Delta v of a trans-Mars injection (TMI)

[yomahabaca]

Hello, I play Ksp Rss and I was wondering why does it only take about 600 m/s more than Earth's escape velocity to have an encounter with Mars while it takes much more Delta v (about 3 km/s) from a solar orbit (same as Earth orbit) to have an encounter with Mars?

What explains this difference?

Is it the Oberth effect or something else that explains this?

Thank you !

[king of nowhere]

21 hours ago, yomahabaca said:

Hello, I play Ksp Rss and I was wondering why does it only take about 600 m/s more than Earth's escape velocity to have an encounter with Mars while it takes much more Delta v (about 3 km/s) from a solar orbit (same as Earth orbit) to have an encounter with Mars?

What explains this difference?

Is it the Oberth effect or something else that explains this?

Thank you !

oberth effect.

yes, I was also surprised when i learned just how strong oberth effect can be

[Leganeski]

This effect also appears in stock, or more generally in any system when ejecting from a relatively large body.

For example, burning to Duna from a solar orbit at the same place as Kerbin's orbit requires 920 m/s, while ejecting to Duna from Kerbin requires only 130 m/s more than reaching Kerbin's escape velocity.

[swjr-swis]

On 9/3/2022 at 9:39 PM, yomahabaca said:

why does it only take about 600 m/s more than Earth's escape velocity to have an encounter with Mars while it takes much more Delta v (about 3 km/s) from a solar orbit (same as Earth orbit) to have an encounter with Mars?

Look at it this way:  when you're in orbit around Earth, you're literally running circles around the planet. So you are in actuality moving through the 'same' orbit -around the sun- faster than the planet itself... just doing so in a very tightly wound spiraling path. That's extra speed/energy that you wouldn't have if you were traveling Earth's orbit all by your lonesome. That extra speed/energy goes with you when you decide to burn to leave Earth, so you need less *extra* energy to get into a Mars transfer orbit than if you were traveling Earth's orbit all by yourself.

(hiding now from the physicists everywhere gnashing their teeth and plotting my impending doom)

 

 

[yomahabaca]

Thanks for the explanation ! : )

Edited September 6, 2022 by yomahabaca

[Leganeski]

On 9/4/2022 at 8:36 PM, swjr-swis said:

Look at it this way:  when you're in orbit around Earth, you're literally running circles around the planet. So you are in actuality moving through the 'same' orbit -around the sun- faster than the planet itself

This is a great explanation!

Conversely, on the opposite side of your planetary orbit, you're moving around the sun slower than the planet. This means you've effectively discarded a bunch of energy for free, helping you lower your solar orbit if for example you want to get to Venus or Eve.

[yomahabaca]

But there is something that I don't understand, as Leganeski said, on the other side of the planetary orbit we move more slowly with respect to the Sun, so how can the Oberth effect work in this situation (trajectory to go to the inner planets)? isn't the Oberth effect supposed to be stronger the faster the satellite/vessel goes? but in this situation the satellite goes slower than the Earth around the Sun.

[Zhetaan]

2 hours ago, yomahabaca said:

But there is something that I don't understand, as Leganeski said, on the other side of the planetary orbit we move more slowly with respect to the Sun, so how can the Oberth effect work in this situation (trajectory to go to the inner planets)? isn't the Oberth effect supposed to be stronger the faster the satellite/vessel goes? but in this situation the satellite goes slower than the Earth around the Sun.

That's because it's technically a gravity assist, as well.  You're moving at high orbital speed with respect to your primary (in this case, Earth), so you get the benefit of the Oberth effect, but you're also using that effect to direct your velocity retrograde against Earth's solar orbit, which is to say, it's a deceleration with respect to the sun.

Many people think of only unpowered gravity assists when confronted with the term gravity assist.  Powered assists also exist.  The end result is that the difference (with caveats, of course) between going to Mars and going to Venus, most everything else being equal, lies in which side of Earth you're on when you start your burn.  However, that looks very similar to what I will call the second half of unpowered assists, where the difference between an acceleration or deceleration is which side of the planet's sphere of influence you exit.

