Differential Equations in
Orbital Mechanics
David Bowling - Nate Youmans - Aidan Wright
The History of Astrodynamics
Since forgotten times, celestial bodies and their orbits have captured the fascination
of astronomers and mathematicians alike. Perhaps this fascination was due to the inexplicably periodic nature of the heavenly realm, with so many patterns and consistencies.
Such consistency made predictions simple, yet the cause of their behavior was often an
enigma. Such widespread curiosity resulted in many influential scientists starting their
careers by studying the stars. Knowledge of orbital operations and the forces involved
in orbital dynamics is essential to fully understand the solar system. We understand
these systems primarily through the work of two scientists. Johannes Kepler developed
laws that describe the motion of planets, while Newton’s laws allow us to understand
why the orbits behave in such a way. Kepler’s laws allow us to visualize and understand
the behavior of orbiting bodies. Newton’s three laws and law of universal gravitation
provides the answer to the why behind Kepler’s laws and the fundamental laws behind
orbital mechanics. The law of universal gravitation states that the force of gravitational
attraction between two objects is directly proportional to the masses of the objects and
inversely proportional to the square of the distance between their two centers and is
one of the most important concepts in orbital mechanics. The tool that we use to apply these laws and understand their behaviors is differential calculus. As current students of mathematics, the tradition of studying the stars is now passed down to us.
As evidenced through scientists such as Pierre-Simon de Laplace (who discovered the
Laplace-Runge-Lenz vector used to describe orbital shapes), Carl Runge (who’s work
allowed for the numerical solutions to orbital equations), Joseph-Louis Lagrange (who
found special solutions now known as Lagrange points) and hundreds of others, many
a mathematician has begun his journey by raising his gaze to the heavens.
In this paper, we will do the same.
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Two-Body Dynamics
The most simple example of orbital mechanics exists as a two-body system, where
there are only two significant masses to
consider. The equations that describe a
two-body system can be solved to determine each body’s position at any given
time. Further constraints on the system
make the solution simpler to reach. If we
assume that one body is fixed in our inertial frame of reference (i.e. has a huge mass
compared to the other body), we can effectively assume that the forces applied to
that fixed body are negligible. The derivation of the equation that describes the state
of a two-body problem starts with Newton’s second law: 𝐹 = 𝑚𝑎. The force acting on the small body (gravity) is only dependent on the distance between the two
bodies. We can derive an equation that describes the center of mass of the system
(sometimes called the barycenter), the position of which is denoted by R:
𝑅=
𝑚1 𝑥1 + 𝑚2 𝑥2
𝑚1 + 𝑚2
The final piece to this puzzle is to utilize
a value called the reduced mass, denoted
with the symbol 𝜇. The reduced mass is the
effective inertial mass of the system and allows us to equate the masses of each body.
Solving for position functions gives us this
equation, where R(t) is the position of the
barycenter, and r(t) is the distance between
the two bodies at a given time.
𝑥1 (𝑡) = 𝑅(𝑡) +
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𝑚2
𝑟(𝑡)
𝑚1 + 𝑚2
𝑥1 (𝑡) = 𝑅(𝑡) −
𝑚2
𝑟(𝑡)
𝑚1 + 𝑚2
One interesting observation that arises
from this derivation is the acceleration of
R, the barycenter, of a two body system.
Adding the two original force equations
together gives us an expression that looks
2 𝑎2
like 𝑚1𝑚𝑎11 +𝑚
+𝑚2 = 0. This is very similar to the
equation for the position of the barycenter.
In fact, this is the equation that describes
the acceleration of the barycenter, which
is equal to zero. This means that in a two
body system, the center of mass drifts with
constant velocity.
Three-Body Dynamics
When an object of negligible mass undergoes the influence of two substantially
massive bodies and attempts to navigate
the system, many interesting situations
occur. We will only look at the circular
restricted three body problem (CR3BP),
which is a lot easier to gain insight than
a normal n-body problem. The restrictions
that we place upon the three body problem are as follows: there are three bodies,
they are all point sources of known mass,
they only interact gravitationally, and the
third body is significantly smaller than the
other two. Our system will model that of
the Earth, Moon, and a man-made satellite. We can assume that the earth and
moon orbit their barycenter in circular orbits, although the barycenter lies below
the earth’s surface. We will study a stable
orbit, known as the Arenstorf orbit, that
appears a solution to the equations for a
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CR3BP. This orbit, discovered by Richard Therefore, our previous equation becomes:
Arenstorf, was used as a basis in the Apollo
missions, and we will derive it.
r̈ 𝑎𝑏𝑠 = a𝑟𝑒𝑙 + Ω × (Ω × r) + 2Ω × v𝑟𝑒𝑙
√
𝑟12 = (𝑥1 − 𝑥2 )2
where:
Ω = Ωk̂
r = 𝑥 ı̂ + 𝑦̂ȷ + 𝑧 k̂
where 𝑥1 is the location of 𝑥1 relative to the
center of gravity.
