Low-Thrust Orbit transfer

[MajorTomtom]

Hi guys,

I want to calculate the total delta V necessary for an electric spacecraft to transfer from one circular orbit to another. I know that in this case the Hohmann transfer is not applicable because of the very-low thrusts at hand, therefore the manoeuver is more or less an ascending (or descending) spiral from one orbit to another, or, speaking more kerbal, you simply burn prograde/retrograde until you reach the desired orbit.

My first guess to calculate the delta V required would be to make the difference between the two orbital speeds, but then I thought about the oberth effect and I suppose it is not that simple.

Does anyone know how to do such a thing, or where to look ?

Thanks in advance, and sorry for my grammar

[p1t1o]

16 minutes ago, MajorTomtom said:

Hi guys,

I want to calculate the total delta V necessary for an electric spacecraft to transfer from one circular orbit to another. I know that in this case the Hohmann transfer is not applicable because of the very-low thrusts at hand, therefore the manoeuver is more or less an ascending (or descending) spiral from one orbit to another, or, speaking more kerbal, you simply burn prograde/retrograde until you reach the desired orbit.

My first guess to calculate the delta V required would be to make the difference between the two orbital speeds, but then I thought about the oberth effect and I suppose it is not that simple.

Does anyone know how to do such a thing, or where to look ?

Thanks in advance, and sorry for my grammar

 

Welcome to KSP & the forums!

 

I think, and someone correct me if Im wrong, that:

The minimum dV required to get from one orbit from another can be obtained by comparing the specific energy of both orbits, no matter how you plan on getting there. 

The specific energy is the sum of the potential energy and kinetic energy, divided by the "reduced mass". Because this and that cancel out, all you need is the semi-major axis of each orbit and the mass of the planet. Because your spacecraft is many billions of times less massive than the planet, it can essentially be ignored, simplifying things somewhat.

Required reading:

https://en.wikipedia.org/wiki/Specific_orbital_energy

 

So now you know the difference in energy, you just need to give that much energy to your spacecraft in kinetic form, some of which will convert to potential energy as your altitude rises.

From that amount of kinetic energy and the known mass of your craft, you can get the minimum required dV.

 

A slow, spiraling ascent is less efficient, so you will require some amount more than this, but its a start.

 

Edited February 13, 2018 by p1t1o

[Steel]

36 minutes ago, MajorTomtom said:

Hi guys,

I want to calculate the total delta V necessary for an electric spacecraft to transfer from one circular orbit to another. I know that in this case the Hohmann transfer is not applicable because of the very-low thrusts at hand, therefore the manoeuver is more or less an ascending (or descending) spiral from one orbit to another, or, speaking more kerbal, you simply burn prograde/retrograde until you reach the desired orbit.

My first guess to calculate the delta V required would be to make the difference between the two orbital speeds, but then I thought about the oberth effect and I suppose it is not that simple.

Does anyone know how to do such a thing, or where to look ?

Thanks in advance, and sorry for my grammar

Unfortunately, once you get beyond simple Hohmann transfers - which can be solved pretty simply and analytically - usually the only way to accurately come up with an answer is to solve the equations numerically.

[PB666]

3 minutes ago, MajorTomtom said:

Hi guys,

I want to calculate the total delta V necessary for an electric spacecraft to transfer from one circular orbit to another. I know that in this case the Hohmann transfer is not applicable because of the very-low thrusts at hand, therefore the manoeuver is more or less an ascending (or descending) spiral from one orbit to another, or, speaking more kerbal, you simply burn prograde/retrograde until you reach the desired orbit.

My first guess to calculate the delta V required would be to make the difference between the two orbital speeds, but then I thought about the oberth effect and I suppose it is not that simple.

Does anyone know how to do such a thing, or where to look ?

The short answer is that in a perfectly spiral exit orbit to the planets SOI uses twice the dV as a infinitely short pulse at the parabolic origin (the developing Pe). The equation is roughly a parabola, neither elliptical or hyperbolic.

I am working or a program to solve the La grangian for the best orbit when  [escape velocity - orbital velocity >> 0.5 * specific thrust * orbital period]. I can tell you that if the Θ of your escape vector then don't burn at all between Θ - 45' to Θ + 45', it only wastes propellant and it does not save time. 

(Θ ± π/4) or using the Pe as the standard reference, don't burn at all between π ± π/4. You can achieve most of your gains between for thrust between  ± π/2. There are problems however,

Assumption 1, you don't have nuclear power. (too heavy)
Assumption 2, power density of your batteries are not infinite.
Assumption 3, your target heliocentric orbit is something other than Earths orbit.
Assumption 4 is you are using solar power.
Assumption 5 is that your starting orbit is minimal (LEO) and that the Earth is opaque.

The earth blocks the sun for the lowest orbits, as a result your escape vector needs to be such that the Pe needs to be place on the sunlit side of the planet.
The oberth effect in the last pass of the Pe before exits is the point in which, if you have additional thrust, then the thrust should be applied, you do not have to reach the target heliocentric orbit, this can be done by burning along the prograde as you are exiting
Many ION drives can reduce the ISP in favor of thrust, if you can reduce the ISP by two fold then applying it at Pe roughly saves fuel because of the oberth effect. You can change the ISP in KSP by altering aspects in the permanent file.
You can use batteries to extend the length of the burn.
You can also use hyperglolics to assist with the Oberth effect during the last pass of Pe.
You can vector down a few degrees approaching Pe  and a few degrees leaving Pe to keep the Pe as low as possible. You can retro at Apo to reduce Pe using highest ISP setting on ION drive.
You can bring a chemical rocket to assist with your escape burn.