Edited September 6, 2022 by Zhetaan

[mhoram]

5 hours ago, yomahabaca said:

But there is something that I don't understand, as Leganeski said, on the other side of the planetary orbit we move more slowly with respect to the Sun, so how can the Oberth effect work in this situation

The problem I see with this argumentation is that you want to take the motion relative to the sun into account, while the Oberth effect is associated with the orbit of the current SOI.

Lets take the example that you are within an Orbit around Kerbin with Periapsis of 75km and Apoapsis of 100km and your goal is to change the Apoapsis to 1000km.

In the initial orbit, your velocity at 75km Periapsis is 2308 m/s

In the final orbit, your velocity at 75km Periapsis is 2713 m/s

So you need to invest 405 m/s Delta-V to change your orbit. For this orbit-change it does not matter, if the Periapsis is near the sun or away from the sun.

In order to transfer to Eve or Duna, you need to leave Kerbins SOI at certain velocities. In order to reach these velocities, just as before you accelerate at Periapsis a certain amount of Delta-V. Again it does not matter, where the Periapsis in relation to the sun is located, because your burn happens within the SOI of Kerbin.

 

If you want to dig deeper into the Oberth topic, there have been several discussions here on the forum:

• https://forum.kerbalspaceprogram.com/index.php?/topic/65676-oberth-effect-vs-gravity-assists/
• https://forum.kerbalspaceprogram.com/index.php?/topic/122445-oberth-effect/
• https://forum.kerbalspaceprogram.com/index.php?/topic/146254-how-much-dv-do-you-lose-with-a-higher-lko-really/

 

[yomahabaca]

Thank you !
I have another question, is the Oberth effect responsible for the fact that an acceleration of 100m/s from the perigee of a 200km/300km orbit will increase the apogee less than an acceleration of 100 m/s applied to the perigee of an orbit of 200km/10000km?

[jimmymcgoochie]

13 hours ago, yomahabaca said:

Thank you !
I have another question, is the Oberth effect responsible for the fact that an acceleration of 100m/s from the perigee of a 200km/300km orbit will increase the apogee less than an acceleration of 100 m/s applied to the perigee of an orbit of 200km/10000km?

I’m no expert on orbital mechanics, but this is not Oberth effect- you’ll see a similar effect orbiting any body at any velocity, from tiny objects like Deimos (orbital velocity of 4m/s or so in RSS with patched conics) to stars orbital velocities in the tens or even hundreds of kilometres per second.

Orbits aren’t linear and gravity decreases exponentially with distance, so applying a linear acceleration at periapsis will have an exponentially greater effect as you get closer to escape velocity until you go too fast for gravity to bring you back again.

[Zhetaan]

On 9/6/2022 at 6:56 PM, yomahabaca said:

Thank you !
I have another question, is the Oberth effect responsible for the fact that an acceleration of 100m/s from the perigee of a 200km/300km orbit will increase the apogee less than an acceleration of 100 m/s applied to the perigee of an orbit of 200km/10000km?

23 hours ago, jimmymcgoochie said:

I’m no expert on orbital mechanics, but this is not Oberth effect- you’ll see a similar effect orbiting any body at any velocity, from tiny objects like Deimos (orbital velocity of 4m/s or so in RSS with patched conics) to stars orbital velocities in the tens or even hundreds of kilometres per second.

Orbits aren’t linear and gravity decreases exponentially with distance, so applying a linear acceleration at periapsis will have an exponentially greater effect as you get closer to escape velocity until you go too fast for gravity to bring you back again.

The Oberth effect does have, well, an effect, at least in part, but the relationship is less obvious because one tends to see the result through the lens of gravitational effects without intuitively recognising the fact that the orbital speed at the periapsis of the 200 x 10,000 orbit is necessarily greater than that at the periapsis of the 200 x 300 orbit.  Instead, one tends to look at it in terms of orbital energy getting closer and closer to escape, but since higher initial velocities are unavoidably part of that, and the Oberth effect is the resultant change of orbital energy that is dependent upon (and attributable to) the initial velocity, the Oberth effect is unavoidably part of that, as well.

To the maths!  (In the spoiler; I'm not a sadist.)