𝑥2 = 𝑥1 + 𝑟12
−𝑚2
𝑟12
𝑥1 =
𝑚1 + 𝑚2
𝑚1
𝑥2 =
𝑟12
𝑚1 + 𝑚2
𝑚1
𝜋1 =
𝑚1 + 𝑚2
𝑚2
𝜋2 =
𝑚1 + 𝑚2
To describe the position of m in relation to
̂
the origin, we use r = 𝑥 ı̂ + 𝑦̂ȷ + 𝑧 k.
r1 = (𝑥 − 𝑥1 )̂ı + 𝑦̂ȷ + 𝑧 k̂
= (𝑥 + 𝜋2 𝑟12 )̂ı + 𝑦̂ȷ + 𝑧 k̂
r2 = (𝑥 − 𝑥2 )̂ı + 𝑦̂ȷ + 𝑧 k̂
= (𝑥 + 𝜋1 𝑟12 )̂ı + 𝑦̂ȷ + 𝑧 k̂
Here, we will define the absolute acceleration where 𝜔 is the initial angular velocity
which is constant. 𝜔 = 2𝜋
𝑇
̇
r̈ 𝑎𝑏𝑠 = a𝑟𝑒𝑙 +a𝐶𝐺 +Ω×(Ω×r)+ Ω×r+2Ω×v
𝑟𝑒𝑙
where, 𝑎𝑟𝑒𝑙 is the rectilinear acceleration
relative to Ω × (Ω × r), which is equal to
centripetal acceleration and 2Ω × v𝑟𝑒𝑙 is the
Coriolis acceleration. Since the velocity of
the center of gravity is constant, a𝐶 𝐺 = 0
and Ω̇ = 0 as the angular velocity of an
orbit is constant.
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ṙ = v𝐶𝐺 + Ω × r + v𝑟𝑒𝑙
̇ ȷ + 𝑧̇ k̂
v𝑟𝑒𝑙 = 𝑥̇ ı̂ + 𝑦̂
̈ ȷ + 𝑧̈ k̂
a𝑟𝑒𝑙 = 𝑥̈ ı̂ + 𝑦̂
Which, when we substitute, becomes:
r̈ = (𝑥̈ − 2Ω𝑦̇ − Ω2 𝑥)̂ı + (𝑦̈ − 2Ω𝑥̇ − Ω2 𝑦)̂ȷ + 𝑧̈ k̂
Newton’s second law of motion is 𝑚a =
𝐹1 + 𝐹2 , where 𝐹1 = − 𝐺𝑚𝑟 31 𝑚 r1 and 𝐹2 =
1
− 𝐺𝑚𝑟 31 𝑚 r2 . If we let 𝜇1 = 𝐺𝑚1 and 𝜇2 = 𝐺𝑚2 :
2
𝑚a = 𝐹1 + 𝐹2
𝑚𝜇1
𝑚𝜇2
𝑚a = − 3 r1 − 3 r2
𝑟1
𝑟2
𝜇1
𝜇2
a = − 3 r1 − 3 r2
𝑟1
𝑟2
(𝑥̈ − 2Ω𝑦̇ − Ω2 𝑥)̂ı + (𝑦̈ − 2Ω𝑥̇ − Ω2 𝑦)̂ȷ + 𝑧̈ k̂
= − 𝜇𝑟 31 r1 − 𝜇𝑟 32 r2
1
2
{ 𝜇
1
̂
= − 3 [(𝑥 + 𝜋2 𝑟12 )̂ı + 𝑦̂ȷ + 𝑧 k]−
𝑟1
}
𝜇2
̂
[(𝑥 + 𝜋1 𝑟12 )̂ı + 𝑦̂ȷ + 𝑧 k]
𝑟23
When we equate coefficients we find that,
𝜇1
𝜇2
𝑥̈ −2Ω𝑦̇ −Ω2 𝑥 = − 3 (𝑥 +𝜋2 𝑟12 )− 3 [(𝑥 +𝜋1 𝑟12 )
𝑟1
𝑟2
𝜇1
𝜇2
𝑦̈ − 2Ω𝑥̇ − Ω2 𝑦 = − 3 𝑦 − 3 𝑦
𝑟1
𝑟2
𝜇1
𝜇2
𝑧̈ = − 3 𝑧 − 3 𝑧
𝑟1
𝑟2
The three equations shown above are a
nonlinear system of ordinary differential
equations. As we are assuming the orbit is
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only on one plane, 𝑧 = 0, which results in
the final two equations.
