Lambert method for in-plane transfer

Question

I'm trying to solve the transfer between a circular orbit with 200km radius and an orbit with 200km perigee and 8000km apogee.

The optimal transfer is hoffman, however, izzo.lambert from poliastro gives different result.

My code:

from astropy import units as u
from poliastro.bodies import Earth
from poliastro.iod import izzo
from poliastro.core.elements import coe2rv
from poliastro.util import norm
import math
import time

Earth_k = Earth.k
Req = Earth.R.to(u.km).value

def keplerian2cartesian(kepler):

    a=(kepler[0]+kepler[1]+Req*2)*1000/2 * u.m
    e=(kepler[0]-kepler[1])/(kepler[0]+kepler[1]+Req*2)

    (R,V)=coe2rv(Earth.k,a,e,kepler[2],kepler[3],kepler[4],kepler[5])

    return [R * u.m] + [V * u.m/u.s]

# Checking different transfer times
def transfer_time(DV_min,vector0,vector,period):

    init_t=10
    t_value=init_t
    (r0,r,v0,v)=(vector0[0],vector[0],vector0[1],vector[1])

    while init_t<period:

        tof = init_t * u.min

        try:
            (f_v0, f_v), = izzo.lambert(Earth_k, r0, r, tof)
        except:
            init_t+=1
            continue

        DV0=norm(f_v0-v0).value*1000
        DV=norm(f_v-v).value*1000

        if(DV0+DV<DV_min):
            t_value=init_t
            DV0_value=DV0
            DV_value=DV
            DV_min=DV0+DV

        init_t+=1

    return (t_value,DV_value,DV0_value)


# Checking different initial and final true anomalies
def transfer_tetta(kepler0,kepler):
    DV_min=100000
    final_tetta=0

    a=(kepler[0]+kepler[1]+Req*2)*1000/2
    period=math.ceil(math.sqrt((a**3/Earth.k.value)*4*(math.pi)**2)/60)

    while final_tetta<360:
        init_tetta=0
        while init_tetta<360:
            vector0=keplerian2cartesian(kepler0 + [init_tetta*math.pi/180])
            vector =keplerian2cartesian(kepler  + [final_tetta*math.pi/180])
            try:
                (init_t,DV,DV0)=transfer_time(DV_min,vector0,vector,period)
            except:
                init_tetta+=5
                continue

            if DV+DV0<DV_min:
                DV_min=DV+DV0
                (t_value,DV0_value,DV_value,init_tetta_value,final_tetta_value)=(init_t,DV0,DV,init_tetta,final_tetta)

            init_tetta+=5
        final_tetta+=5

    print("t: ", t_value, "DV_init: ", DV0_value,"DV_final: ", DV_value)
    print("DV: ", DV_min)
    print("Init tetta: ", init_tetta_value, "Final tetta: ", final_tetta_value)

# 1->2
transfer_tetta([200,  200, 64.3*math.pi/180, 0, 300*math.pi/180],[8000, 200, 64.3*math.pi/180, 0, 300*math.pi/180])

Answer 1

You need to take into account that the Hohmann transfer and the Lambert solution receive different inputs and are therefore not equivalent.

• The Hohmann transfer is known to be the optimal two-impulse transfer between two coplanar, circular orbits. There are several proofs to this. The initial and final states are defined by position and velocity. Therefore, we already know what are the initial and final velocities we want to achieve, and the Hohmann transfer just gives us the two impulses needed to reach the final state.
• The Lambert problem "is concerned with the determination of an orbit from two position vectors and the time of flight". In other words: the initial velocity before the transfer and the final velocity after the transfer are not specified. Therefore:
  • When the transfer angle is 180 degrees, there is an infinite number of trajectories with the same departure and arrival velocity that take us from the initial to the final position.
  • The algorithm doesn't know whether the initial and final orbits are coplanar or not, because there are no such orbits. Remember: we only have the position vectors.
  • The algorithm will tell us which velocities does the transfer orbit have at the initial and final position. But it's on us to compute the total cost of the transfer!
  • The time of flight is an input, not an output.

In my opinion they can't be compared, because they don't have the same amount of information about the problem. They serve different purposes. I might be wrong about this, but I would need you to describe in detail what exactly do you want to achieve.

If the objective is just checking that "given two positions separated 180 degrees and the time of flight that results from the Hohmann transfer, I want to see whether the Lambert method gives me the same solution", the question makes no sense, and the problem is ill-posed anyway.

Answer 2

If both orbits are in the same plane, then the optimal solution will be called hohmann transfer where you need to start from perigee of the inner orbit towards an apogee of the outer orbit. So you will start from 200 km perigee and end at 8000 km apogee.

I hope this is clear now.