• 0

# orbital insertion economics 2: more complicated

## Question

Posted (edited)

ok, so we all know that if you want to insert in a low orbit to land somewhere, you're better off entering the body's SOI with a periapsis close to it, to use oberth effect. and also so your periapsis will already be low so you can circularize the orbit in one single burn.

but! assume we instead actually want to reach a high orbit. because we are placing a satellite contract, or because we want to rendez-vous with a space station in high orbit. we don't want to land or to move lower. so you have two options:

a) enter with high periapsis, then circularize. one single manuever, you're already there

b) enter with low periapsis, put your apoapsis on intended orbit, then raise your periapsis. it takes two manuever, but you get to use oberth effect to save fuel on the insertion.

things are even more complicated if you must rendez-vous with a space station: unless you timed your trip perfectly to circularize orbit already on intercept, you will need to get to a parking orbit first, and from there intercept the station.

so you can

c) enter with a high periapsis, don't circularize but stay in an elliptical orbit, just short of leaving the SoI. circularize when this would bring you on an intercept with the station. it's harder to do so on an elliptical orbit, but doable. you also can fix small errors to your inclination on a high orbit, where it's cheap to do so.

d) enter with a low periapsis, leave periapsis intercepting the space station. raise your periapsis when it's time.

again, the second option requires to move around the orbit more, as you need an additional manuever (technically you don't; but in high orbit raising/lowering orbit is very cheap, so case c lowering the ellipse to an intercept is going to be cheaper, and the deltaV for circularizing at intercept is also going to be cheaper). but again, option d saves fuel on intercept due to oberth effect. and orbit insertion is the most expensive manuever, so saving fuel there matters more.

what do you think, in the two case, would be the most efficient manuever?

i would assume it depends (higher speed of orbital insertion, it's more important to save on insertion with oberth effect. higher mass of the celestial body, changing orbit around it is more expensive, it is more important to save fuel on ohmann transfer), but in this case, what would be some useful parameters to go by? how much would be the difference anyway?

Edited by king of nowhere

## Recommended Posts

• 0

I think the cost of having to raise your Pe back up again is going to outweigh any savings from Oberth. I would choose a) for the first choice.

For the second situation I would choose a variation of c) for the same reason. Ideally, come in on the same plane as the station and set your Pe on its orbit. Then when you do your capture burn you could set your Ap so that you intercept the station the next time you swing back down to Pe. You don't really have to calculate to pull it off, just set the station as the target and burn until the two position markers come together. There's no reason you have to do it that way, but it feels like the most elegant solution.

##### Share on other sites
• 0
36 minutes ago, Grogs said:

For the second situation I would choose a variation of c) for the same reason. Ideally, come in on the same plane as the station and set your Pe on its orbit. Then when you do your capture burn you could set your Ap so that you intercept the station the next time you swing back down to Pe. You don't really have to calculate to pull it off, just set the station as the target and burn until the two position markers come together. There's no reason you have to do it that way, but it feels like the most elegant solution.

yes, that would be not only most elegant, but also most efficient, because all the thrust you use is equalizing your speed to that of the station. doing it with a modified hohmann would require more deltaV to reach the same speed of the station.

i'm not convinced, however, that the saving from oberth effect is always inferior to what you'd spend to move around the orbit. consider for example a moho mission: it takes 2500 m/s for the capture burn. if oberth effect lets you save even just 10% of that, it would fully pay for periapsis raising.

i was considering also the case of the smaller moons, where you only spend very little deltaV to raise your orbit; but in those cases, you also gain less from oberth effect. so, perhaps scenarios b and d are cconvenient only for moho and eeloo.

maybe i will try one of the next days

##### Share on other sites
• 0

I would be very surprised if there was any situation where burning lower and then raising your Pe to the target orbit was better. I strongly suspect that it is always better to enter with a Pe at the target orbit, and then do the circularize in one single burn.

I've been surprised before and expect to be surprised again, so I invite anybody to show proof that I am wrong.

