An Introduction to Newtonian Celestial

Mechanics, and a comparison of Hohmann and

Bi-elliptic transfers

John T-R

March 9, 2019

1

Contents

1 Introduction 3

2 Basis of Newtonian Mechanics 32.1 Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Newton’s Law of Gravity . . . . . . . . . . . . . . . . . . . . . . 42.3 Formulation of the N-Body Problem . . . . . . . . . . . . . . . . 4

3 Constants of the System 63.1 Centre of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.2 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . 73.3 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

4 Ellipses 84.1 Equation for an ellipse in Cartesian coordinates . . . . . . . . . . 84.2 Equation for an ellipse in Polar coordinates centred at one focus 104.3 Area of an ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

5 Kepler’s Laws 125.1 Foundational Work . . . . . . . . . . . . . . . . . . . . . . . . . . 125.2 Kepler’s First Law . . . . . . . . . . . . . . . . . . . . . . . . . . 145.3 Kepler’s Second Law . . . . . . . . . . . . . . . . . . . . . . . . . 155.4 Kepler’s Third Law . . . . . . . . . . . . . . . . . . . . . . . . . . 16

6 Useful Orbital Equations 166.1 Equation linking orbital velocity, position, and semi-major radius 17

7 Transfer Orbits 187.1 Hohmann Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . 197.2 Bi-Elliptic Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . 19

8 A Qualitative Comparison via Matlab 198.1 Region A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228.2 Region B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228.3 Region C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228.4 Region D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238.5 Region E . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

9 Summary of this comparison 23

2

Figure 1: Kepler’s Laws demonstated (test image)

1 Introduction

Throughout all of human history, our ancestors have observed the heavenly bod-ies. Amongst the background of fixed points of light, seven ’plantai’, or ”wan-derers”, followed fixed yet seemingly arbitrary paths through the sky. However,these bodies were widely interpreted as the Seven Heavens, which has differentconnotations in different religions. Some examples include the layers of paradise(Christianity), the residences of the Prophets (Islam), or the seats of the Lord(Judaism). As such, the geocentric view of the universe was the most widelyaccepted; the Earth lay at the centre of the universe with all of the other bodiesorbiting around it along very complicated curves.However, throughout the late medieval period, Heliocentrism became more andmore widely accepted. This culminated in Newton using it as the foundationfor his theory of universal gravitation which accurately described the motionsof the planets. I will also show that this theory also satisfies Kepler’s Laws - aset of laws that were derived from observations of the planets.In this short essay, I will describe the foundations of Newtonian Celestial Me-chanics: Newton’s Laws of Motion, and a formulation of the N-Body problem. Iwill then show how Newton’s Law of Gravity upholds all of the laws of motion,as well as maintaining the constraints assigned to a closed system - constantenergy, momentum, and angular momentum.I will then take a slight detour to describe ellipses before demonstrating thatNewton’s Law of Gravity upholds all of Kepler’s Laws. Finally, I will usethis work to describe two methods of transferring between circular orbits - theHohmann and Bi-elliptic transfers - alongside a comparison between them.

2 Basis of Newtonian Mechanics

[1]

2.1 Laws

Newtonian mechanics is founded upon three fundamental laws:

3

1. Any body in motion remains moving at a constant velocity unless actedon by an exterior force

2. F = ṗ where p = mṙ

3. Every action has an equal and opposite reaction

2.2 Newton’s Law of Gravity

If two bodies A and B have masses mA and mB respecively, and the distancebetwee them is r, then the force acting on both of them has a magnitude of

Gmambr2

where G is a constant, and the direction of the force on A is towards B and viceversa.

2.3 Formulation of the N-Body Problem

Let

M =

n∑i=1

mi

be the total mass of the system, where mi is the mass of the ith particle, and

let

R =

n∑i=1

miri

M

be the Centre of Mass of the system, where ri is the position of the ith particle.

