Oberth - Interpretation of additional energy & velocity neutral burns

[AaronLS]

So looking at a formula for calculating the resulting kinetic energy from a change in dV.

Craft mass == m == 1kg

Starting Velocity Scenario A == v_a == 1000 m/s

Starting Velocity Scenario B == v_b == 1100 m/s

If I burn to add 500 m/s of velocity in both scenarios, I would calculate the change in energy as

Delta Energy A == ( (1kg * 1500 m/s)² - (1kg * 1000 m/s)² ) /2 == 1250000 J

Delta Energy B == ( (1kg * 1600 m/s)² - (1kg * 1100 m/s)² ) /2 == 1350000 J

Assertions:

1) In both scenarios, the amount of fuel consumed to obtain a 500 m/s change in velocity is identical.

2) In the second scenario, I've gained more kinetic energy for the same amount of fuel consumption.

3) Conservation of energy is not violated. In the first scenario, 100 kJ more kinetic energy was imparted to exhaust than the second. So the same amount of kinetic energy was produced, just a different portion was divided among exhaust and ship.

If I've misunderstood anything in my assertions, please let me know.

Now how do I interpret extra kinetic energy?

This looks like only a small difference, but it seems like the inverse square law of gravity to distance allows it to have a huge affect. As I move away from the planet, my kinetic energy is being converted to gravitational potential energy. So I can essentially I can go further. 1250 kJ takes me the same distance in both cases, but in the first that will be the apoapsis(assuming non-escape velocity, and a infinitely thin ellipsis such that I am 0 m/s at apoapsis) and in the second I have 100 kJ left over and will be able to go much further. Since gravity is inverse square to distance, that small 100 kj of extra kinetic energy will translate to a huge amount of extra distance, correct? My velocity would be much lower at that point, so I'd be covering less distance over time, but I would speculate that the inverse square of gravity means that it will take exponentially longer for that kinetic energy to be converted to gravitational potential energy.

I could determine my velocity when 100 kJ is left by solving for velocity:

100000 J == (1kg * v)² / 2

v = 447 meters/sec

Without calculating exactly how far from the planet I am at that point, I know the influence of gravity is exponentially smaller(maybe wrong phrasing) due to inverse square law. Therefore it seems like with a big chunk of my velocity remaining, I will be able to go much much further.

So in summary, while an extra 10% starting velocity only ended up with an extra 8% J kenetic energy (100 kJ / 1250 kJ), the effect in terms of distance would probably be amplified much more, maybe exponentially. It seems like in practice this is what happens, but not sure if my speculative "why" that happens is on the right track.

Velocity Neutral Burns

I understand why retrograde burns also benefit from this effect, since it is still a change in velocity and so energy change is governed by the formula. But what about a burn that doesn't change velocity such as a radial or normal burn? Mathematically it seems like the oberth affect wouldn't apply, since there is no change in velocity during the burn. (I thought these didn't change velocity, but it seems in practice they do, but I suspect that's only due to the radial/normal stability not keeping me perfectly aligned, correct? A perfectly aligned normal or radial burn should not change velocity?)

[K^2]

1. Your assertions are essentially correct. You just need to keep in mind that exhaust starts out having kinetic energy already. So in the extreme case of the ship moving faster than exhaust velocity, the exhaust is slowed down due to engine operation, and it's kinetic energy is reduced. That means the ship's kinetic energy gain can be higher than chemical energy available in burnt fuel.

2. Because kinetic energy is quadratic in velocity, it's easy to convince yourself with the help of Pythagoras' Theorem that Oberth effect for radial and normal burns is zero. Only pro/retrograde component of dV benefits from it.

[arkie87]

Velocity Neutral Burns

I understand why retrograde burns also benefit from this effect, since it is still a change in velocity and so energy change is governed by the formula. But what about a burn that doesn't change velocity such as a radial or normal burn? Mathematically it seems like the oberth affect wouldn't apply, since there is no change in velocity during the burn. (I thought these didn't change velocity, but it seems in practice they do, but I suspect that's only due to the radial/normal stability not keeping me perfectly aligned, correct? A perfectly aligned normal or radial burn should not change velocity?)

Burning normal or radial does increase velocity magnitude (root sum of squares) but it does not benefit from oberth (as you stated).

