

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17 hours ago, sevenperforce said:

@Mad Rocket Scientist So here's the kicker.

Adding 100 m/s of dV adds 100 m/s of dV. That doesn't change. 100 m/s of dV is the same instantaneous change in velocity regardless of where it is applied, at that moment.

However, if you add 100 m/s of dV to your velocity while your fuel is moving very very fast, then you end up leaving with far more velocity than if you added that 100 m/s while your fuel is moving relatively slow, because fuel that is moving fast has a greater amount of inertia/momentum/impulse than fuel that is moving slow.

That's the way to think about it, conceptually. The math is...the math. The math works out on its own.

So it seems like the Oberth effect would come into play somewhat in this situation.

Although it's not the main reason orbits are more efficient than going straight up, orbits get you going sideways (and thus faster) sooner, providing a small compounding bonus relative to going straight up thanks to the Oberth effect.

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One thing to consider is how the vectors line up.  Thrust produces a force prograde to your rocket.  Gravity produces a force equal to your weight towards the CoM of the planet beneath you.

For planets with an atmosphere or cases where your TWR is relatively low, you have to start pretty much straight up.  It should be pretty clear that for normal (nowhere near suborbital flight), any model rocket launched straight up will go higher than one launched at an angle.  It is only when the sum of the vectors start to "miss the planet" that going sideways really begins to help (that and the Oberth effect thanks to the larger overall magnitude of the combined force).  For something like Minmus, I think you can use wheeled takeoff and landing on the "seas" for optimum direction (just make sure you miss any slopes/mountain ranges).

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13 hours ago, YNM said:

... but not "speed", energy or (magnitude of) angular momentum.

So? We're just talking about the fact that dV applied as an impulse is always the same change in velocity. Velocity being a vector is the only bit that applies. How all other quantities transform is up to details. Inclination change produces zero change in energy. But it's still the same velocity change for the dV applied. Whereas prograde burn will increase energy.

Angular momentum is a lot more like velocity here. An inclination change produces the vector change in angular momentum that has the same magnitude as increase in angular momentum due to prograde burn. In fact, dL/m = r x dV, so only a radial burn will produce no change in angular momentum. (But is still the same change in velocity!)

Edited by K^2

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3 hours ago, K^2 said:

We're﻿ just talking about the fact that dV applied as an impulse is ﻿always the same change in velocity.﻿

True. That's how you calculate the required "dV" for plane changes.

But before and after you still have the same energy.

Tying dV to energy is futile. And so is tying reaction engines with power output.

Edited by YNM

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