The total force on the ith particle can be given as

Fi = −Gn∑

j=1,j 6=i

mimj|ri − rj|2

· ri − rj|ri − rj|

Fi = −Gn∑

j=1,j 6=i

mimj|ri − rj|3

· (ri − rj)

From Newton’s Second Law, when we take all masses as constants,

Fi =d

dtmiṙ = mir̈i

Therefore

mir̈i = −Gn∑

j=1,j 6=i

mimj|ri − rj|3

· (ri − rj)

r̈i = −Gn∑

j=1,j 6=i

mj|ri − rj|3

· (ri − rj)

4

Let rij = ri − rj and rij = |rij|. Then we get

r̈i = −Gn∑

j=1,j 6=i

mjrijr3ij

If ri = xiex + yiey + ziez where ex, ey, and ez make up a basis, then wecan define the grad function as

∇i = ex∂

∂xi+ ey

∂

∂yi+ ez

∂

∂zi

Define the gravitational potential energy of a body within a system as theenergy required to move said body from infinity to its current position. Thiscan be expressed for one body as

V (r) = −∫ r∞

F · dr = −∫ r∞−GMm

r2· r · dr

V (r) = −∫ r∞−GMm

r2· dr = −GMm

r

Extrapolating this method out for all bodies in a system, we get

V (r) = −Gn∑j=1

n∑k=j+1

mjmkrjk

= −12G

n∑j=1

n∑k=1,k 6=j

mjmkrjk

Since everything besides rjk is constant,

∇iV (r) = −1

2G

n∑j=1

n∑k=1,k 6=j

mjmk · ∇i(

1

rjk

)

The only time where ∇i(

1rjk

)is nonzero is when either j or k is equal to i.

Thereforewe can cancel all non-zero terms out of our summation to get

∇iV (r) = −Gn∑

j=1,j 6=i

mimj · ∇i(

1

rij

)where

∇i( 1rij

)= ∇i((xi − xj)2 + (yi − yj)2 + (zi − zj)2)−

12

= − (xi − xj)ex + (yi − yj)ey + (zi − zj)ezr3ij

= − rijr3ij

This then gives

∇iV (r) = Gn∑

j=1,j 6=i

mimj ·rijr3ij

5

Comparing this with

r̈i = −Gn∑

j=1,j 6=i

mjrijr3ij

we note that∇iV (r) = −mir̈i

3 Constants of the System

[2]There are three constants in any closed system:

1. The Centre of Mass should move at a constant velocity (This is the sameas conservation of Linear Momentum)

2. The Angular Momentum should remain constant

3. The total energy should remain constant

I will now demonstrate that Newton’s Law of Gravity upholds these constants.

3.1 Centre of Mass

By summing all forces,

n∑i=1

mir̈i = −G∑

(i,j)≤n,i6=j

mimjrijr3ij

where the left hand side is the sum of the overall force on each body, and theright hand side is the sum of all gravitational forces. By Newton’s Third Law,we know that the right hand side consists of pairs of forces, which cancel out.Therefore

n∑i=1

mir̈i = 0

Integrating both sides twice gives

n∑i=1

miri = at+ b

where a and b are constants of integration. This results in

R =

n∑i=1

miri

M=

at+ b

M

Therefore the Centre of Mass moves at a constant velocity.

6

3.2 Angular Momentum

We can express the total angular momentum as

L =

n∑i=1

(miri × ṙi)

Differentiating both sides gives

L̇ =d

dt

( n∑i=1

(miri × ṙi))

L̇ =

n∑i=1

(miṙi × ṙi) +n∑i=1

(miri × r̈i)

All the terms in the left hand side are zero (Since the two vectors are the sametherefore parallel to each other). This then gives

L̇ =

n∑i=1

(miri × r̈i)

Substituting the value of r̈i in, we get

L̇ =

n∑i=1

miri ×(−G

n∑j=1,j 6=i

mjrijr3ij

)= −G

∑(i,j)≤n,i 6=j

mimj · (ri × rij)r3ij

L̇ = −G∑

(i,j)≤n,i 6=j

mimj · (ri × (ri − rj))r3ij

= G∑

(i,j)≤n,i 6=j

mimj · (ri × rj)r3ij

All the terms in the right hand sum come in equal and opposite pairs, so theycancel out. Therefore

L̇ = 0

Therefore the angular momentum is constant.

