In this document, I derive the equations of motion for an ion-thruster powered spacecraft and use numerical methods to calculate its optimal trajectory to Saturn. I did this work within 48 hours for the University Physics Competition in 2020.

1. Problem A: Optimal Trajectory to Saturn for Ion-Thruster
Powered Spacecraft
Kristopher Kerames
November 2020
Figure 1. Spacecraft Propelled by an Ion Thruster [2]

2. Kerames 2
Abstract
Ion propulsion uses too low a thrust to be able to perform Hohmann transfer like most rocket-
powered spacecraft do. Instead of solving for a Hohmann transfer trajectory, a system of second
order partial differential equations were derived to define the required motion of the spacecraft
from Euler’s laws of motion. Numerical methods were used to solve those equations. A constant
force of thrust was used throughout the entire trajectory. It was found that the most fuel-efficient
trajectory involved using a constant thrust of 0.1N, despite that the spacecraft was capable of
0.4N of thrust. This trajectory took 25.9 years, and only used 2,085kg of fuel. A 7.5 year
trajectory was the shortest one calculated and could be achieved using only about 330kg more
fuel than the most fuel-efficient trajectory. Some of the assumptions made in the development of
the solution prevented the final trajectory from being as efficient as it could be. The main ways
in which the solution may have been improved would be by using elliptical rather than circular
orbits, and using a gravity assist maneuver.

3. Kerames 3
Problem Definition
In order to travel from Earth to Saturn, a propulsion system is needed. The problem states
that an Ion Thruster spacecraft is in a circular orbit around the Earth with an orbital period of 90
minutes and needs to enter a circular orbit around Saturn with an orbital period of 40 hours. The
spacecraft has relatively low thrust in comparison to chemical propellants. The ion thrusters can
produce 400mN of thrust and have a specific impulse of 4000 seconds. The minimum amount of
fuel required, duration and control strategy for the thruster need to be analyzed to reach Saturn in
the most fuel efficient manner possible.
Analysis
Assumptions
Several assumptions were made in this analysis. The first assumption is that Saturn and
Earth orbit in the same plane. It was found that Saturn’s and Earth’s orbital planes only varied by
2.485° In reality, planetary orbits are elliptical however, in order to simplify the problem, the
assumption was made that planetary orbits are circular due to their low eccentricity. Earth’s
eccentricity is 0.0167 and Saturn’s eccentricity is 0.0565 according to the Goddard Space Flight
Center [1]. Another assumption is that the Sun was fixed and did not move with respect to time.
The Sun was used as a fixed reference point. Furthermore, the assumption was made that
standard gravity is 9.8 m/s2 Standard gravity is used in our specific impulse calculations and by
convention is 9.8m/s2
. The next assumption is that the trajectory is free from obstructions and
gravity other than that of Earth and Saturn. For instance, it was assumed that the spacecraft
would not get hit by an asteroid in the asteroid belt and that the spacecraft would be launched at
a position where gravitational influences from other planets would be small along its trajectory.
The effects of general relativity were also disregarded. The final assumption made is that each
planet can be approximated as a point particle, containing all of the planet’s mass, positioned at
their center of mass.
General Approach
The general approach is to use Euler’s laws of motion to write a governing equation for
acceleration of the spacecraft. This equation is a non-linear, two-dimensional differential
equation. Thus, a numerical method was used to calculate the time it would take to reach Saturn
with the speed that results in an orbital period of 40 hours. Namely, Euler’s method was used.
This governing equation accounts for the varying phase angle between the planets, and varying
thrust. The magnitude of thrust will remain constant throughout the trajectory, but will be varied
in each iteration. By leaving these quantities variable, the launch point and thrust force that
produce the most efficient trajectory can also be found. The most efficient way to use fuel in a
low-power thruster is to minimize thrust while keeping it constant. For this reason, FT will be
varied [6]. Given the thrust of 400mN (0.4N) that the spacecraft is capable of, the thrust will be
varied from 0–0.4N.
Generally, Hohmann transfer is used to launch rockets to other planets. However, a large
amount of thrust is required during Hohmann transfer in order to propel the spacecraft to escape
velocity, then later slow it down causing it to trace the orbit of the destination planet. The ion
thruster produces relatively low thrust. Therefore, the thruster will need to be left on for a longer
duration, rather than applying quick pulse of thrust that propels the spacecraft to escape velocity.

