Analyzing the Parker Solar Probe flybys

1. Modulus of the exit velocity, some features of Orbit #2

First, using the data available in the reports, we try to compute some of the properties of orbit #2. This is not enough to completely define the trajectory, but will give us information later on in the process.


In [1]:
from astropy import units as u

In [2]:
T_ref = 150 * u.day
T_ref


Out[2]:
$150 \; \mathrm{d}$

In [3]:
from poliastro.bodies import Earth, Sun, Venus

In [4]:
k = Sun.k
k


Out[4]:
$1.3271244 \times 10^{20} \; \mathrm{\frac{m^{3}}{s^{2}}}$

In [5]:
import numpy as np
$$ T = 2 \pi \sqrt{\frac{a^3}{\mu}} \Rightarrow a = \sqrt[3]{\frac{\mu T^2}{4 \pi^2}}$$

In [6]:
a_ref = np.cbrt(k * T_ref ** 2 / (4 * np.pi ** 2)).to(u.km)
a_ref.to(u.au)


Out[6]:
$0.55249526 \; \mathrm{AU}$
$$ \varepsilon = -\frac{\mu}{r} + \frac{v^2}{2} = -\frac{\mu}{2a} \Rightarrow v = +\sqrt{\frac{2\mu}{r} - \frac{\mu}{a}}$$

In [7]:
energy_ref = (-k / (2 * a_ref)).to(u.J / u.kg)
energy_ref


Out[7]:
$-8.0283755 \times 10^{8} \; \mathrm{\frac{J}{kg}}$

In [8]:
from poliastro.ephem import Ephem
from poliastro.util import norm

from astropy.time import Time

In [9]:
flyby_1_time = Time("2018-09-28", scale="tdb")
flyby_1_time


Out[9]:
<Time object: scale='tdb' format='iso' value=2018-09-28 00:00:00.000>

In [10]:
r_mag_ref = norm(Ephem.from_body(Venus, flyby_1_time).rv()[0])
r_mag_ref.to(u.au)


Out[10]:
$0.72573132 \; \mathrm{AU}$

In [11]:
v_mag_ref = np.sqrt(2 * k / r_mag_ref - k / a_ref)
v_mag_ref.to(u.km / u.s)


Out[11]:
$28.967364 \; \mathrm{\frac{km}{s}}$

2. Lambert arc between #0 and #1

To compute the arrival velocity to Venus at flyby #1, we have the necessary data to solve the boundary value problem.


In [12]:
d_launch = Time("2018-08-11", scale="tdb")
d_launch


Out[12]:
<Time object: scale='tdb' format='iso' value=2018-08-11 00:00:00.000>

In [13]:
r0, _ = Ephem.from_body(Earth, d_launch).rv()
r1, V = Ephem.from_body(Venus, flyby_1_time).rv()

In [14]:
r0 = r0[0]
r1 = r1[0]
V = V[0]

In [15]:
tof = flyby_1_time - d_launch

In [16]:
from poliastro import iod

In [17]:
((v0, v1_pre),) = iod.lambert(Sun.k, r0, r1, tof.to(u.s))

In [18]:
v0


Out[18]:
$[9.5993373,~11.298552,~2.9244933] \; \mathrm{\frac{km}{s}}$

In [19]:
v1_pre


Out[19]:
$[-16.980821,~23.307528,~9.1312908] \; \mathrm{\frac{km}{s}}$

In [20]:
norm(v1_pre)


Out[20]:
$30.248465 \; \mathrm{\frac{km}{s}}$

3. Flyby #1 around Venus

We compute a flyby using poliastro with the default value of the entry angle, just to discover that the results do not match what we expected.


In [21]:
from poliastro.threebody.flybys import compute_flyby

In [22]:
V.to(u.km / u.day)


Out[22]:
$[648499.74,~2695078.4,~1171563.7] \; \mathrm{\frac{km}{d}}$

In [23]:
h = 2548 * u.km

In [24]:
d_flyby_1 = Venus.R + h
d_flyby_1.to(u.km)


Out[24]:
$8599.8 \; \mathrm{km}$

In [25]:
V_2_v_, delta_ = compute_flyby(v1_pre, V, Venus.k, d_flyby_1)

In [26]:
norm(V_2_v_)


Out[26]:
$27.755339 \; \mathrm{\frac{km}{s}}$

4. Optimization

Now we will try to find the value of $\theta$ that satisfies our requirements.