If you dont believe me, get into an orbit as far out as minmus and burn normal/anti-normal. You will leave Kerbin

I have two ways i like to think about Oberth effect:

Logically:

(a) since gravity accelerates you back towards the planet at a rate (it kills m/s per second), you want to climb away as fast as possible to minimize the amount of time you spend in the gravitational influence

Mathematically:

( since gravity wells are ENERGY wells rather than (delta) VELOCITY wells, you want to escape with as much energy as possible since kinetic energy is quadratic-- you will end up with more velocity left over.

Some applications where Oberth effect has no bearing:

basically, anywhere you need a change in velocity (and not a change in energy). For instance, landing. Burning at peripasis provides no boost in efficiency. If your surface velocity is 1000 m/s, you need at least 1000 m/s to land. Oberth effect wont help you.

If that makes sense, great. If not, ignore it!

[AaronLS]

Burning normal or radial does increase velocity magnitude (root sum of squares) but it does not benefit from oberth (as you stated).

Ah, if I burn normal, and viewed my velocity as the sum of two vectors: the existing vector pointing prograde, and the vector pointing normal, then my new vector is the previous prograde vector (p), plus the normal vector (n). Of course in practice, as I burn locked normal then my angle is changing, but if assumed that I burned just for a moment normal. This would be a right angle and you would calculate the new vector as (p^2 + n^2)^.5, which means that very little of the burn goes towards increasing the magnitude.

If p == 1000, and I perform the same burn norm(without tracking normal since that would make things complicated) that would be 500 m/s in the normal direction:

(1000^2 + 500^2)^.5 == 1118

Delta Energy A == ( (1kg * 1118 m/s)² - (1kg * 1000 m/s)² ) /2 == 124632 J

If instead initial p == 1100

(1100^2 + 500^2)^.5 == 1208

Delta Energy B == ( (1kg * 1208 m/s)² - (1kg * 1100 m/s)² ) /2 == 124632 J

So yeh, the fact that only a small portion of the normal burn contributes to magnitude, completely cancels the oberth effect. If my initial prograde speed is higher, that means as per Pythagoran thereom, less of of my normal burn contributes to the magnitude. I imagine there's a way formula wise to substitude the pathagorem formula in for 1118 and 1208 and somehow show that they are always equal regardless of what the value of p is, the initial velocity.

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Thanks to both for helping me understand.

[arkie87]

Ah, if I burn normal, and viewed my velocity as the sum of two vectors: the existing vector pointing prograde, and the vector pointing normal, then my new vector is the previous prograde vector (p), plus the normal vector (n). Of course in practice, as I burn locked normal then my angle is changing, but if assumed that I burned just for a moment normal. This would be a right angle and you would calculate the new vector as (p^2 + n^2)^.5, which means that very little of the burn goes towards increasing the magnitude.

If p == 1000, and I perform the same burn norm(without tracking normal since that would make things complicated) that would be 500 m/s in the normal direction:

(1000^2 + 500^2)^.5 == 1118

Delta Energy A == ( (1kg * 1118 m/s)² - (1kg * 1000 m/s)² ) /2 == 124632 J

If instead initial p == 1100

(1100^2 + 500^2)^.5 == 1208

Delta Energy B == ( (1kg * 1208 m/s)² - (1kg * 1100 m/s)² ) /2 == 124632 J

So yeh, the fact that only a small portion of the normal burn contributes to magnitude, completely cancels the oberth effect. If my initial prograde speed is higher, that means as per Pythagoran thereom, less of of my normal burn contributes to the magnitude. I imagine there's a way formula wise to substitude the pathagorem formula in for 1118 and 1208 and somehow show that they are always equal regardless of what the value of p is, the initial velocity.

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Thanks to both for helping me understand.

Interesting.

Specific Kinetic Energy: KE = V^2/2

Velocity magnitude: V = sqrt(p^2+n^2)

Thus:

delta KE = ((p^2+n^2)^0.5)^2/2 - p^2/2 =n^2/2

So yeah, burning normal does not benefit from oberth

[Red Iron Crown]

Now how do I interpret extra kinetic energy?

The greater change in orbital energy for a prograde or retrograde burn is reflected in a greater change in the orbit's semi-major-axis.

Burns which do not change speed (i.e. inclination changes or eccentricity changes) are more efficient when speeds are lower rather than greater.