Caveat: This is of course assuming you don't need to modify your orbital tilt. Things can get a bit complicated for that.

##### Share on other sites
• 0

It depends, on how fast your are coming into the SOI, and how high the target orbit is.

You can plan a maneuver to go directly into the target orbit, and compare the delta-V to the resulting orbital velocity.

If the required delta-V is greater than the orbital velocity (coming in very fast and aiming for a high orbit)
then you might save some fuel by doing the capture burn very low, where your fuel is much more effective at removing your excess energy,
and then paying to get most of the orbital velocity back when you are back at the target-orbit height.

There is a concept of a "gate orbit" that comes up when people talk about the reverse problem: what orbital heights are good from which to launch an interplanetary voyage.
The gate-orbit height for any given destination is the dividing line between which of your two options is more efficient when returning from that destination.

##### Share on other sites
• 0
Posted (edited)

I did a quick test and I'm going to have to admit that I'm wrong on this one, at least in a specific case like Moho. I chose a scenario where we're inbound to Moho with a target orbit of 1 million meters (1 Mm). I launched from Kerbin, then had to make a correction burn to get a Moho intercept. I chose 2 different Pe's: 15 km and 1 Mm. In both cases, the capture was set for an Ap of 9 Mm, which is close to the Moho SOI. For the 15 km case, I then added a burn at Ap to raise the Pe up to 1 Mm. At that point, both orbits are equal and any subsequent maneuvering would be, too.

Case 1: Inital Pe @ 15 km: Capture burn is 3004.3 m/s. The burn to raise Pe to 1 Mm is 34.0 m/s. Total dV = 3038.3 m/s.

Case 2: Initial Pe is 1 Mm. Capture burn (and total) is 3504.2 m/s. This is about 466 m/s more than case 1.

Some caveats:

- This is a pretty extreme case dealing with such a fast approach speed and a very high target orbit. I suspect it does get into the gate orbit concept @OHara mentioned and that there's a cutoff Pe where it's more efficient to burn at the target Pe instead of as low as possible. I recall a really old thread where someone had calculated these orbits for each pair of bodies, but I can't find it now.

- In practical terms, you also have to consider that the maneuver calculator assumes an instantaneous burn. Coming in hot to Moho a lot of the burn will be far from Pe unless you have a very high TWR. Still, we're talking a savings of 13% of dV so even a fairly long burn should realize significant savings.

Edited by Grogs

##### Share on other sites
• 0
13 hours ago, king of nowhere said:

what do you think, in the two case, would be the most efficient manuever?

It depends on the insertion altitude.  Escape velocity is dependent on orbital altitude, which means that the inverse of it, what I will call capture-to velocity, is dependent, as well.  This means that you can't simply choose an apsis just above the planetary surface, compare it to an apsis at the extreme edge of the sphere of influence, and assume that there is a linear relationship between the two.  It may work out that an extreme is also an ideal (and, with utmost respect to @Grogs, Moho is an extreme-enough destination that I suspect it works out that way there), but in most cases, using two extremes is only extremely inconclusive; you can't interpolate from that.  We have to go farther into the data.

@OHara has the correct idea with respect to gate orbits; as there is an ideal altitude from which to make a dive to get maximum gain for the Oberth effect on ejection burns, so too, is there an ideal insertion altitude for capture.  This part does require that the periapsis be as close to the planet as possible, but the point is that, assuming that you can start at a given altitude (so this does ignore the cost of launching to that orbit in the first place), the best altitude from which to make the dive is not necessarily at either extreme of the sphere of influence because there is a balance to be struck between the savings of going faster for the ejection and the costs of de-circularising.  Starting closer to minimum periapsis means lower costs for de-circularising but less speed, and thus less Oberth effect, at the ejection burn, but starting closer to the edge of the sphere of influence means spending so much to dive that it eclipses the Oberth effect savings.  Since this is true for the ejection, it must be equally true for the capture, because physically, the two manoeuvres are identical.

