Tether Fundamentals J. Peláez - IEECT G D Advanced Topics In Astrodynamics – Tethered Systems...

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DG Advanced Topics In Astrodynamics – Tethered Systems

Tether Fundamentals

J. Peláez

Grupo de Dinámica de Tethers (GDT)

ETSI Aeronáuticos

Universidad Politécnica de Madrid, 28040 Madrid, Spain.

Barcelona, 2004

Friday, 9 July

z

Tether Fundamentals – p.1/48

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What is a tether?

Definition of the wordtetherin theMerriam-Websterdictionary:

1. something (as a rope or chain) by which an animal is fastened so that it can range only

within a set radius

2. the limit of one’s strength or resources «at the end of my tether »

• A tetheris a cable used in space to tie together two different spacecrafts.

• Tethersare long. Tether lengths of the order of kilometers

• Tethersare thin. Tether diameters of the order of millimeters

• Tethersare strong. They are made of resistant materials


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What does a tether do?

On Earth, ropes and cables have been used in many different scenarios. Sailing, suspension

bridges, high voltage lines, overhead power cables of the railroads...; there are plenty of

examples where cables turned out to be extremely useful. From the beginning of XX century,

there were persons thinking about the use of tethers in space. At present, and this opinion is

shared by many people, tethers would become, at least potentially, as useful in space as they

have always been on Earth.

The number of applications of tethers in space is large:creation of artificial gravity, generation

of thrust, maneuvers and exchange of angular momentum, atmospheric studies, etc...The key

that makes feasible the use of tethers islightness. In space, the forces needed to keep objects

together using a tether aresmall. Thus, very thin cables can be used to tie satellites, and small

sections mean small weights, an essential requirement of space operation.

• Tethers in Space Handbook, by M. L. Cosmo and E. C. Lorenzini, NASA Marshall

Space Flight Center, (Third Edition) December 1997 (Can be downloaded inPDF

format from the Internet sitehttp://cfa-www.harvard.edu/~spgroup/handbook.html)


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16 Tether Missions Have Flown Since 1966

NAME DATE ORBIT LENGTH COMMENTS

Gemini XII 11 1966 LEO 30 m Spin stable, 0.15 RPM

Gemini XII 12 1966 LEO 30 m Local vertical, stable swing

H-9M-69 1980 Suborbital 500 m Partial deployment

S-520-2 1981 Suborbital 500 m Partial deployment

Charge-1 1983 Suborbital 500 m Full deployment

Charge-2 1984 Suborbital 500 m Full deployment

ECHO-7 1988 Suborbital ? Magnetic field aligned

Oedipus-A 1989 Suborbital 958 m Spin stable, 0.7 RPM

Charge-2B 1992 Suborbital 500 m Full deployment

TSS-1 1992 LEO < 1 km Electrodynamic, partial deploy, retrieved

SEDS-1 1993 LEO 20 km Downward deploy, swing and cut

PMG 1993 LEO 500 m Electrodynamic, upward deploy

SEDS-2 1994 LEO 20 km Local verticle stable, downward deploy

Oedipus-C 1995 Suborbital 1 km Spin stable, 0.7 RPM

TSS-1R 1996 LEO 19.6 km Electrodynamic, severed

TiPS 1996 LEO 4 km Long-life tether on orbit > 2 year

Data from Les Johnson, NASA MSFC (2000)


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Summary of the tether mission history and near future projects (year 2000)


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Most of the photographs in this presentation have been takenfrom the JSC (http://www.jsc.nasa.gov/) through the Internet

View of the Agena rocket last stage and the tether from the Gemini XII craft.

A 100 ft. tether line connects the Agena Target Docking Vehicle with the Gemini 12 spacecraft during its 32nd revolution

of the Earth. Clouds over Pacific Ocean are in the background.


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View of the Agena rocket last stage and the tether from the Gemini XII craft.

Mexico, Arizona and New Mexico as seen from the Gemini XII spacecraft.