3.3 Energy

The total work done on each body can be expressed as

n∑i=1

mir̈i · ṙi = −n∑i=1

ṙi · ∇iV = −n∑i=1

(ẋi∂V

∂xi+ ẏi

∂V

∂yi+ żi

∂V

∂zi

)V is a function dependent on all the positions of every particle, therefore by thechain rule

n∑i=1

mir̈i · ṙi = −dV

dt

7

Figure 2: An example of an ellipse

Integrating both sides gives∫ n∑i=1

mir̈i · ṙidt = −V + E

1

2

n∑i=1

miṙi · ṙi = −V + E

where E is a constant of the integration Note that the left hand side is the sumof all the kinetic energy. Therefore we can write

T =1

2

n∑i=1

mi|ṙi|2 = −V + E

E = T + V

Therefore the total energy of the system is constant.

4 Ellipses

A circle can be constructed by ensuring that the distance to a point called afocus remains constant. In the same way, we can construct an ellipse by choos-ing two foci and ensuring that for any point on your curve, the distance to onefocus plus the distance to the other focus remains constant. If we label thesedistances as d1 and d2, we are therefore keeping d1 + d2 constant.

4.1 Equation for an ellipse in Cartesian coordinates

Without loss of generality, in figure 2, we begin by placing our two foci along thehorizontal axis at points (−c, 0) and (c, 0). We then denote the semi-majorradius as a, and the semi-minor radius as b. Choose a point on the ellipsewith co-ordinates (x, y). This then gives:

d1 = d((−c, 0), (x, y)) =√

(x+ c)2 + (y2)

d2 = d((c, 0), (x, y)) =√

(x− c)2 + (y2)

d1 + d2 is constant, so choosing (x, y) = (a, 0), we have

d1 + d2 =√

(a+ c)2 + (02) +√

(a− c)2 + (02) = (a+ c) + (a− c) = 2a

for any choice of (x, y). We can then say

d1 + d2 =√

(x+ c)2 + (y2) +√

(x− c)2 + (y2) = 2a

8

√(x+ c)2 + (y2) = 2a−

√(x− c)2 + (y2)

Squaring both sides gives

(x+c)2+(y2) =(

2a−√

(x− c)2 + (y2))2

= 4a2−2(

2a√

(x− c)2 + (y2))

+(x−c)2+(y2)

x2 + 2cx+ c2 = 4a2 − 4a√

(x− c)2 + (y2) + x2 − 2cx+ c2

4cx = 4a2 − 4a√

(x− c)2 + (y2)√(x− c)2 + (y2) = a− cx

a

Squaring both sides again gives

(x− c)2 + (y2) =(a− cx

a

)2= a2 − 2cx+ c

2x2

a2

x2 − 2cx+ c2 + y2 = a2 − 2cx+ c2x2

a2

x2(

1− c2

a2

)+ y2 = a2 − c2

x2(a2 − c2

a2

)+ y2 = a2 − c2

Dividing through by a2 − c2 gives

x2

a2+

y2

a2 − c2= 1

Note that if we choose (x, y) = (b, 0), then the two triangles formed by thispoint, the origin, and the two foci are congruent (as the shape is symmetrical),meaning that d1 = d2 (in this instance). Further, we know that

d1 = d2 =√c2 + b2

Therefore we know that

d1 + d2 = 2√c2 + b2 = 2a

Squaring both sides gives the following relationship between the position of thefoci, the semi-major radius and the semi-minor radius:

a2 = b2 + c2

This allows us to rewrite the equation of our ellipse to be:

x2

a2+y2

b2= 1

9

4.2 Equation for an ellipse in Polar coordinates centredat one focus

However, all of our observations of orbits are measured in polar co-ordinates,and are measured from one of the foci (i.e. the position of a satellite as seenfrom Earth). Without loss of generality, let us choose the focus at (c, 0) to bethe focus that we are measuring from. Therefore we have to substitute x and yout for functions of r and θ, where r is the distance of the point from our focus,and θ is the angle formed between the positive horizontal axis and the line tothe point.More specifically,

x− c = r cos θ

y = r sin θ

This gives us(r cos θ + c)2

a2+

(r sin θ)2

a2 − c2= 1

(r2 cos2 θ + 2cr cos θ + c2)(a2 − c2) + a2r2 sin2 θ = a2(a2 − c2)

a2r2 cos2 θ+2a2cr cos θ+a2c2−c2r2 cos2 θ−2c3r cos θ−c4+a2r2 sin2 θ = a4−a2c2

a2r2 + 2cr cos θ(a2 − c2)− c2r2 cos2 θ = a4 − 2a2c2 + c4 = (a2 − c2)2

This gives us a quadratic in (a2 − c2):