Therefore, diving to low altitude and capturing to an orbit with an apoapsis of some specific altitude for purposes of later circularisation can be more efficient, provided that the target apoapsis is close to that ideal altitude.

There was an extremely illuminating post from @OhioBob on this subject five years ago.  I can't explain it better than he can, and he also provided helpful graphs, so here's a link to the post:

##### Share on other sites
• 0
Posted (edited)

As several previous answers have already mentioned, this is the gate orbit effect, but reversed.

TL;DR Is it cheaper to directly circularize at desired altitude? Non-precise answer: No for Moho, slightly for Duna and probably not for other planets.
EDIT: Actual data: Yes for Jool and Eve! No for Dres and Eeloo.

Anyways, let's do some math!

Our 2 strategies are:

1. Brake directly into circular orbit at desired altitude.
2. Brake into elliptical orbit with PE as low as possible and AP at desired altitude then circularize at AP in a second burn.

We'll be making heavy use of the vis-viva equation: v = sqrt(μ*(2 /r - 1/ a))

Define:

• v = Orbital velocity at edge of target SOI, approximately 4,100 m/s for a direct Kerbin -> Moho transfer.
• α = r1 = Lowest possible radius for braking e.g. 15km above Moho's surface = 265,000m
• β = Radius of target SOI, approx 9.6 million meters for Moho
• μ = Standard gravitational parameters for target, 1.69E11 for Moho

Approach 1:

Hyperbolic intercept velocity at desired altitude (r2): vh2 = sqrt(v2 + 2μ * (1 / r2 - 1 / β))
Circular orbital velocity at desired altitude: v2 = sqrt(μ / r2)
Delta-v: vh2 - v2

Approach 2:

Hyperbolic intercept velocity at desired altitude (α = r1) =  vh1 = sqrt(v2 + 2μ * (1 / α - 1 / β))
Hohmann transfer orbit velocity at r1: vt1 = sqrt(2μ * (1 / α - 1 / (α + r2)))
Hohmann transfer orbit velocity at r2: vt2 = sqrt(2μ * (1 / r2 - 1 / (α + r2)))
Circular orbital velocity at desired altitude: v2 = sqrt(μ / r2)
Delta-v = vh1 - vt1 + v2 - vt2

Graph:
x-axis is desired altitude.
y axis is approach 1 delta-v minus approach 2 delta-v: vh2 -vh1 + vt1 + vt2 - 2v2
If y is negative then approach 1 is cheaper.

Moho

Clearly it's always cheaper to use approach 2! At radius of 1,250,000 = altitude of 1,000,000 the figure of ~400 m/s agrees with @Grogs empirical findings.

Duna

This is more interesting. Below a radius of 725000 meters it's more efficient to circularize directly at Duna, although not by much (~10 m/s).

In general the higher the velocity at SOI edge then the lower the altitude at which approach 2 becomes cheaper. Since Duna is the closest planet to Kerbin, other planets will have a higher intercept velocity at SOI edge and hence a lower altitude for the cut-over point.

EDIT:
Jool
The massive SOI means Strategy 1 is cheaper until 225,000,000 meters (past the AP of Pol)

Eve
Strategy 1 is good until around 5,750,000 meters.

Edited by ManEatingApe

##### Share on other sites
• 0
4 hours ago, ManEatingApe said:

As several previous answers have already mentioned, this is the gate orbit effect, but reversed.

TL;DR Is it cheaper to directly circularize at desired altitude? Non-precise answer: No for Moho, slightly for Duna and probably not for other planets.
EDIT: Actual data: Yes for Jool and Eve! No for Dres and Eeloo.

Anyways, let's do some math!@Grogs

wow! it's very rare that i see someone making more math spreadsheet for a game than i do.

and it even turned out to be useful. i was expecting the differences to be small, since the two opposite effects mostly cancel each other, but picking the right insertion can save upwards to 20% deltaV for the trip.

could you put the calculation sheet available for download? i would like to try it for the various moons too.

##### Share on other sites
• 0
45 minutes ago, king of nowhere said:

could you put the calculation sheet available for download? i would like to try it for the various moons too.