4. Kerames 4
Therefore, escape velocity will not need to be reached. Instead, the spacecraft will accelerate
continuously, spiraling away from the Sun until it reaches the final desired position and final
speed, Vdest. The spacecraft will be launched at the point where its tangential velocity, with
respect to Earth’s center, is in the same direction as the tangential velocity of Earth’s center. This
is the point where the craft is moving fastest with respect to the Sun, and its kinetic energy is
maximized.
Derivation of Governing Equation
Figure 2. Free Body Diagram of the Spacecraft.
The specific impulse in seconds can be defined by the following equation:
(1)
where 𝐹! is the thrust force in Newtons (N), 𝑔" = 9.8 𝑚/𝑠2 is standard gravity. 𝐼#$ is the
specific impulse measured in seconds. 𝑚
̇ is the mass flow rate in kg/s.
By dividing both sides of the equation above by 𝑚
̇ , the exhaust velocity, 𝑣%& is obtained:
(2)
The thrust force can be expressed as the product of the mass flow rate, ṁ and exhaust velocity:
(3)
𝐹
⃑! = −𝑔" ∗ 𝐼#$ ∗ 𝑚
̇
𝑣%& = 𝐼#$ ∗ 𝑔"
𝐹
⃑! = −𝑚
̇ ∗ 𝑣%&

5. Kerames 5
After substituting equation (2) into (3) and plugging in the given values for 𝐼#$ and 𝑔", the mass
flow rate can be expressed as:
(4)
Where m(t) represents mass as a function of time. In order to find m(t), we integrate the mass
flow rate over time:
(5)
In addition, the rocket equation including external forces and constant thrust is:
(6)
Where a(t) is acceleration as a function of time, and ∑ 𝐹 is the sum of external forces on the
rocket, which also vary with time. From equations (3) and (6), we get:
(7)
Where 𝑅
4⃑#'(𝑡) is the position vector of the spacecraft with respect to the Sun.
Aside from thrust, the external forces on the spacecraft are only the forces of gravity due to Sun,
Earth, and Saturn. The equation for the force due to gravity is:
(8)
Where: G=6.674x10-11
Nm2
/kg2
, M is the mass of the gravitational body, r is the magnitude of
the position vector of the gravitational body with respect to the spacecraft, and 𝑟̂ is the unit
vector corresponding to r. The masses for M will be defined as follows:
• Mass of Earth: Me = 5.9724 E24 kg
• Mass of Saturn: Ms = 568.34 E24 kg
• Mass of Saturn: Msun = 1.9885 E30 kg
[3] [4] [5]
To obtain the position vectors of the planets with respect to the moving spacecraft, we will
define those vectors in terms of their positions in a fixed Cartesian coordinate system where the
Sun is fixed at the origin.
𝑚
̇ =
𝑑𝑚(𝑡)
𝑑𝑡
=
−𝐹
⃑!
39,200
> 𝑑𝑚
((*)
,"""
= >
−𝐹
⃑!
39,200
𝑑𝑡
*!
"
𝑚(𝑡) = 5000 −
𝐹
⃑!
39,200
@ 𝐹
⃑ − 𝑚
̇ ∗ 𝑣%& = 𝑚(𝑡) ∗ 𝑎
⃑(𝑡)
𝑎
⃑(𝑡) =
𝑑-
𝑅
4⃑#'(𝑡)
𝑑𝑡-
=
∑ 𝐹
⃑ − 𝐹
⃑!
𝑚(𝑡)
𝐹
. = 𝐺
𝑀𝑚(𝑡)
𝑟-
𝑟̂