(a2 − c2)2 − 2cr cos θ(a2 − c2) + (c2r2 cos2 θ − a2r2) = 0

We can solve this to get

(a2 − c2) =2cr cos θ ±

√(−2cr cos θ)2 − 4(1)(c2r2 cos2 θ − a2r2)

2

(a2 − c2) = cr cos θ ±√c2r2 cos2 θ − c2r2 cos2 θ + a2r2

(a2 − c2) = cr cos θ ± ar

Note that by construction, c < a, therefore cr cos θ < ar. If we were to usecr cos θ − ar, we would then get

(a2 − c2) = b2 < 0 =⇒ ⊥

Therefore we have(a2 − c2) = cr cos θ + ar

Rearranging to get r gives

r =a2 − c2

a+ c cos θ=

a− c2

a

1 + ca cos θ

10

Let p = a − c2

a and define the eccentricity of our ellipse to be given as e =ca .

This gives us

r =p

1 + e cos θ

We can define the periapsis rp and the apoapsis ra to be the closest andfurthest points on our curve. By construction, we know that

rp = a− c

andra = a+ c

Therefore we havera − rp = 2c

ra + rp = 2a

Therefore we can write

e =c

a=

2c

2a=ra − rpra + rp

This gives us a way to calculate the eccentricity of orbits via observations. Theeccentricity of an ellipse can be thought of as a measure of how elongated anellipse is. For example, the eccentricity of the Earth’s orbit is 0.0167 (rp =0.9832AU, ra = 1.0167AU), whilst the eccentricity of the orbit followed byHalley’s Comet is 0.9671 (rp = 0.586AU, ra = 35.082AU).

4.3 Area of an ellipse

Consider our Cartesian equation for our ellipse:

x2

a2+y2

b2= 1

Note that the values of x are between −a and a. We can also rearrange theabove equation to get

y2 = b2(

1− x2

a2

)This means that our values of y lie in the range

−b√

1− x2

a2≤ y ≤ b

√1− x

2

a2

Therefore we can say that the region, R, inside our ellipse is

R =

{(x, y)

∣∣∣∣∣x ∈ [−a, a], y ∈[− b√

1− x2

a2, b

√1− x

2

a2

]}

11

This then gives us the area, A, as the integral

A =

∫ a−adx

∫ b√1− x2a2

−b√

1− x2a2a

dy =

∫ a−a

2b

√1− x

2

a2dx

We can integrate this by using the substitution x = a sin(u) This gives dx =a cos(u)du and our integral is now between −π2 and

π2 . Therefore we have

A =

∫ π2

−π22b

√1− a

2 sin2(u)

a2a cos(u)du = ab

∫ π2

−π22√

1− sin2(u) cos(u)du = ab∫ π

2

−π22 cos2(u)du

Note that 2 cos2(u)− 1 = cos(2u), therefore we can write

A = ab

∫ π2

−π2[cos(2u) + 1]du = ab

[1

2sin(2u) + u

]π2

−π2

= πab

5 Kepler’s Laws

Kepler managed to generate three laws of motion under gravity based on hisobservations of the planets:

1. All orbits are ellipses with the parent body as one focus. Thisgives the expected equation for the motion of the planets (in polar co-ordinates) as

r =p

1 + e cos θ

where p is a fixed constant, and e is the eccentricity of the ellipse. Note θis measured from the planet’s closest approach.

2. A line joining a planet and the parent body sweeps out equalareas in an equal amount of time.

3. The square of the orbital period is proportional to the cube ofthe orbit’s semi-major radius, or T 2 ∝ a3, where a is the semi-majorradius.

We will now proceed to show that these laws hold under Newtonian Me-chanics.