Spoiler

Instantaneous orbital speed is given by the following:

v = √(μ [(2 / r) - (1 / a)])

Where:
v = orbital speed
μ = standard gravitational parameter of Earth (3.98600 x 10¹⁴ m³/s²)
r = the radial distance of the point to be calculated (in this case, the perigee of the orbit--but it must be measured from Earth's centre, not surface), and
a = the semi-major axis of the orbit in question

At the perigee of a 200 km x 300 km orbit of Earth, with a pre-calculated perigee of 200,000 + 6,371,000 = 6,571,000 m and semi-major axis of [(200,000 + 6,371,000) + (300,000 + 6,371,000)] /2 = (6,571,000 + 6,671,000) / 2 = 6,621,000 m, we get the following orbital speed:

v = √(3.98600 x 10¹⁴ [(2 / 6,571,000) - (1 / 6,621,000)])
v = √(3.98600 x 10¹⁴ [3.043676762 x 10^-7 - 1.510345869 x 10^-7])
v = √(3.98600 x 10¹⁴ x 1.533330893 x 10^-7)
v = √(61,118,569.39)
v = 7,818 m/s

Please note that I am showing rounded-off values while calculating with much more accurate ones.

For the larger 200 x 10,000 orbit (same r, a = 11,471,000 m), we get the following, instead:

v = √(3.98600 x 10¹⁴ [(2 / 6,571,000) - (1 / 11,471,000)])
v = √(3.98600 x 10¹⁴ [3.043676762 x 10^-7 - 8.717635777 x 10^-8])
v = √(3.98600 x 10¹⁴ x 2.171913184 x 10^-7)
v = √(86,572,459.53)
v = 9,304 m/s

Rearranging the vis-viva equation a bit, we get a value for a, the semi-major axis, from the following:

a = 1 / ([2 / r] - [v² / μ])

From a, the periapsis, and known Earth radius, the apoapsis is simply:

r_a = 2a - r_p - 2(6,371,000)

Where:
r_a = apoapsis
r_p = periapsis
a = semi-major axis

Let's see what happens when we increase the orbital speed at periapsis for both of those orbits by 100 m/s.

Orbit 1 periapsis speed:  7,817.84 + 100 = 7,917.84 m/s
Orbit 2 periapsis speed:  9,304.43 + 100 = 9,404.43 m/s

Orbit 1 semi-major axis:

a = 1 / ([2 / 6,571,000] - [(7,917.841)² / 3.98600 x 10¹⁴])
a = 1 / ([3.043676762 x 10^-7] - [62,692,190.27 / 3.98600 x 10¹⁴])
a = 1 / (3.043676762 x 10^-7 - 1,572809590 x 10^-7)
a = 1 / 1.470867172 x 10^-7
a = 6,798,704 m

And the Orbit 1 apoapsis:

r_a = 2(6,798,710.442) - 200,000 - 2(6,371,000)
r_a = 13,597,420.88 - 200,000 - 12,742,000
r_a = 655,408 m

Note that the apoapsis is doubled from only 100 m/s of added speed; this is actually an illusion, and it is incorrect to consider this a doubling.  The increase should be viewed from the perspective of the focus of the orbit, in which case it increased from 6,671 to 7,026.4 km.

Now, for Orbit 2:

Semi-major axis:

a = 1 / ([2 / 6,571,000] - [(9,404.437)² / 3.98600 x 10¹⁴])
a = 1 / ([3.043676762 x 10^-7] - [88,443,303.62 / 3.98600 x 10¹⁴])
a = 1 / (3.043676762 x 10^-7 - 2.218848561 x 10^-7)
a = 1 / 8.248282014 x 10^-8
a = 12,123,752 m

And the Orbit 2 apoapsis:

r_a = 2(12,123,736.78) - 200,000 - 2(6,371,000)
r_a = 24,247,473.56 - 200,000 - 12,742,000
r_a = 11,305,504 m

This gives us a total apoapsis increase for Orbit 1 of 355.4 km, and for Orbit 2 of 1,305.5 km.