6. Kerames 6
The planets’ average distances relative to the Sun (semi-major axis) are defined as follows:
• Earth’s semi-major axis Re = 149.6 E9 m
• Saturn’s semi-major axis: Rs = 1,433.5 E9 m
First, the positions of each object relative to the Sun must be defined:
Figure 3. Angle Diagram. Figure is not to scale.
𝑅
4⃑# , 𝑅
4⃑% , 𝑅
4⃑#/0, and 𝑅
4⃑#' are the position vectors of Saturn, Earth, Sun, and the spaceship with
respect to the Sun, respectively. Let θi = 0° be the position of the Earth when the spacecraft is
launched. 𝜙 is the phase angle between Earth and Saturn. Δ𝜙 is the change in angle of
Saturn over some time interval.
The angular speeds of Earth and Saturn about the Sun are ωe and ωs, respectively. These can be
related to θ and Δ𝜙 with:
𝑅
4⃑# and 𝑅
4⃑% can be rewritten in terms of θ and Δ𝜙. The position vectors become:
(9)
𝑅
4⃑#
𝑅
4⃑%
ωet= θ and ωst= Δ𝜙
𝑅
4⃑% = 𝑅% cos(ω%𝑡) 𝚤̂ + 𝑅% sin(ω%𝑡) 𝚥̂
𝑅
4⃑# = 𝑅# cos(ω#𝑡 + ϕ) 𝚤̂ + 𝑅# sin(ω#𝑡 + ϕ) 𝚥̂
𝑅
4⃑#/0 = 0𝚤̂ + 0𝚥̂
𝑅
4⃑#' = 𝑥𝚤̂ + 𝑦𝚥̂

7. Kerames 7
Earth’s angular speed of revolution is
ωe = 2 * π / T e
Similarly, Saturn’s revolution speed is,
ωe = 2 * π / T s
Where T e = 365.25 days is the period of Earth’s rotation around the Sun, and T s = 29.46 years
is the period of Saturn’s rotation around the Sun.
Now, position vectors of the gravitational bodies with respect to the spacecraft can be developed.
These will be used to describe the forces acting on the spacecraft.
Figure 4. Position Vector Diagram (not to scale). The spacecraft is drawn in a moving, body-
fixed coordinate system.
𝑅
4⃑%/#' , 𝑅
4⃑#/#' , and 𝑅
4⃑#/0/#' are the position vectors of Earth, Saturn, and the Sun with respect to
the spacecraft, respectively. They can be written in terms of the other vectors:
𝑅
4⃑#
𝑅
4⃑%
𝑅
4⃑#'
𝑅
4⃑#/#'
𝑅
4⃑%/#'

8. Kerames 8
(10)
By plugging in equation set (9) into equation set (10) we get equations for the position vectors,
with respect to the craft’s body-fixed coordinate system, in terms of x and y:
(11)
The net forces on the craft can now be written in terms of these vectors. In general, from
equation (8),
𝐹
. = 𝐺
𝑀𝑚(𝑡)
𝑟-
𝑟̂
r can be calculated as e𝑅
4⃑e, where 𝑅
4⃑ is a general variable to represent each of the position vectors
in equation set (10). 𝑟̂ can be calculated as
2
3⃑
52
3⃑5
Thus, Fg for each gravitational body becomes:
(12)
Where e𝑅
4⃑e = f𝑅&
-
+ 𝑅6
-
⇒ e𝑅
4⃑e
8
= h𝑅&
-
+ 𝑅6
-
i
8/-
𝑅
4⃑%/#' = 𝑅
4⃑% − 𝑅
4⃑#'
𝑅
4⃑#/#' = 𝑅
4⃑# − 𝑅
4⃑#'
𝑅
4⃑#/0/#' = −𝑅
4⃑#'
𝑅
4⃑%/#' = (𝑅% cos(ω%𝑡) − 𝑥)𝚤̂ + (𝑅% sin(ω%𝑡) − 𝑦) 𝚥̂
𝑅
4⃑#/#' = (𝑅# cos(ω#𝑡 + ϕ) − 𝑥) 𝚤̂ + (𝑅# sin(ω#𝑡 + ϕ) − 𝑦) 𝚥̂
𝑅
4⃑#/0/#' = −𝑥𝚤̂ − 𝑦𝚥̂
𝑟̂
𝑟-
=
𝑅
4⃑
e𝑅
4⃑e
÷ e𝑅
4⃑e -
=
𝑅
4⃑
e𝑅
4⃑e
8
𝐹
. = 𝐺𝑀𝑚(𝑡)
𝑅
4⃑
e𝑅
4⃑e
8