5.1 Foundational Work

In order to prove the laws, we must begin by defining a new set of basis vectors:er, eθ, and ez, where

er =r

r

eθ = ez × er

12

Under out previous basis of ex, ey, and ez, the two new vectors can beexpressed as

er = (cos θ, sin θ, 0)

eθ = (− sin θ, cos θ, 0)

Therefore we can writer = rer

v = ṙ =d

dt(rer) = ṙer + rėr

To get ėr, we use our old basis:

ėr =d

dt(cos θ, sin θ, 0) = (−θ̇ sin θ, θ̇ cos θ, 0) = θ̇(− sin θ, cos θ, 0) = θ̇eθ

Thereforev = ṙer + rθ̇eθ

Differentiating again gives

a =d

dt(ṙer + rθ̇eθ) = r̈er + ṙėr + ṙθ̇eθ + rθ̈eθ + rθ̇ėθ

To get ėθ, we again use our old basis:

ėθ =d

dt(− sin θ, cos θ, 0) = (−θ̇ cos θ,−θ̇ sin θ, 0) = −θ̇(cos θ, sin θ, 0) = −θ̇er

Therefore we haveėr = θ̇eθ

andėθ = −θ̇er

Substituting these into our equation of acceleration gives

a = r̈er + ṙθ̇eθ + ṙθ̇eθ + rθ̈eθ − rθ̇2er

Thereforea = er(r̈ − rθ̇2) + eθ(2ṙθ̇ + rθ̈)

In our context, the only force present is gravity, which is a radial force. Thereforewe can say that

a =−GMr2

er

Which gives us−GMr2

er = er(r̈ − rθ̇2) + eθ(2ṙθ̇ + rθ̈)

er and eθ are orthogonal by construction, therefore we can solve this equationcomponent-wise, giving us

−GMr2

= r̈ − rθ̇2

13

and0 = 2ṙθ̇ + rθ̈

Multiplying this second equation by r gives

0 = 2rṙθ̇ + r2θ̈ =d

dt(r2θ̇)

Therefore r2θ̇ is constant. Note that angular momentum is given by L = r×v.

L = r× v = rer × (ṙer + rθ̇eθ) = (rer × ṙer) + (rer × rθ̇eθ) = r2θ̇ez

This gives |L| = r2θ̇, which we know is a constant. Therefore the angularmomentum of this system is constant.

5.2 Kepler’s First Law

Consider the first equation given at the end of our foundation work:

−GMr2

= r̈ − rθ̇2

along withh := |L| = r2θ̇

Therefore we have that

θ̇ =h

r2

Substituting this into our first equation gives

−GMr2

= r̈ − h2

r3

Let u = 1r so that u = u(θ) and θ = θ(t). Note that real orbits do not passthrough the body they are orbiting around, so u is always well defined. Usingthis, we can now write

ṙ =dr

dt=d( 1u )

dt=d( 1u )

du

du

dt= − 1

u2du

dt= −r2 du

dt= −r2 du

dθ

dθ

dt= −hdu

dθ

h is a constant, so we can now write

r̈ = −h ddt

(du

dθ

)= −h d

dθ

(du

dθ

)· dθdt

= −hθ̇ d2u

dθ2

We already know that

θ̇ =h

r2= hu2

Therefore we can write

r̈ = −h2u2 d2u

dθ2

14

Substituting this into the first equation gives

−GMu2 = −h2u2 d2u

dθ2− h2u3

GM = h2d2u

dθ2+ h2u

GM

h2=d2u

dθ2+ u

This is solvable using our work from ”Differential Equations”, giving us thesolution

u(θ) =GM

h2[1 + e cos(θ − θ0)]

where both e and θ0 are arbitrary constants. Since θ0 represents a rotation of

our system, without loss of generality we can let θ0 = 0. Let p =h2

GM , so thatwe can write

u(θ) =1

r(θ)=

1

p[1 + e cos(θ)]

giving

r(θ) =p

1 + e cos(θ)

which gives shapes dependent on the value of e (note e > 0 in reality):e < 1 =⇒ Ellipsee = 1 =⇒ Parabolae > 1 =⇒ Hyperbola

Note that the latter two of these do not give orbits, instead they give thepath followed by an object on an escape trajectory, so we shall only considere < 1. Therefore all orbits are elliptical.