If you account for the oberth effect in exit burns, the effect of spiraling (performance reduction) increases with the absolute value of the difference of specific potential energy between the source heliocentric orbit and the target specific potential energy orbit. So that spiraling out and pushing out the Pe makes hohmann transfers less efficient than if you have a short pulse burn.

An example a burn to Pluto or mercury might use 3 times as much propellant a burn to mars might only use 2.3 times as much propellant compared to an infinitely powerful ION drive with same ISP.

SME =  SKE - SPE (frequently called the Hamiltonian equation) when SME is such that SKE>SPE any positive SME is multiplied by 2 and SQRT of which is velocity at escape, that is the transfer velocity. S = specific, M = mechanical, E = energy, K = kinetic, P = potential.

SPE = µ/r and r is to be calculated at the semimajor axis a. Since KE = v2/2 the change of KE, SME is greatest when dV is applied at the highest possible velocity in an orbit . . .this is the core basis of the oberth effect. And any SKE you gain in excess of SPE at Pe is yours to keep when you escape. As you approache Pe your velocity can be calculated with the Vis-visa equation. This can be compared with the PE of  a circular orbit of same radius (r), from this you can calculate the SKE you have, that you  need and the SPE that you need to overcome.

For that reason Spiral orbits and at  the last pass of the planet power inefficient burns are both counterproductive.

 

 

[K^2]

7 hours ago, Steel said:

Unfortunately, once you get beyond simple Hohmann transfers - which can be solved pretty simply and analytically - usually the only way to accurately come up with an answer is to solve the equations numerically.

Low-thrust spiral transfer has a very simple analytical solution. Albeit, derivation is slightly more involved than Hohmann. dE/dt = v.F always, and E = -mv²/2 for circular orbit. Taking derivative of later to simplify the former, we have dv/dt = -F/m. Integral on the left side is the difference in velocities between two orbits. Integral on the right side is the total delta-V. So delta-V required is literally sqrt(μ/r₁) - sqrt(μ/r₂).

Which, by the way, @PB666, comes out to only sqrt(2) more delta-V for burn to SOI, unless I missed a factor somewhere in my math.

[PB666]

1 hour ago, K^2 said:

Low-thrust spiral transfer has a very simple analytical solution. Albeit, derivation is slightly more involved than Hohmann. dE/dt = v.F always, and E = -mv²/2 for circular orbit. Taking derivative of later to simplify the former, we have dv/dt = -F/m. Integral on the left side is the difference in velocities between two orbits. Integral on the right side is the total delta-V. So delta-V required is literally sqrt(μ/r₁) - sqrt(μ/r₂).

Which, by the way, @PB666, comes out to only sqrt(2) more delta-V for burn to SOI, unless I missed a factor somewhere in my math.

Its correct, but only from the starting radius, if the Pe increases significantly then its more. The key point is that if you burn only at Pe to achieve SOI+ you never have to pay the cost circularizing the orbit, which just before SOI would entail a burn from almost zero velocity to circularize. In a spiral orbit as you increase orbital momentum the speed with which you burn goes down as you rise, and thus simply stated

Given a prograde thrust*time of defined total length (in dV) the higher average speed of your burn contributes more energy toward escape than any set of situations where the average speed over all moments of burn is lower.

 

[K^2]

That shouldn't matter. At infinity, your orbital speed is zero, so direction and eccentricity become irrelevant. Id est, no circularization at infinity. It's the same reasoning for why escape velocity is direction-independent. So your claim of double the delta-V requirement and mine of sqrt(2) increase are at odds.

I'll check the math later with a simulation, because I'm curious. I just wanted to see if you can spot any errors in my derivations, perhaps.

[PB666]

3 hours ago, K^2 said:

That shouldn't matter. At infinity, your orbital speed is zero, so direction and eccentricity become irrelevant. Id est, no circularization at infinity. It's the same reasoning for why escape velocity is direction-independent. So your claim of double the delta-V requirement and mine of sqrt(2) increase are at odds.

I'll check the math later with a simulation, because I'm curious. I just wanted to see if you can spot any errors in my derivations, perhaps.

Thats true but a spiral orbit can be seen as semi-ciruclarizing at each point in the process, its not key, its just a notation. What is key is adding prograde energy at the less that maximal speed (as close to the tolerable atmosphere as possible), which is profoundly made worse in burns beyond the parabola because the energy in excess of SPE you get to keep as SKE. and that translates as V = SQRT(2*SKE) and so it makes. It is certainly true that if you burn to say r = 10,000, circularlize, then burn again to exit, you will have spent more dV. 

 

[Delusional Poet]

I don’t know if it applies in KSP, but in the real world between 2 circular orbits dV=V2-V1.  For an orbit radius increase about 400%, this is about 11% less efficient than a Hohmann Transfer.