How much of those increases is attributable to the Oberth effect?  To find out, we need to consider the problem, not in terms of speed, but in terms of energy.  To put it simply (okay, more simply), total orbital energy is a constant.  That should make sense, especially for closed orbits:  the speed change from apoapsis to periapsis and back again results from the trade between potential energy, which depends on altitude, and kinetic energy, which depends on velocity, that repeats over time.  Thus, to increase altitude on an unpowered trajectory, one must decrease speed, and vice versa.  Neglecting perturbation, that trade back and forth will repeat perpetually.  To return to the same periapsis and apoapsis again and again, the energy trade must be exactly the same each time, which can only happen if the total energy remains constant.

The Oberth effect is an additional increase in energy that results from the influence of the initial velocity on a powered manoeuvre--i.e., it's tied to the kinetic energy.  Practically, it stems from the fact that the mass system of a powered rocket is composed of two parts:  the rocket itself, and its exhaust.  The Oberth effect is the practical result of the rocket taking for itself some of the kinetic energy that, in other circumstances, would go into the exhaust.

For the following calculations, I will work in terms of specific energy, which is energy per unit mass.  This means that we can neglect the mass of the rocket and its exhaust, and it simplifies some of the other calculations:  for example, specific kinetic energy is simply v² / 2.

For a general velocity change, we have an initial velocity v_i and the velocity change, Δv.  The specific kinetic energy change, Δe_k, is the difference between initial and final specific kinetic energy, e_ki and e_kf, as follows:

e_ki = (1 / 2) * v_i²
e_ki = (v_i² / 2)

e_kf = (1 / 2) * (v_i + Δv)²
e_kf = (1 / 2) * (v_i + Δv)(v_i + Δv)
e_kf = (1 / 2) * (v_i² + 2v_iΔv + Δv²)
e_kf = (v_i² /2) + v_iΔv + (Δv² / 2)

Δe_k = e_kf - e_ki
Δe_k = [(v_i² /2) + v_iΔv + (Δv² / 2)] - [(v_i² / 2)]
Δe_k = v_iΔv + (Δv² / 2)

From this, we can see that the specific energy change depends on the velocity change, which we expect, but it also depends on the initial velocity.  In fact, from some quick algebra, we can see that when the initial velocity is greater than half of the velocity change, the initial velocity term is the dominant term of the equation.

Let's look at the burns from before and see what we can find out.

For Orbit 1, the initial specific kinetic energy at periapsis is this:

e_ki = (7,817.841)² / 2
e_ki = 61,118,622.27 / 2
e_ki = 30,559,319 J/kg

... Which I will round to 30,559 kJ/kg.

For Orbit 1's 100 m/s burn, we get this:

Δe_k = (7,817.84) * (100) + ([100]² / 2)
Δe_k = (781,784) + (10,000 / 2)
Δe_k = 781,784 + 5,000
Δe_k = 786,784 J/kg
Δe_k = 787 kJ/kg

For Orbit 2, the initial specific kinetic energy:

e_ki = (9,304.437)² / 2)
e_ki = (86,572,417.62) / 2)
e_ki = 43,286,278 J/kg

Which I will round to 43,286 kJ/kg.

For the 100 m/s burn:

Δe_k = (9,304.437) * (100) + ([100]² / 2)
Δe_k = 930,444 + 5,000
Δe_k = 935,444 J/kg
Δe_k = 935 kJ/kg

Orbit 2 gets an additional 148 kJ from the same burn, solely on account of its higher initial velocity. (Of course, if we wanted to be exhaustive, then we could also say that Orbit 1 gets 782 kJ over the 5 kJ that it would have starting from rest, since the only way to have no Oberth effect at all is to have no initial velocity.)

Where do we go from here?  Let's look at the total orbital energy and see what we can learn.

Total specific orbital energy is given by the following:

e = e_k + e_p
e = (v² / 2) - (μ / r) = -(μ / 2a)

And the only new variables here are e, the total specific orbital energy, and e_p, the specific potential energy.

Orbit 1's total specific orbital energy, pre-burn, is this:

(v² / 2) - (μ / r) = -(μ / 2a)
([7,817.84]² / 2) - (3.9860044188 x 10¹⁴ / [6,571,000]) = -(3.9860044188 x 10¹⁴ / 2 * [6,621,000])
(61,118,637.12 / 2) - (60,660,545.1) = -(3.9860044188 x 10¹⁴ / 13,242,000)
30,559,318.6 - 60,660,545.1 = -30,101,193.2
-30,101,227 = -30,101,193

Which, for our purposes, is equivalent:  34 joules of rounding error is not worth mentioning.  We have -30,101 kJ/kg against -30,101 kJ/kg.