9. Kerames 9
Equation (7) can now be rewritten as:
(13)
As this equation is two-dimensional, it must be split into two partial differential equations in the
x and y directions. If the spacecraft is always firing its ion thruster in the direction of its velocity,
then 𝐹
⃑&! and 𝐹
⃑6! can be defined as:
(14)
Complete Set of Governing Equations
By plugging in equations (14), (11), and (5) into (13), the final set of governing equations
becomes:
(15)
𝑑-
𝑅
4⃑#'(𝑡)
𝑑𝑡-
=
∑ 𝐹
⃑ + 𝐹
⃑!
𝑚(𝑡)
=
𝐺𝑀%𝑚(𝑡)
𝑅
4⃑%/#'
e𝑅
4⃑%/#'e
8 + 𝐺𝑀#𝑚(𝑡)
𝑅
4⃑#/#'
e𝑅
4⃑#/#'e
8 + 𝐺𝑀#/0𝑚(𝑡)
𝑅
4⃑#/0/#'
e𝑅
4⃑#/0/#'e
8 + 𝐹
⃑!
𝑚(𝑡)
= 𝐺 k
𝑀%
e𝑅
4⃑%/#'e
8 𝑅
4⃑%/#' +
𝑀#
e𝑅
4⃑#/#'e
8 𝑅
4⃑#/#' +
𝑀#/0
e𝑅
4⃑#/0/#'e
8 𝑅
4⃑#/0/#'l +
𝐹
⃑!
𝑚(𝑡)
𝐹
⃑&! = 𝐹!
𝑑𝑥
𝑑𝑡
f(
𝑑𝑥
𝑑𝑡
)- + (
𝑑𝑦
𝑑𝑡
)-
𝐹
⃑6! = 𝐹!
𝑑𝑦
𝑑𝑡
f(
𝑑𝑥
𝑑𝑡
)- + (
𝑑𝑦
𝑑𝑡
)-
𝑑"
𝑥
𝑑𝑡"
= 𝐺 &
𝑀#(𝑅# cos(ω#𝑡) − 𝑥)
0𝑅
1⃑#/%&0
' +
𝑀%(𝑅% cos(ω%𝑡 + ϕ) − 𝑥)
0𝑅
1⃑%/%&0
' +
𝑀%()(−𝑥)
0𝑅
1⃑%()/%&0
'5 +
𝐹
11⃑
𝑥𝑇
5000 −
𝐹*
39,200
𝑑"
𝑦
𝑑𝑡"
= 𝐺 &
𝑀#(𝑅# cos(ω#𝑡) − 𝑦)
0𝑅
1⃑#/%&0
' +
𝑀%(𝑅% cos(ω%𝑡 + ϕ) − 𝑦)
0𝑅
1⃑%/%&0
' +
𝑀%()(−𝑦)
0𝑅
1⃑%()/%&0
'5 +
𝐹
11⃑
𝑦𝑇
5000 −
𝐹*
39,200

10. Kerames 10
Results
MATLAB was used to perform Euler’s method using the governing equations. A
convergence study was conducted and revealed that a 300-second time increment was sufficient
for convergence of results. The minimum amount of fuel necessary to arrive at Saturn with the
desired velocity is 2,085kg of fuel. This is equal to 41.7% of the rocket’s initial mass. The
magnitude of constant thrust required was 0.1N. The phase angle between the Earth and Saturn
was -13π/30 radians. It took 25.9 years to arrive at Saturn given these conditions. Alternative
conditions resulted in trajectories as short as 7.5 years using 0.4N of thrust, and consuming
2,415kg of fuel. A graph of the optimal trajectory can be seen below (Fig. 5). The speed of the
craft increases as it approaches Saturn indicating that it is using the planet’s momentum to
increase its speed (Fig. 6). The distance of the craft from the Sun increases at an accelerated rate
over time as spirals get larger (Fig. 7).
Figure 5. Graph of most efficient trajectory. The orange dot is the sun. The yellow dot is Saturn.
The blue line traces the trajectory of the spacecraft.