There's no sheet, just the formulas mentioned previously, plugged into a grapher.
As best as I can tidy it the combined formula is:

The values for α, β and μ are obtained from the KSP wiki page for each body. The value for "v" I found empirically for each body by running a quick test mission, however you could probably work this out analytically.

##### Share on other sites
• 0
9 hours ago, king of nowhere said:

wow! it's very rare that i see someone making more math spreadsheet for a game than i do.

Math is very important in KSP but we don't do it ourselves so often for 2 reason:

2.There is a mod/tool.

##### Share on other sites
• 0
4 hours ago, Spricigo said:

Math is very important in KSP but we don't do it ourselves so often for 2 reason:

2.There is a mod/tool.

I can't mark a difference between 1 and 2.

13 hours ago, ManEatingApe said:

The values for α, β and μ are obtained from the KSP wiki page for each body. The value for "v" I found empirically for each body by running a quick test mission, however you could probably work this out analytically.

The value that you refer to as v is noted in a lot of places as vinf or v and called the hyperbolic excess velocity.  It results from breaking the transfer injection into two parts:  the first is escape from the origin body, and the second is the remainder of the transfer--this second part is what v is.  Because of its nature, it is specific not only to the planets involved in the transfer, but also to particular transfer windows.  It absolutely can be figured analytically, but that problem is functionally identical to the one solved by a transfer planner.

##### Share on other sites
• 0
30 minutes ago, Zhetaan said:

I can't mark a difference between 1 and 2.

Now that you mention it, I realise 2 is a subset of 1.

##### Share on other sites
• 0
Posted (edited)
20 hours ago, ManEatingApe said:

We'll be making heavy use of the vis-viva equation: v = sqrt(2 /r - 1/ a)

But Vis-Viva is based on the amount of kinetic energy and gravitational potential energy at given points, so doesn't take in to account the Oberth effect does it? Oberth effects the amount of energy you gain from a given quantity of fuel doesn't?  So does a burn from 100 to 200m/s use more fuel than a burn from 500 to 600 m/s?  If so this would push the altitude of the gate orbit up a bit.

Edited by RizzoTheRat

##### Share on other sites
• 0
Posted (edited)
1 hour ago, RizzoTheRat said:

But Vis-Viva is based on the amount of kinetic energy and gravitational potential energy at given points, so doesn't take in to account the Oberth effect does it?

Why not? The Oberth Effect is nothing more than a result of how much kinetic energy the craft have at the moment it executes a maneuver.

Edited by Spricigo

##### Share on other sites
• 0
2 hours ago, RizzoTheRat said:

But Vis-Viva is based on the amount of kinetic energy and gravitational potential energy at given points, so doesn't take in to account the Oberth effect does it?

The vis-viva equation emerges from the interplay of potential and kinetic energy in a gravity well and covers anything in any Keplerian orbit.  Incidentally, the gravitational parameter is missing from the vis-viva equation that you quoted; it is not quite correct.  The Oberth effect has to do with the kinetic energy of a vessel at speed but does not necessarily require an orbit.  Insofar as they both relate to kinetic energy, they have a relationship with one another.

2 hours ago, RizzoTheRat said:

Oberth effects the amount of energy you gain from a given quantity of fuel doesn't?  So does a burn from 100 to 200m/s use more fuel than a burn from 500 to 600 m/s?

Let me try a somewhat different illustration for this.  Let's say that you are in a circular 100 km orbit about Kerbin, moving at 2,246 m/s, the orbital speed for that altitude.  Let's say that you execute a thirty-second burn at some point on this orbit:  your apoapsis rises, let's say from 100 to 200 km for this burn with this engine.  This requires a delta-V of 74 m/s, so we're probably using an Ant for this for a burn to take thirty seconds, but the point still stands.