Post-burn, the specific orbital energy is this:

(v² / 2) - (μ / r) = -(μ / 2a)
([7,917.84]² / 2) - (3.9860044188 x 10¹⁴ / [6,571,000]) = -(3.9860044188 x 10¹⁴ / 2 * [6,798,704.13])
(62,692,205.31 / 2) - (60,660,545.1) = -(3.9860044188 x 10¹⁴ / 13,597,408.26)
(31,346,102.66) - (60,660,545.1) = -29,314,442.5
-29,314,442.4 = -29,314,442.5

Again, this agrees, this time to less than a joule.  The total specific orbital energy is -29,314 kJ/kg.  Note that as the orbit size increases, the energy increases.  Also note that the difference between this energy and the pre-burn energy is the same as before:  787 kJ/kg.  That should make sense, because we're using two forms of the same equation.

Now, let's do the same for Orbit 2:

Pre-burn:

(v² / 2) - (μ / r) = -(μ / 2a)
([9,304.44]² / 2) - (3.9860044188 x 10¹⁴ / [6,571,000]) = -(3.9860044188 x 10¹⁴ / 2 * [11,471,000])
(86572555.48 / 2) - (60,660,545.1) = -(3.9860044188 x 10¹⁴ / 22,942,000)
43,286,277.74 - 60,660,545.1 = -17,374,267.36
-17,374,267.4 = -17,374,267.4

Which comes to -17,374 kJ/kg.

Post-burn:

(v² / 2) - (μ / r) = -(μ / 2a)
([9,404.44]² / 2) - (3.9860044188 x 10¹⁴ / [6,571,000]) = -(3.9860044188 x 10¹⁴ / 2 * [12,123,752.01])
(88,443,442.96 / 2) - (60,660,545.1) = -(3.9860044188 x 10¹⁴ / 24,247,504.02)
44,221,721.48 - 60,660,545.1 = -16,438,823.62
-16,438,823.6 = -16,438,823.6

Which gives us -16,439 kJ/kg, for a difference of 935 kJ/kg, as we expect.

Orbit 2 enjoys the benefit of an additional 148 kJ, as mentioned before.  Let's pretend that we can subtract that much from the orbital energy and see where that takes us.  This isn't the most accurate: technically, to subtract off the Oberth effect, we'd need to take everything influenced by the initial velocity, which for a 100 m/s burn, means everything but 5 kJ.  Instead, let's see what the Oberth effect does for Orbit 2 by limiting ourselves to the energy gain of Orbit 1.

In this case, we start with the same pre-burn Orbit 2 and add only the 787 kJ/kg that Orbit 1 added.

Orbit 2 specific orbital energy:  -17,374 kJ/kg
Orbit 1 energy addition:  787 kJ/kg

Reduced Orbit 2 specific energy:  -17,374 + 787 = -16,587 kJ/kg

This equates to the following:

-16,587,000 = -(μ / 2a)
-16,587,000 = -(3.9860044188 x 10¹⁴ / 2a)
2a = -3.9860044188 x 10¹⁴ / -16,587,000
2a = -24,030,506.84 m

Having solved for the major axis (not semi-major axis!), we can figure the reduced apoapsis easily:

r_a = 2a - 2 * (6,371,000) - 200,000
r_a = 24,030,506.84 - 12,742,000 - 200,000
r_a = 11,088,506.84 m

Which is equivalent to 11,089 km.  This means that, of the real apoapsis of 11,306 km, 217 km of it is directly attributable to the Oberth effect on this burn over the burn in Orbit 1.  In other words, because the vessel is moving approximately 1,487 m/s faster in Orbit 2, it enjoys an additional apoapsis increase of 217,000 m , or 16.6% of the total increase.

However, that's not anything like the complete picture.  Remember, the Oberth effect applies anywhere that there is a velocity, because anywhere that there is a velocity, there is kinetic energy.