11. Kerames 11
Figure 6. Speed of the craft over time.
Figure 7. Distance of the craft from the Sun over time.

12. Kerames 12
Conclusion
The most efficient trajectory was found to use only 2,085kg of fuel. However another
trajectory was found to last 7.5 years using only 330kg more fuel than the most efficient
trajectory did. This shorter trajectory may be more practical, as time constraints may outweigh
the cost of fuel in some contexts. Results confirmed that keeping the ion thrusters on a low thrust
for the entire duration of the trip was the most efficient way to travel.
Results may have been improved by avoiding some of the assumptions that were made.
Using elliptical orbits, rather than circular, would have improved the accuracy of results.
Depending on how the elliptical orbits are oriented, the shortest distance between them is
considerably shorter than that of circular orbits. This could have reduced the overall distance of
the trip, and saved fuel. Another thing that would have made the most material impact on results
would have been to use a gravity assist maneuver. Mars and Jupiter are positioned such that they
could be used for this maneuver. That would have allowed the spacecraft to use some of those
planets’ angular momenta in order to speed itself up without using additional fuel. By adding the
forces due to Mar’s and Jupiter’s gravity to the governing equations, and performing more phase
angle iterations in MATLAB, the optimal trajectory using gravity assist could be found.

13. Kerames 13
References
[1] Bombardelli, C., Baù, G. & Peláez, J. Asymptotic solution for the two-body problem
with constant tangential thrust acceleration. Celest Mech Dyn Astr 110, 239–256
(2011). https://doi-org.lib-proxy.fullerton.edu/10.1007/s10569-011-9353-3
[2] Dunbar, B. (n.d.). Ion Propulsion: Farther, Faster, Cheaper. Retrieved November 08,
2020, from http://www.nasa.gov/centers/glenn/technology/Ion_Propulsion1.html
[3] Earth Fact Sheet. (n.d.). Retrieved November 08, 2020, from
https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html
[4] Planetary Fact Sheet. (n.d.). Retrieved November 08, 2020, from
https://nssdc.gsfc.nasa.gov/planetary/factsheet/
[5] Saturn Fact Sheet. (n.d.). Retrieved November 08, 2020, from
https://nssdc.gsfc.nasa.gov/planetary/factsheet/saturnfact.html
[6] S. X. staff, “Researcher calculates optimal trajectories to Mars and Mercury for a
spacecraft with electric propulsion,” Phys.org, 12-Sep-2019. Available:
https://phys.org/news/2019-09-optimal-trajectories-mars-mercury-spacecraft.html.
[Accessed: 08-Nov-2020].

14. Kerames 14
Appendix
MATLAB Code:
%Author: Team 124
%This script uses Euler's method to solve a governing equation for the
%trajectory of a spacecraft, powered by an ion thruster, traveling
%from Earth to Saturn.
clc; clear; close;
%The following lines store the required constants.
Re=149.6e9; %[m] Average distance from Sun to Earth
Rs=1433.5e9; %[m] Average distance from Sun to Saturn
Vsrev=9680.678558; %[m/s] Velocity with which Saturn revolves around Sun
we=1.991e-7; %[rad/s] Angular velocity of Earth about the Sun
ws=6.7584e-9; %[rad/s] Angular velocity of Saturn about the Sun
Me=5.9724e24; %[kg] Mass of Earth
Ms=568.34e24; %[kg] Mass of Saturn
Msun=1.9885e30; %[kg] Mass of Sun
Vdest=21509.10786; %[m/s] Desired velocity when entering Saturn's orbit
G=6.674e-11; %[Nm^2/kg^2] Gravitational constant
phi=(-78)*pi/180:-pi/3500:-15175*pi/35000; % [rad] This varies the phase
angle, phi,
%for each trajectory.
FT=0.1:0.1:0.4; %0:0.1:0.4; [N] This varies the constant thrust for each
trajectory
dt=300; %[s] This is the time increment of ten days
tend=25.9*365.25*24*3600; %[s] This is the max possible time for spaceship's
trajectory
t=0:dt:tend; %[s] Varies time
%Turns variables into matrices with as many elements as time steps to
%store values from each iteration.
x=zeros(1,length(t));
y=zeros(1,length(t));
Vx=zeros(1,length(t));
Vy=zeros(1,length(t));
Vmag=zeros(1,length(t));
xddot=zeros(1,length(t));
yddot=zeros(1,length(t));
FTx=zeros(1,length(t));
FTy=zeros(1,length(t));
Msc=zeros(1,length(t));
Rmag=zeros(1,length(t));
for l=FT
%The following are initial conditions
X0=1.4961e11; %[m] X component of initial position
Y0=0; %[m] Y component of initial position
Vx0=0; %[m/s] X component of initial velocity
Vy0=we*Re+7740.584; %[m/s] Y component of initial velocity. Made up of
%tangential speed of Earth with respect to Sun plus tangential speed of
%spacecraft with respect to Earth.
FTx0=0; %[N] x component of initial thrust
FTy0=l; %[N] y component of initial thrust
Msc0=5000; %[kg] Initial mass of spacecraft
Rmag0=(X0^2+Y0^2)^(1/2);