Now let's say that you're on a 100 x 300 km orbit and you execute a thirty-second burn at the periapsis.  This time, you're moving at 2,382 m/s, and an addition of 74 m/s results in an orbit of 440 km.  The engine produces thrust from mass flow of its propellant; you develop the same thrust from the same burn time and the same amount of propellant.  And yet your apoapsis rises by 40 km more than it did for the first burn.

*BAMF*, Oberth effect.  The rocket did not burn more propellant, but it was in a different situation; namely, the rocket was moving faster, i.e., had more kinetic energy, at the beginning of the burn.

The Oberth effect is not a free energy device:  it is simply an observation that if you have more kinetic energy to begin, then you can do more with it.  Of course, that's true regardless of the energy type:  you get more delta-V from carrying more propellant, too.  If you kick the ball harder, then it goes farther, but if the ball is rolled to you faster, then you can kick it farther in that case, as well.  Energy and momentum must be conserved, but if you start with more of it, then the conservation results in the individual pieces of the system all getting a larger share.

##### Share on other sites
• 0
9 minutes ago, Zhetaan said:

Incidentally, the gravitational parameter is missing from the vis-viva equation that you quoted; it is not quite correct.

Good catch! I have corrected the original post.

##### Share on other sites
• 0
On 6/30/2020 at 9:12 PM, Superfluous J said:

I would be very surprised if there was any situation where burning lower and then raising your Pe to the target orbit was better.

On 7/1/2020 at 7:11 AM, Grogs said:

Case 1: Inital Pe @ 15 km: Capture burn is 3004.3 m/s. The burn to raise Pe to 1 Mm is 34.0 m/s. Total dV = 3038.3 m/s.

Case 2: Initial Pe is 1 Mm. Capture burn (and total) is 3504.2 m/s. This is about 466 m/s more than case 1.

I am officially surprised, thank you for doing the legwork!

##### Share on other sites
• 0
11 hours ago, Spricigo said:

Math is very important in KSP but we don't do it ourselves so often for 2 reason:

2.There is a mod/tool.

3 reasons

3. Fuel is very cheap and we don't need to keep in super tight budgets.

##### Share on other sites
• 0
4 hours ago, Zhetaan said:

Now let's say that you're on a 100 x 300 km orbit and you execute a thirty-second burn at the periapsis.  This time, you're moving at 2,382 m/s, and an addition of 74 m/s results in an orbit of 440 km.  The engine produces thrust from mass flow of its propellant; you develop the same thrust from the same burn time and the same amount of propellant.  And yet your apoapsis rises by 40 km more than it did for the first burn.

*BAMF*, Oberth effect.  The rocket did not burn more propellant, but it was in a different situation; namely, the rocket was moving faster, i.e., had more kinetic energy, at the beginning of the burn.

actually that's wrong. the extra height you gain is not caused by oberth effect (well, not much). it's simply that the further you go from the planet, the weaker its gravity, so it takes less thrust to gain an equal amount of height. in other words, the amount of energy needed to gain the same altitude is lower.

##### Share on other sites
• 0
On 7/1/2020 at 3:42 PM, king of nowhere said:

All the cool kids are using Jupyter notebooks these days, or so I'm told :