Let me, finally, attempt to illustrate this idea with a comparative analogy.  An engine is perfectly capable of firing during a static test (or, for that matter, before you release the launch clamps in KSP).  In this case, the velocity is fixed at zero; therefore, the kinetic energy and Oberth effect are also fixed at zero.  Nevertheless, something happens:  energy is liberated from the propellant, and it has to go somewhere.  Well, I can answer that:  it goes into the exhaust.  Some goes into the engine body as heat, but that generally just heats the exhaust.  Let's consider the opposite situation:  let's invent a speed, which I will unimaginatively call critical speed, at which the rocket velocity is exactly equal to the exhaust velocity.  The rocket fires, and the exhaust thus has zero velocity (and zero kinetic energy).  Where did the kinetic energy go?  I can answer that one, too:  it went into the rocket.  As the rocket's initial kinetic energy increases, the energy it extracts from the propellant also increases, because the propellant is moving with the rocket and thus has a total energy related to the sum of its chemical energy and also its existing kinetic energy.  Note that this does not require the presence of a gravitational field:  the Oberth effect appears in free space, too.  Of course, you get more out of it if you can use gravity to get a free speed increase at the periapsis, but it still appears.  You see this in engine design and propellant choices, too.  Efficient engines have hot combustion temperatures and low-mass combustion products.  This results in higher exhaust velocities and equivalently, high kinetic energies, which means that the Oberth effect can appear inside a rocket engine's combustion chamber, too.

I hope that helps a bit.

[mhoram]

Thanks @Zhetaan for the detailed explanation. I tried to put the explanation into a graphic.

Have a look at this picture. (sourcecode)

 

Each coordinate denotes an orbit with:
x axis = initial Apoapsis/Periapsis (Altitude above Earth surface in km, SOI of Earth is about 0.929e6km)
y axis = initial Periapsis/Apoapsis (Altitude above Earth surface in km)
(Apoapsis and Periapsis get switched in the top left half of the graph, so that for the calculation Apoapsis is always larger than Periapsis)
The Orbits on the diagonal from lower left over A1 to top right are the circular orbits.

The color at each coordinate denotes the change of Apoapsis caused by a 100m/s burn at Periapsis (the more red, the more change) (dark red is about 0.929e6km, white is about 0km, difference between two threshold lines is about 23225km)

Ignore the two white areas B1 and B2, they are caused by hyperbolic orbits which have a negative Apoapsis.

Lets have a look at few datapoints:
A1 (circular orbit at altitude of about 440000km) and A2 (orbit with Periapsis of about 40000km and Apoapsis of about 350000km)
The threshold lines indicates that if you burn 100m/s at their respective Periapsis, you change the Apoapsis by the same amount.
This means, that for highly eccentric orbits, it is easier to raise Apoapsis than it is for circular orbits.

Comparing A1 to A3 (orbit with Periapsis of about 280000km and Apoapsis of about 440000km) which have both the same Apoapsis:
Even though you are faster at the Periapsis of orbit A3 than at the Periapsis of orbit A1, a 100m/s burn raises the Apoapsis of orbit A3 less than the Apoapsis of orbit of A1.
This shows, as stated in the preivous post, that the Oberth effect is not the only thing that affects the amount by which Apoapsis gets raised and is linked to the energy of the orbits.

Edited September 10, 2022 by mhoram
added link to source code

[yomahabaca]

Thanks a lot ! it really helps me : )

Edited September 9, 2022 by yomahabaca

[yomahabaca]

If I understand correctly, the A1 orbit has less energy because it is less in the gravitational well of the Earth and therefore it is easier to exit (raised the apoastre) while the A3 orbit has its periapsis closer to Earth (deeper in the gravitational well) then its average orbital speed is higher and its orbital energy is higher so it is more difficult to raise the apoastre. The A3 orbit benefits more from the Oberth effect than the A1 orbit, but the effect is minor compared to the orbital energy, it becomes the majority for more eccentric orbits such as the A2 orbit.