15. Kerames 15
%Stores initial values in matrices
x(1)=X0;
y(1)=Y0;
Vx(1)=Vx0;
Vy(1)=Vy0;
FTx(1)=FTx0;
FTy(1)=FTy0;
Msc(1)=Msc0;
Vmag(1)=Vy0;
Rmag(1)=Rmag0;
%The following for loop iterates through different values of t to find the
%one that maximizes final mass.
for k=phi
for j=2:length(t)
if Rmag(j)<Rs
n=(j-2)*dt; %This is the time elapsed
Rxsunsc=-x(j-1); %x coordinate of Sun (sun) position with respect to
spacecraft (sc)
Rysunsc=-y(j-1); %y coordinate of Sun (sun) position with respect to
spacecraft (sc)
Rxesc=Re*cos(we*n)-x(j-1); %x coordinate of Earth (e) position with respect
to spacecraft (sc)
Ryesc=Re*sin(we*n)-y(j-1); %y coordinate of Earth (e) position with respect
to spacecraft (sc)
Rxssc=Rs*cos(ws*n+k)-x(j-1); %x coordinate of Saturn (s) position with
respect to spacecraft (sc)
Ryssc=Rs*sin(ws*n+k)-y(j-1); %y coordinate of Saturn (s) position with
respect to spacecraft (sc)
Mag3Rsunsc=(Rxsunsc^2+Rysunsc^2)^(3/2); %Cubed magnitude of position of
%Sun (sun) with respect to spacecraft (sc)
Mag3Resc=(Rxesc^2+Ryesc^2)^(3/2); %Cubed magnitude of position of
%Earth (e) with respect to spacecraft (sc)
Mag3Rssc=(Rxssc^2+Ryssc^2)^(3/2); %Cubed magnitude of position of
%Saturn (s) with respect to spacecraft (sc)
%The following equation is the governing equation for motion in
%x-direction. xddot is the x component of acceleration of the spacecraft.
xddot(j-1)=G*(Me*Rxesc/Mag3Resc+Ms*Rxssc/Mag3Rssc+Msun*Rxsunsc/Mag3Rsunsc)...
+FTx(j-1)/Msc(j-1);
%Similarly, the equation in the y-direction is below.
yddot(j-1)=G*(Me*Ryesc/Mag3Resc+Ms*Ryssc/Mag3Rssc+Msun*Rysunsc/Mag3Rsunsc)...
+FTy(j-1)/Msc(j-1);
%The following are equations for velocity, position, and thrust respectively.
Vx(j)=Vx(j-1)+xddot(j-1)*dt;
Vy(j)=Vy(j-1)+yddot(j-1)*dt;
x(j)=x(j-1)+Vx(j-1)*dt;
y(j)=y(j-1)+Vy(j-1)*dt;
Vmag(j)=(Vx(j)^2+Vy(j)^2)^(1/2);
Msc(j)=Msc0-l*n/39200;
Rmag(j)=(x(j)^2+y(j)^2)^(1/2);
FTx(j)=l*Vx(j-1)/Vmag(j-1);
FTy(j)=l*Vy(j-1)/Vmag(j-1);
end
end
plot(x, y); %Plots spacecraft's trajectory around sun, and final position of
Saturn.
hold on

16. Kerames 16
plot(0, 0, 'marker', '.', 'markersize', 50);
plot(Rs*cos(ws*(length(t)-2)*dt+k), Rs*sin(ws*(length(t)-2)*dt+k),...
'marker', '.', 'markersize', 25);
title('Optimal Trajectory');
xlabel('x');
ylabel('y');
end
end