5.3 Kepler’s Second Law

Suppose that between times t and t + δt, an angle of δθ is swept out. Byapproximating the area of this segment as the area of a triangle, we get that

δA ≈ 12· r · rδθ = 1

2r2δθ

Dividing through both sides by δt, we get

δA

δt≈ 1

2r2δθ

δt

Taking the limit as δt→ 0 gives the equationdA

dt=

1

2r2dθ

dt=

1

2r2θ̇ =

1

2|L|

which we know is constant. This gives that equal areas are swept out in equalamounts of time, proving Kepler’s Second Law. Furthermore, it shows that thislaw is a direct result of the conservation of angular momentum.

15

5.4 Kepler’s Third Law

From our proof of Kepler’s First law, we have that

r(θ) =p

1 + e cos(θ)

We know that the area is given by A = πab. Also, since dAdt which is constant,we know that

A

T=dA

dt

where T is the orbital period. Therefore we have

T =AdAdt

=πab12h

=2πab

h

Note that a2 − c2 = b2, therefore

b =√a2 − c2 = a

√1− c

2

a2= a

√1− e2

Therefore

T =2πa2

√1− e2h

Also recall that

p = a− c2

a= a(1− c

2

a2) = a(1− e2)

Therefore √1− e2 = p 12 a− 12

Substituting this into our equation for T gives

T =2πa

32 p

12

h

Squaring both sides gives

T 2 =4π2a3p

h2

Resulting inT 2 ∝ a3

as required.

6 Useful Orbital Equations

For the next section, I will be using a variety of orbital formula. As such, thissection will be focussed on generating these formula.

16

6.1 Equation linking orbital velocity, position, and semi-major radius

We will now use the fact that energy and angular momentum are conservedto get a very useful result. Consider the energy at an orbit’s periapsis andapoapsis (in this section, anything with a in the subscript is a value measuredat the apoapsis, and similarly with p for the periapsis):

E =1

2mv2a −

GMm

ra=

1

2mv2p −

GMm

rp

Some simple algebra of the last two equations gives

1

2

(v2a − v2p

)= GM

(1

ra− 1rp

)Also consider the angular momentum at an orbit’s periapsis and apoapsis:

h = mrava = mrpvp

We then haverava = rpvp

giving

vp =ravarp

Substituting this into our first equation gives

1

2

(v2a −

(ravarp

)2)= GM

(1

ra− 1rp

)

v2a2

(1−

(rarp

)2)= GM

(rp − rararp

)v2a2

(r2p − r2ar2p

)= GM

(rp − rararp

)v2a2

= GM

(rp − rararp

)(r2p

r2p − r2a

)= GM

(rp

ra(rp + ra)

)Note that 2a = rp + ra, therefore

v2a2

= GM

(rp

ra(2a)

)= GM

(ra + rp − rara(2a)

)= GM

(2a− rara(2a)

)= GM

(1

ra− 1

2a

)Finally, consider the orbital energy at an arbitrary point, r, where our objecthas a velocity of v:

E =1

2mv2 − GMm

r=

1

2mv2a −

GMm

ra

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Rearranging the last two equations gives

v2

2− GM

r=v2a2− GM

ra

v2

2− GM

r= GM

(1

ra− 1

2a

)− GM

ra= −GM

2a

Therefore we havev2

2= GM

(1

r− 1

2a

)v2 = GM

(2

r− 1a

)

7 Transfer Orbits

When companies are positioning satellites, they typically begin by launchingthe satellite into a Low Earth Orbit (LEO), which is any orbit with an apoapsislower than 2,000km (1,200mi). They then position the satellite into the desiredhigher orbit via a series of ’burns’, or instantaneous changes of your velocity.In this section, I will outline two methods of transferring a satellite from a lowcircular orbit to a higher one, before providing a model detailing which to usein order to minimise the amount of energy the satellite would expend.The energy expended by a spacecraft primarily comes in the form of changesto kinetic energy. In order to achieve higher orbits, burns are performed inthe direction of travel (or prograde), and to achieve lower orbits you burn inthe opposite direction (or retrograde). Therefore we have that changes to theenergy of the spacecraft caused by burns are just changes in its kinetic energy,or

∆E ≈ ∆EkFurther, we know that changes to the kinetic energy are proportional to changesto the square of the velocity, so we then have

∆E ∝ (v + ∆v)2 − v2 = 2v∆v + ∆v2

where v is our starting velocity, and ∆v is the instantaneous change of velocity.Notice that our change in velocity ∆v is multiplied by our starting velocity,but the energy expended by the spacecraft are only proportional to ∆v2. Thismeans that changes in velocity that occur at higher speeds are more efficient.Therefore performing burns when more of our energy is stored as kinetic energymakes the spacecraft more efficient - this is known as the Oberth effect, andthis effect is what allows the Bi-elliptic transfer to be more energy efficient (insome cases).