[mhoram]

2 hours ago, yomahabaca said:

If I understand correctly, the A1 orbit has less energy because it is less in the gravitational well of the Earth and therefore it is easier to exit (raised the apoastre) while the A3 orbit has its periapsis closer to Earth (deeper in the gravitational well) then its average orbital speed is higher and its orbital energy is higher so it is more difficult to raise the apoastre.

There seems to be a misunderstanding. The A1 orbit has more energy than the A3 orbit.

Lets have a look at it qualitatively:

• The higher the altitude, the higher the potential energy
• The faster the velocity, the higher the kinetic energy.

For a ship in a Kepler elliptical orbit, it is valid, that kinetic energy + potential energy is always the same. That is why the formula e = e_k + e_p = (v² / 2) - (μ / r) can be used to specify the energy of an orbit, since it is constant for every point in that orbit.

The ship at Apoapsis in orbit A1 is faster than the ship at Apoapsis in orbit A3. Since Apoapsis is equal for both orbits (and so also the potential energy), A1 must have a higher energy than A3.

This graphic shows the energy of orbits (sourcecode):

x axis = orbit Apoapsis/Periapsis (Altitude above Earth surface in km, SOI of Earth is about 0.929e6km)
y axis = orbit Periapsis/Apoapsis (Altitude above Earth surface in km)
(Apoapsis and Periapsis get switched in the top left half of the graph, so that for the calculation Apoapsis is always larger than Periapsis)

Color denotes the energy of the orbit. The more green, the higher the energy. (two adjacent threshold lines have the same energy difference)

The fact that the threshold lines are farer apart, when the altitudes are higher, shows, that the energy difference becomes smaller for higher orbits.

2 hours ago, yomahabaca said:

The A3 orbit benefits more from the Oberth effect than the A1 orbit, but the effect is minor compared to the orbital energy, it becomes the majority for more eccentric orbits such as the A2 orbit.

I agree with that.

Edited September 10, 2022 by mhoram

[Leganeski]

1 hour ago, mhoram said:

3 hours ago, yomahabaca said:

If I understand correctly, the A1 orbit has less energy because it is less in the gravitational well of the Earth and therefore it is easier to exit (raised the apoastre) while the A3 orbit has its periapsis closer to Earth (deeper in the gravitational well) then its average orbital speed is higher and its orbital energy is higher so it is more difficult to raise the apoastre.

There seems to be a misunderstanding. The A1 orbit has more energy than the A3 orbit.

An important point to remember here is that the gravitational potential energy is always negative, decreasing from zero as the position gets closer to the body from far away. In this case, the A3 orbit is deeper than the A1 orbit in Earth's gravitational well, causing its total energy to be more negative. This means that it is in fact a lower-energy orbit than A1.

 

1 hour ago, mhoram said:

This graphic shows the energy of orbits (sourcecode):

The straightness of the threshold lines demonstrates the fact that total energy is dependent only on the semi-major axis (the average of periapsis and apoapsis), and not eccentricity.

[yomahabaca]

So if this time I understood correctly, the more one is in a high energy orbit (apoastre and periapsis high) the easier it is to go up its apoastre (ignoring the Oberth effect).
If we ignore the Oberth effect, an acceleration of 100 m/s at the periapsis of orbit A2 would increase the apoastre less than the same acceleration on orbit A1 because the orbital energy of orbit A2 is lower than that of the A1 orbit. The lower an orbit has its periapsis, the lower its orbital energy. (If the Oberth effect is taken into account, the A2 orbit would increase its apoastre more than the A1 orbit.)

It is as if, the lower an orbit has its periapsis, the more it is held by gravity, which reduces the increase in apoastre for the same delta v applied to the periapsis.

Edited September 10, 2022 by yomahabaca

[mhoram]

19 hours ago, yomahabaca said:

If the Oberth effect is taken into account, the A2 orbit would increase its apoastre more than the A1 orbit

I would disagree with this statement. If you burn 100m/s at the Periapsis of A1 and A2, then the Apoapsis change of both orbits would be the same. I chose A1 and A2 specifically so that a 100m/s burn results in the same Apoapsis change.

Edited September 11, 2022 by mhoram
fix typo

[yomahabaca]

It's true, I was wrong about that

Edited September 11, 2022 by yomahabaca