Spoiler
```
# Eve
µ = 8.1717e12; # m³/s²
a = 790_000; # m minimum radius
b = 85_109_365; # m radius of SOI
v_inf = 900; # m/s  the asymptotic velocity of a hyperbolic orbit; v_inf^2 = v_SOI^2 − 2µ/b

def f1(r):
return math.sqrt(v_inf**2 + 2*µ/r) - math.sqrt(µ/r)

def f2(r):
return (abs(math.sqrt(v_inf**2 + 2*µ/a)
- math.sqrt(2*µ/a - 2*µ/(a+r)))
+ abs(math.sqrt(µ/r)
- math.sqrt(2*µ/r - 2*µ/(a+r))))

import matplotlib.pyplot as plt

r = [a + (0.5*b-a)*(i/120) for i in range(1,120)]
dv1 = [f1(x) for x in r]
dv2 = [f2(x) for x in r]
x = [q/1e6 for q in r] # radius in Mm

fig = plt.figure()
ax = plt.axes()
ax.plot(x, dv1, label='single burn at final radius')
ax.plot(x, dv2, label='low burn, then circularise')
ax.plot(x, [math.sqrt(µ/x) for x in r], '--', label='final orbital velocity')
ax.set_ylabel('total delta-V, m/s')
ax.set_ylim(ymin=0)
ax.grid()
ax.legend();```

The radius of orbit, where we switch from one approach to the other being most efficient,
exactly where the single-burn delta-V is the same as the final orbital velocity.
This allows a reasonably efficient way to find which approach is more efficient in KSP by setting just one maneuver.

I have some intuition for this crossover, in the case where the parent body is very dense and heavy :
the Oberth effect removes your interplanetary energy for negligible delta-V in that case,
and then you need pay only the orbital velocity of your target.

In the general case, I can prove it from @ManEatingApe's math with some (very ugly) algebra.

But I have no intuition for why the crossover point is so simply related to the final orbital velocity, in the general case, and that drives me crazy.

##### Share on other sites
• 0
Posted (edited)
5 hours ago, OHara said:

All the cool kids are using Jupyter notebooks these days, or so I'm told.

TIL Juypter Notebooks exist, good call on the recommendation. Now if only there was a Kerboscript option...

5 hours ago, OHara said:

The radius of orbit, where we switch from one approach to the other being most efficient, exactly where the single-burn delta-V is the same as the final orbital velocity.

Nice insight!

Edited by ManEatingApe

##### Share on other sites
• 0
3 minutes ago, ManEatingApe said:

TIL Juypter Notebooks exist. Now if only there was a Kerboscript option...

Well, there's KRPC - it's not Kerboscript but you can retrieve data from your vessels and control them with a Python script - no reason why it wouldn't work inside Jupyter Notebooks.

##### Share on other sites
• 0
Posted (edited)
On 7/2/2020 at 6:31 PM, king of nowhere said:

actually that's wrong. the extra height you gain is not caused by oberth effect (well, not much). it's simply that the further you go from the planet, the weaker its gravity, so it takes less thrust to gain an equal amount of height. in other words, the amount of energy needed to gain the same altitude is lower.

You are correct in that the Oberth effect does not explain all of the altitude gain, and I was in error to imply that it was.  However, I believe the Oberth contribution to be greater than you suggest, but I confess to admit that I do not know by how much that is the case.  The only thing I can think to do is to work through the entire calculation from a standpoint of kinetic energy and see where that takes us.

Also please accept my apologies for the text wall in the spoiler.  I had a difficult time determining which parts to hide and which to leave out, so I hid it all.

Spoiler

Orbital energy, defined as the sum of the potential and kinetic energies, remains constant for an object in orbit (at least when you simplify it to a two-body problem), but that object in orbit consists of both its own mass (and energy) and also the mass (and energy) of its propellant.  When the rocket completes a burn, the propellant does not disappear--it serves as either the reservoir that supplies or the repository that takes away the 'excess' energy seen in the Oberth effect.  Consider a rocket that makes a prograde burn at periapsis:  the rocket is lifted to a higher orbit, but the exhaust is in fact sapped of its energy and driven to a lower orbit.  Thus, one may imagine a cloud of exhaust that, provided the rocket is burning in low-enough orbit, falls back to re-enter the atmosphere of the planet (or crash on it in the case of an airless body).  Similarly, a rocket that burns retrograde to re-enter dumps its 'excess' kinetic energy into a cloud of exhaust that goes to higher orbit--high enough, if circumstances are right, to launch that exhaust to an escape trajectory.