[Iskierka]

One key thing: energy is dependent on the relative perspective, and in both cases you impart the same energy to the exhaust, as you change only by a certain velocity, and energy exchanges involved in producing force are only considered in the local reference frame. This is why you can appear to, in extremes, gain far more energy than the fuel actually contained - as far as the fuel is concerned, you didn't, as you gained exactly Ve relative to any given bit of expended fuel. The Oberth effect comes from what the local reference frame's effects look like to an outside observer - similar to gravity assists. As far as the planet-orbiter system is concerned, nothing particularly special happened, as they met and left as normal with no net change in energy, it only becomes significant when given an outside perspective.

[K^2]

It's not so much an "outside perspective," as you need an inertial reference frame to consider energy changes. That's why Oberth effect is relevant when you take a massive body as the origin of your coordinate system, rather than the ship that orbits it. From perspective of another orbiting body, energy need not be conserved, as it is an accelerated frame of reference.

[Lukaszenko]

Some applications where Oberth effect has no bearing:

basically, anywhere you need a change in velocity (and not a change in energy). For instance, landing. Burning at peripasis provides no boost in efficiency. If your surface velocity is 1000 m/s, you need at least 1000 m/s to land. Oberth effect wont help you.

I don't get what you're saying with this part. It's clearly better to do a suicide burn when landing (which is as close to periapsis as possible) than to try to kill the speed elsewhere, no?

[GoSlash27]

AaronLS,

You've already hit on how the Oberth effect works, but the gravity well doesn't actually have anything to do with it.

We think of our budget to go places as a "delta vee", but in reality it's a "delta Ek". We think of it as an "effect" because we intuitively think in terms of velocity, but it's not actually an "effect" so much as a discrepancy.

Since (as you point out) adding velocity when you're going fast adds more kinetic energy than when you're going slow, you can generate your required kinetic energy more efficiently at periapsis than you can at apoapsis.

Best,

-Slashy

[arkie87]

I don't get what you're saying with this part. It's clearly better to do a suicide burn when landing (which is as close to periapsis as possible) than to try to kill the speed elsewhere, no?

When escaping any body, burning 100 m/s at periapsis above escape velocity results in more than 100 m/s upon leaving the planet's SoI, due to Oberth effect.

When landing on any body, burning at periapsis does not reduce deltaV to below craft's surface velocity.

Incidentally, a "suicide burn" is not the most efficient way to land. What's more efficient is to drop periapsis low, and burn retrograde at periapsis, adjusting the angle to cancel gravity until all of your horizontal velocity is gone (the inverse of the most efficient way to take off in atmosphereless bodies).

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but the gravity well doesn't actually have anything to do with it.

We think of our budget to go places as a "delta vee", but in reality it's a "delta Ek".

The reason our budget to go places is measured in delta KE and not deltaV is because of the way gravity wells work: they are energy wells and not deltaV wells.

[Red Iron Crown]

Incidentally, a "suicide burn" is not the most efficient way to land. What's more efficient is to drop periapsis low, and burn retrograde at periapsis, adjusting the angle to cancel gravity until all of your horizontal velocity is gone (the inverse of the most efficient way to take off in atmosphereless bodies).

You're describing a low orbit suicide burn.

[arkie87]

You're describing a low orbit suicide burn.

I might be wrong, but a "suicide" burn is burning vertically as a craft rapidly descends vertically, such that the craft is at altitude 0 m when craft velocity hits 0 m/s.

It's not a suicide burn if you are burning horizontally (at least, i dont see why it could be "suicidal")

[Red Iron Crown]

I might be wrong, but a "suicide" burn is burning vertically as a craft rapidly descends vertically, such that the craft is at altitude 0 m when craft velocity hits 0 m/s.

It's not a suicide burn if you are burning horizontally (at least, i dont see why it could be "suicidal")

I consider a suicide burn to be one continuous maximum-thrust burn almost until touchdown. There will always be both a horizontal and vertical component to the velocity to be killed, the case you describe is on the end of the spectrum with more horizontal and less vertical.

[arkie87]

I consider a suicide burn to be one continuous maximum-thrust burn almost until touchdown. There will always be both a horizontal and vertical component to the velocity to be killed, the case you describe is on the end of the spectrum with more horizontal and less vertical.

I understand why vertical suicide burns are referred to as such since a slight miscalculation in distance will result in slamming into the ground and instant death.

Horizontal "suicide" burns less so, unless an unexpected tall mountain comes up over the horizon.

Edited May 20, 2015 by arkie87