In [27]:
from poliastro.twobody import Orbit

In [28]:
def func(theta):
    V_2_v, _ = compute_flyby(v1_pre, V, Venus.k, d_flyby_1, theta * u.rad)
    ss_1 = Orbit.from_vectors(Sun, r1, V_2_v, epoch=flyby_1_time)
    return (ss_1.period - T_ref).to(u.day).value

There are two solutions:


In [29]:
import matplotlib.pyplot as plt

In [30]:
theta_range = np.linspace(0, 2 * np.pi)
plt.plot(theta_range, [func(theta) for theta in theta_range])
plt.axhline(0, color="k", linestyle="dashed");



In [31]:
func(0)


Out[31]:
-9.142672330001195

In [32]:
func(1)


Out[32]:
7.09811543934556

In [33]:
from scipy.optimize import brentq

In [34]:
theta_opt_a = brentq(func, 0, 1) * u.rad
theta_opt_a.to(u.deg)


Out[34]:
$38.598709 \; \mathrm{{}^{\circ}}$

In [35]:
theta_opt_b = brentq(func, 4, 5) * u.rad
theta_opt_b.to(u.deg)


Out[35]:
$279.3477 \; \mathrm{{}^{\circ}}$

In [36]:
V_2_v_a, delta_a = compute_flyby(v1_pre, V[0], Venus.k, d_flyby_1, theta_opt_a)
V_2_v_b, delta_b = compute_flyby(v1_pre, V[0], Venus.k, d_flyby_1, theta_opt_b)

In [37]:
norm(V_2_v_a)


Out[37]:
$29.978799 \; \mathrm{\frac{km}{s}}$

In [38]:
norm(V_2_v_b)


Out[38]:
$29.491925 \; \mathrm{\frac{km}{s}}$

5. Exit orbit

And finally, we compute orbit #2 and check that the period is the expected one.


In [39]:
ss01 = Orbit.from_vectors(Sun, r1, v1_pre, epoch=flyby_1_time)
ss01


Out[39]:
0 x 1 AU x 18.8 deg (HCRS) orbit around Sun (☉) at epoch 2018-09-28 00:00:00.000 (TDB)

The two solutions have different inclinations, so we still have to find out which is the good one. We can do this by computing the inclination over the ecliptic - however, as the original data was in the International Celestial Reference Frame (ICRF), whose fundamental plane is parallel to the Earth equator of a reference epoch, we have change the plane to the Earth ecliptic, which is what the original reports use.


In [40]:
ss_1_a = Orbit.from_vectors(Sun, r1, V_2_v_a, epoch=flyby_1_time)
ss_1_a


Out[40]:
0 x 1 AU x 24.0 deg (HCRS) orbit around Sun (☉) at epoch 2018-09-28 00:00:00.000 (TDB)

In [41]:
ss_1_b = Orbit.from_vectors(Sun, r1, V_2_v_b, epoch=flyby_1_time)
ss_1_b


Out[41]:
0 x 1 AU x 13.3 deg (HCRS) orbit around Sun (☉) at epoch 2018-09-28 00:00:00.000 (TDB)

In [42]:
from poliastro.frames import Planes

In [43]:
ss_1_a.change_plane(Planes.EARTH_ECLIPTIC)


Out[43]:
0 x 1 AU x 3.4 deg (HeliocentricEclipticIAU76) orbit around Sun (☉) at epoch 2018-09-28 00:00:00.000 (TDB)

In [44]:
ss_1_b.change_plane(Planes.EARTH_ECLIPTIC)


Out[44]:
0 x 1 AU x 12.8 deg (HeliocentricEclipticIAU76) orbit around Sun (☉) at epoch 2018-09-28 00:00:00.000 (TDB)

Therefore, the correct option is the first one.


In [45]:
ss_1_a.period.to(u.day)


Out[45]:
$158.75975 \; \mathrm{d}$

In [46]:
ss_1_a.a


Out[46]:
$85839412 \; \mathrm{km}$

And, finally, we plot the solution:


In [47]:
from poliastro.plotting import StaticOrbitPlotter

frame = StaticOrbitPlotter(plane=Planes.EARTH_ECLIPTIC)

frame.plot_body_orbit(Earth, d_launch)
frame.plot_body_orbit(Venus, flyby_1_time)
frame.plot(ss01, label="#0 to #1", color="C2")
frame.plot(ss_1_a, label="#1 to #2", color="C3");