𝜇1
𝜇2
𝑥̈ −2Ω𝑦̇ −Ω2 𝑥 = − 3 (𝑥 +𝜋2 𝑟12 )− 3 [(𝑥 +𝜋1 𝑟12 )
𝑟1
𝑟2
𝜇2
𝜇1
𝑦̈ − 2Ω𝑥̇ − Ω2 𝑦 = − 3 𝑦 − 3 𝑦
𝑟1
𝑟2
The initial velocity is 10.9 km/s, with a
flight path angle of 20◦ (The initial conditions of the Apollo mission). In order
to solve these equations, we must numerically integrate them. When plotted, the
Arenstorf orbit looks like this:
+2𝑤𝑒 𝑉𝑦 + 𝑥̈
𝑑𝑉𝑦
𝐺𝑀
3 0 𝐺𝑀𝑎𝑒2
𝑧2
= − 3 𝑦 − 𝐽2
𝑦(1 − 5 2 ) + 𝑤𝐸2 𝑦
5
𝑑𝑡
𝑟
2
𝑟
𝑟
+2𝑤𝑒 𝑉𝑥 + 𝑦̈
𝐺𝑀
3 0 𝐺𝑀𝑎𝑒2
𝑧2
𝑑𝑉𝑧
= − 3 𝑧 − 𝐽2
𝑧(1 − 5 2 ) + 𝑥̈
𝑑𝑡
𝑟
2
𝑟5
𝑟
Two of the solving methods for the orbital
differential equations are the Runge-Kutta
and Adam-Bashforth-Moulton numerical
solving methods. They are both valid for
solving the numerical integration of orbit determination and prediction, however
the Runge-Kutta method is single-step and
the Adam-Bashforth-Moulton method is
multi-step. The Runge-Kutta focuses on
the applications of differential equations of
the physics equations for position and velocity to calculate an initial value and then
solve an IVP differential equations problem.
1
𝑎𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 = (𝐾1 + 2𝐾2 + 2𝐾3 + 𝐾4 )
6
𝐾1 = 𝐹 (𝑃0 , 𝑉0 , 𝑎3𝑟𝑑 )
Numerical Techniques
The basic idea of numerical solving
methods is that we take real world
data and use equations to create predictions/approximations for the different
properties; such as position or velocity.
The below equation shows the derivatives
of the velocity vectors for satellites in the
x, y, and z planes.
𝐾2 = 𝐹 (𝑃𝐾1 , 𝑉𝐾1 , 𝑎3𝑟𝑑 )
𝐾3 = 𝐹 (𝑃𝐾2 , 𝑉𝐾2 , 𝑎3𝑟𝑑 )
𝐾4 = 𝐹 (𝑃𝐾3 , 𝑉𝐾3 , 𝑎3𝑟𝑑 )
1
𝑃𝐾𝑛+1 = 𝑃0 + 𝑉0 𝐾𝑛 𝑡 + 𝐾𝑛 × 𝛼(𝛽𝑡)2
2
𝑉𝐾𝑛+1 = 𝑉0 + 𝐾𝑛 𝛽𝑡
It can be expanded to larger orders by
increasing the number of Runge-Kutta co𝑑𝑉𝑥
𝐺𝑀
3 0 𝐺𝑀𝑎𝑒2
𝑧2
2
= − 3 𝑥 − 𝐽2
𝑥(1 − 5 2 ) + 𝑤𝐸 𝑥 efficients used.
𝑑𝑡
𝑟
2
𝑟5
𝑟
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For 𝛼
𝛽
𝐾2
1
4/27
𝐾3
4
1/18
𝐾4
4
1/12
𝐾5
4
1/8
𝐾6 36
1/54
𝐾7 720 1/4320
𝐾8 20
1/20
𝐾9 480 1/288
𝐾10 820 1/820
The Adams-Bashforth-Moulton numerical method is a multi-step process divided into a predictor and a corrector. The
predictor is called the Adams-Bashforth
method and the corrector is AdamsMoulton. One benefit of this method is
faster correction calculation. The AdamsBashforth method is a variation of Euler’s
Method, specifically the first degree algorithm. The Adams-Moulton’s first degree
method is called backwards Euler’s and
the second degree is trapezoidal rule. This
means when Adam-Bashforth-Moulton is
used for these calculations, it is used from
the third degree. For the third degree, three
initial values are required and these are
typically solved for by Runge-Kutta or similar methods. The predictor is then applied
with these values.
The corrector algorithm is then run and
adjusts the values.
𝑛
𝑎𝑐 = ℎ ∑ 𝐶𝑀 𝑎𝑖
𝑖=1
While these two methods have their respective advantages and disadvantages,
they can both be used to calculate positional data effectively.
Looking Forward
Orbital mechanics is one of the earliest calculated applications of differential
equations, yet it also remains a modern
field of incredible complexity. This duality
comes from the breadth and depth of the
field. Some people require a high degree
of precision for landing rovers on Mars
while others use these equations simply
to tell which planets they are looking at.
Both of these applications however, still require basic understanding of the differential equations involved in orbital mechanics. So the next time you solve an equation
or look up into the night sky, remember the
footsteps you walk in on your journey towards understanding the heavens.
𝑛
𝑎𝑝 = ℎ ∑ 𝐶𝐵 𝑎𝑖
𝑖=1
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