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Figure 3: An example of a Hohmann transfer

Figure 4: An example of a Bi-Elliptic transfer

7.1 Hohmann Transfer

The Hohmann Transfer is arguably the simplest transfer method; first make aburn to increase the apoapsis of your orbit to be at the same height as yourdesired new orbit. Then once you reach your apoapsis, make a secondary burnto circularise / bring your periapsis up to the same value as your apoapsis.

7.2 Bi-Elliptic Transfer

Similar to the Hohmann Transfer, the Bi-elliptic Transfer begins by raisingyour apoapsis to some height above the desired orbital radius. Then you makea second burn at apoapsis to raise your periapsis to the orbital radius desired,before finally circularising once your reach periapsis. Note that whilst thisappears to use more energy (since you are reaching a higher altitude), this canstill lead to a lower energy expenditure. Note that this maneuver requires 3burns instead of the 2 used for Hohmann transfers. Some types of enginescan only be ignited a limited number of times, so this extra burn requirementneeds to be considered. Also note that this method takes significantly longer toexecute, so whilst in some situations it is more energy viable, you might choosea Hohmann transfer in order to save time.

8 A Qualitative Comparison via Matlab

In order to make this comparison, I have created some Matlab code to graphthe Bi-elliptic and Hohmann transfer energy requirements. Note that in thiscode, I have assumed that the mass remains unchanged throughout, and allchanges in velocity occur instantaneously. Considering the mass change causedby fuel consumption requires knowledge of the ”wet” and ”dry” mass of thespacecraft (the mass with and without the fuel respectively), and the engineefficiency. However this would then make the graph constructed less general,and in practice does not change the plot drastically. The assumption of instan-taneous burns also does not alter our calculations a lot; usually burn times arefar shorter than any orbital periods involved. In this graph, note that the areasshown in red are the areas where the bi-elliptic transfer is more efficient, withr2 and r3 chosen according to the position on the graph. The other areas arewhere the Hohmann transfer is more efficient.

%Drawing a graph for both the Hohmann and Bi-elliptic transfer

%r2 - height of transfer orbit’s apoapse

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%r3 - desired height

%Note the Hohmann transfer is the case where r2 = r3, and is shown in red

%Begin by defining our mesh for points to sample

step = 0.1;

start = 0.3;

endp = 10;

r2mesh = [start:step:endp];

r3mesh = [start:step:endp];

[r2, r3] = meshgrid(r2mesh,r3mesh);

%Draw the graph for the Bi-elliptic transfer

hold off

BE_E = arrayfun(@(a,b) energy(a,b),r2,r3); %Gives imaginary values when r2

end

%Deriving the energy required for a Bi-elliptic transfer

function E = energy(R2,R3);

v1 = vel(1,1,1);

dv1 = vel(1,R2,1) - v1;

v2 = vel(R2,R2,1);

dv2 = vel(R2,R2,R3) - v2;

v3 = vel(R3,R2,R3);

dv3 = vel(R3,R3,R3) - v2;

E = abs(2.*v1.*dv1 + dv1.^2) + abs(2.*v2.*dv2 + dv2.^2) + abs(2.*v3.*dv3 + dv3.^2);

end

In the above graphic, the multicoloured area denotes the varies energy require-ments of performing Bi-elliptic transfers, where r2 is the ”middle” orbit apoapse(aka transfer orbit), and r3 is the desired orbital radius. The area shaded inred is the region where performing the Hohmann transfer is less efficient thanthe Bi-elliptic. For the sake of clarity, I will split the regions of the graph into5 areas: A, B, C, D, and E. A is the lower right-hand blue section, and B is thelarger red section. C is the top left blue section, and then D and E are the 2 redsections above and below the 45◦line respectively. Note that this entire sectionis going to be qualitative, and most bounds will simply be referred to ”upper”or ”lower”, since whilst all these curves are distinct, they are unnamed since as

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far as I can tell, I am the first to generate this graph.