The overall effect of the Oberth effect, if you'll pardon the repetition, is to require less propellant to achieve the same delta-V because the propellant moving at high speed has more usable energy than propellant not moving at high speed.  This does not result in free delta-V for the rocket because the delta-V needed is measured in terms of the orbit, whereas the delta-V applied is measured in terms of the rocket.  Increasing velocity reduces the delta-V needed, which in turn reduces the delta-V applied, and the result is less expended propellant.  This is functionally identical to starting from a lower velocity and using a high-energy propellant or a more efficient engine; the source of the energy is not so important as its application, but it's important for rockets because their limited energy reserves provide strong incentive to find and eke out every possible joule.

Incidentally, what we see as specific impulse in the part description, Isp, when multiplied by the g0 that we find in some iterations of the Tsiolkovsky rocket equation, results in a quantity known as the engine's exhaust velocity.  More efficient engines have higher exhaust velocities; I leave it to the astute to consider the reason why a higher exhaust velocity results in more efficient burns.

The amount of delta-V required to achieve a particular orbital change is a (near-)constant for any given orbit; this is why delta-V maps are useful (as opposed to, for example, a map that gave figures in tonnes of propellant).  But delta-V is not kinetic energy, and it is a mistake to conflate the two.

Gravity isn't necessary to explain the Oberth effect, either; it would appear in completely free space, as well.  A rocket in free space travelling at 100 m/s and accelerating to 200 m/s would enjoy a specific kinetic energy increase (specific kinetic energy is kinetic energy per unit mass--it allows us to ignore the mass of the vessel) from 5,000 J/kg to 20,000 J/kg, for an increase of 15,000 J/kg.  A rocket accelerating from 200 m/s to 300 m/s would see an increase from 20,000 J/kg to 45,000 J/kg, for an increase of 25,000 J/kg, or 10 kJ/kg more for the same delta-V.  If you turn the rocket about and thrust retrograde, then you reduce from 300 J/kg to 200 J/kg, for an increase of -25,000 J/kg (because it's really a decrease)--but in that case, rather than gaining an 'extra' 10,000 J/kg, it loses an 'extra' 10,000 J/kg.

Mathematically:

eki = (1/2) vi2

Where eki is the initial specific energy and vi is the initial velocity.  After this, there are two ways to view the change:  one is to consider the total energy change in terms of final energy, and the other is to consider the change in velocity.  Therefore:

ekf = (1/2) (vf2)
Δek = ekf - eki
Δek = (1/2) (vf2) - (1/2) (vi2)
Δek = (1/2) (vf2 - vi2)

Where Δek is the specific energy change, ekf is the final specific energy, and vf is the final velocity.

Alternatively:

ekf = (1/2) (vi + Δv)2
ekf = (1/2) (vi2 + 2viΔv + Δv2)

So the energy difference, Δek:

Δek = ekf - eki
Δek = (1/2) (vi2 + 2viΔv + Δv2) - (1/2) vi2
Δek = (1/2) (vi2 + 2viΔv + Δv2 - vi2)
Δek = (1/2) (2viΔv + Δv2)

Please note the 2viΔv term.  If we distribute the (1/2), then we get this:

Δek = viΔv + (1/2) Δv2

This suggests that the total energy change is dependent on the initial velocity, but not in the part that processes the 'raw' energy change, for lack of a more technically accurate term.  Independent energy change may be more appropriate.  That second term is quite obviously reminiscent of the kinetic energy formula, but the first is something of an oddity.  It's still a v2 term, but the velocities are different from one another.  This is where the Oberth effect appears.

After putting it in numbers for burns from 100 to 200 and from 200 to 300 m/s:

Δek = 100*100 + (1/2) * 1002