8.1 Region A

This area is blue since it is (almost) always more efficient to go to a higherorbit via either a direct route (Hohmann transfer), or a Bi-elliptic transfer thattakes your apoapsis above your desired height. This also means it is alwaysmore efficient to perform a single Hohmann transfer instead of two separateones, each raising your orbital radius. Note that the 45◦line bounding A above(almost everywhere) is caused by the equivalence of the Hohmann and Bi-Elliptictransfers along it.

8.2 Region B

In this region, the Bi-elliptic transfer wins out. However, note that the curvebounding the region from above implies that the Bi-elliptic transfer is moreefficient provided that your transfer orbit apoapsis is not raised too high. Thecurve bounding this region above is caused by simply going into such a highorbit that the gains made back from the Oberth effect are not as much as thelosses due to completing the initial radius raising burn.

8.3 Region C

Raising the transfer apoapsis too high results in the maneuver becoming ineffi-cient. Note that this region is only joined to region A due to the resolution of theplot - in reality the two regions are only joined at (1,1), where the Hohmann andBi-elliptic transfers are equivalent. Furthermore, the section of this region thatis below r2 = 1 also implies that it is not always efficient to perform a burn as a

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bi-elliptic transfer when decending. If this were the case, it would actually implya continuous solution - if splitting the burn up was always more efficient, wewould be encouraged to split the burn into infinitely many infinitesimal burns.

8.4 Region D

When your desired orbital radius is less than half your starting radius, it isactually more efficient to have a transfer orbital radius slightly higher or lowerthan your original (Note that along the line r2 = 1, the Bi-elliptic and Hohmanntransfers are again equal, but are shaded as to fit to their surroundings). Havinga transfer radius slightly higher than your initial position allows for a greaterpercentage of the total δv to be expended lower down, and having a transferradius lower than your start allows for a percentage of your burn normallyconducted at a radius of 1 to be completed lower. Both of these strategies takeadvantage of the Oberth Effect, where the former is maximising the amountburnt lower down and the latter is minimising the amount burnt higher up.The curve bounding this region above is caused by the same effect as whatcauses the upper bound of B, whereas the lower bound is caused by the gainsmade from the Oberth effect being less than the energy expended to lower yourorbit by that much.

8.5 Region E

This region is implying that whilst setting your transfer radius slightly higherthan your target is inefficient, setting it slightly lower can yield a more efficientsolution. Again, this is caused by the Oberth effect. This also means thatthere are more efficient solutions where going below your target radius is moreefficient, but only for maneuvers that aim to decrease your orbital radius.

9 Summary of this comparison

Whilst our simplifications and assumptions make this graph unsuited for findingsolutions directly, the various divisions of the graph imply that there do exist sce-narios where the bi-elliptic transfer is the more efficient method. Furthermore,this graph suggests where these solutions might exist, and the ’smoothness’ ofthe various curves involved suggests that it is likely possible to obtain thesevarious regions as regions bounded by known curves. If this can be done, oursimplifications regarding the mass of the parent body and the value of G wouldjust result in a linear transformation of these equations. However, since satel-lites have different dry and wet masses, different engine efficiencies, and differentthrust outputs, dropping our assumptions on the constant mass of the spacecraftand the instantaneous changes in velocity would result in a unique graph beinggenerated for every spacecraft. Between this and the added effects from factorssuch as other gravitationally attractive bodies, air resistance (which is certainly

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not negligible in low orbits), and radiation pressure, it rapidly becomes appro-priate to use iterative methods and simulations. This has the affect of being farmore accurate for specified satellites, but losing the general solutions providedby the graph.In conclusion, whilst this method does have its flaws, this graph demonstratesthe general regions where bi-elliptic and Hohmann transfers should be used. Assuch, it is a useful resource when trying to get a quick idea as to which methodto use.

References

[1] Mauri Valtonen and Hannu Kattunen. The three-body problem. CambridgeUniversity Press, pages 20–23, 2005.

[2] Mauri Valtonen and Hannu Kattunen. The three-body problem. CambridgeUniversity Press, pages 25–27, 2005.

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