Δek = 10,000 + 5,000
Δek = 15,000 J/kg

Δek = 200*100 + (1/2) * 1002
Δek = 20,000 + 5,000
Δek = 25,000 J/kg

It should be readily obvious that the difference between the first and the second equations is the 10,000 J/kg difference in the first terms.  What these equations describe is, in part, an energy contribution related to the change in velocity that has nothing to do with the initial velocity--that would be the 5,000 joules per kilogram that is liberated by increasing velocity by 100 m/s.  This part is universal--any velocity change of 100 m/s will either liberate or confine 5,000 J/kg of specific kinetic energy.  But the equations also include a contribution that relates to the energy's action upon an already-moving mass:  the rocket has specific kinetic energy by virtue of its motion, and the velocity change uses that energy in addition to the new energy that we see in the latter term.

Now, going from 2,246 to 2,320 m/s, and from 2,382 to 2,456 m/s as in my previous orbit examples:

Δek = 2,246 * 74 + (1/2) * 742
Δek = 166,204 + 2,738
Δek = 168,942 J/kg

Δek = 2,382 * 74 + (1/2) * 742
Δek = 176,268 + 2,738
Δek = 179,006 J/kg

This means a specific energy difference of 10,064 J/kg.  Let's see what this energy will do to our orbit.

Specific orbital energy is given by the following:

e = ek + ep
e = (v2 / 2) - (μ / r) = -(μ / 2a)

Where v is the velocity, μ is the standard gravitational parameter, r is the orbital distance, and a is the semimajor axis.  You may note the similarity to the vis-viva equation; the result from this equation is sometimes called vis-viva energy.  You may also note that the energy is given as a negative number; this is a convention that reflects the closed nature of the orbit.  Until one reaches escape velocity, the potential energy dominates.  Energies of zero denote parabolic orbits (and escape velocity) and positive energies denote hyperbolic ones.  (This also means that a hyperbolic orbit has a negative semimajor axis; you may have noticed that in Kerbal Engineer.)

For the 2,246 m/s, 100 x 100 km orbit:

(2,2462 / 2) - (3.5316 x 1012 / 700,000) = -(3.5316 x 1012 / [2 * 700,000])
2,522,000 - 5,045,000 = -3.5316 x 1012 / 1,400,000
-2,523,000 J/kg = -2,523,000 J/kg

Within the limits of rounding error.

For the 2,320 m/s, 100 x 200 km orbit:

(2,3202 / 2) - (3.5316 x 1012 / 700,000) = -(3.5316 x 1012 / [2 * 750,000])
2,691,000 - 5,045,000 = -2,354,000
-2,354,000 J/kg = -2,354,000 J/kg

And the difference in energy between the two orbits is 169,000 J/kg, which fits, again within the limits of rounding error.

For the 2,382 m/s, 100 x 300 km orbit:

(2,3822 / 2) - (3.5316 x 1012 / 700,000) = -(3.5316 x 1012 / [2 * 800,000])
2,837,000  - 5,045,000 = -(3.5316 x 1012 / 1,600,000)
-2,208,000 J/kg = -2,207,000 J/kg

... There is a slight discrepancy, but again, it fits within rounding error.

Finally, for the 2,456 m/s, 100 x 440 km orbit:

(2,4562 / 2) - (3.5316 x 1012 / 700,000) = -(3.5316 x 1012 / [2 * 870,000])
3,016,000 - 5,045,000 = -(3.5316 x 1012 / 1,740,000)
-2,029,000 J/kg = -2,030,000 J/kg

... I'll not belabour the point, but suffice to say that the orbital energy difference in these two orbits is approximately equal to 179,000 J/kg.

This means that we can substitute the energy reduced by the 'extra' from the Oberth effect and see what kind of orbit we get.

-2,029,000 J/kg - 10,000 J/kg (again, please excuse the rounding) equals -2,039,000 J/kg.

-2,039,000 = -(3.5316 x 1012 / [2 * a])
2 * a = -3.5316 x 1012 / -2,039,000
2 * a = 1,732,026 m

This is the major axis; we don't need to solve for a.

Subtract the periapsis and Kerbin's diameter:

1,732,026 - 100,000 - 1,200,000 = 432,026 m

440 km - 432 km gives 8 km attributable to the Oberth effect, which is 20% of the gain.  It is perhaps small, but it's definitely not negligible.

So, to conclude, I submit this:

*BAMF*, Oberth effect.  But it's still there.

Edited by Zhetaan

##### Share on other sites
• 0

@Zhetaan that's by far the best explanation of the Oberth Effect I recall to see in those forums.

I suggest to keep a bookmark for when questions about it arise again.

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.