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View of the Agena Target Docking Vehicle as seen from Gemini 12 spacecraft.

Side view of the Agena Target Docking Vehicle as seen from theGemini 12 spacecraft during rendezvous and docking

mission in space. The two spacecraft are 50 ft. apart.


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Artist concept of U.S./Italian program calledTethered Satellite. Real view of the mission TSS-1 (1992).


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STS-46 Tethered Satellite System 1 (TSS-1) satellite deployment from OV-104

The satellite is reeled out via its thin Kevlar tether into the blackness of space during deployment operations from the payload bay of Atlantis.

At the bottom of the frame is the satellite upper boom including (bottom to top) the 12-m deployment boom, tip can, the docking ring, and

concentric ring damper. The Langmuir probe and the dipole-field antenna are stowed at either side of the TSS-1 satellite.


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TSS-1R. The Tethered is seen as it is reeled out during early stages of deployment operations

The crew deployed the TSS, which later broke free. The seven member crew was launched aboard the Space Shuttle Columbia on

February 22, 1996 and landed on March 9, 1996.


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View of the TSS-1 satellite.∼ 518 Kg, diameter∼ 1.6 m


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TSS-1R. The Tether is partially deployed


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Tether broken. The tether was made of a core ofNomexwrapped withcopper wire, insulated withTeflon

and covered withbraided Kevlarandbraided Nomex. This construction offers electrical conductivity and

strength.


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To begin with....

rvc

Associated with a circular orbit there are two significant para-

meters:

Circular velocity vc =

√

µ

r

Angular frequency ω =

√

µ

r3

Typical values, when orbiting the Earth, are

r = RE ≃ 6378 Km ⇒

v1c =√

gRE ≃ 8 Km/s First cosmic velocity

ω1c = 1.25431 · 10−3 rad/s

ω = ωE ≃ 7.29 · 10−5 rad/s ⇒

T = 1 sideral day geosynchronous

rg = 42166 Km geostationary, vg ≃ 3.07 Km/s


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Space elevator

Original fromTsiolkovskii

andJerome Pearson


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In a tethered system, two masses orbiting at different heights share a common orbital

frequencyΩ0 ⇒ The third Kepler law is broken by the tether tension

m1

m2

r1

r2

r0

T

T

G

g1

g2


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Assuming a massless tether

m1

m2

T

T

G

g1

g2

µm1

r2

1

− T = m1Ω2

0r1

µm2

r2

2

+ T = m2Ω2

0r2

⇒

Two equations with

two unknowns:T andΩ0

Solving these equations provides:

Ω2

0 =µ

r2

1r2

2

·m1r

2

2 + m2r2

1

m1r1 + m2r2

T =µm1m2

r2

1r2

2

·r3

2 − r3

1

m1r1 + m2r2

Ω2 ≤ Ω0 ≤ Ω1

which are functions of(r1, r2, m1, m2).


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Instead of(r1, r2, m1, m2) the parameters(rG, ǫ,m, φ) which are defined by

m = m1 + m2, ǫ =L

rG

, cos2 φ =m1

m

rG =1

m(m1r1 + m2r2), sin2 φ =

m2

m

are more appropriated. Typically, the value ofǫ is very small

(ǫ ∼ 0.015 for a tether about 100 Km length in LEO.)

With the help of(rG, ǫ,m, φ) we get interesting results introducing the relations

r1= rG(1 − ǫ sin2 φ), m1= m cos2 φ

r2= rG(1 + ǫ cos2 φ), m2= m sin2 φ

in the expressions forΩ0 andT , and expanding forǫ aboutǫ = 0.


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Useful tether features

Tethers are thin ...

In effect, weight is one of the most important parameters of an spatial mission. To raise to LEO a mass of

1 Kg costs, in approximation: about 6000 — 12000e. To reach higher orbits or inter-planetary

trajectories is obviously more expansive.


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Useful tether features

Tethers are thin ...

In effect, weight is one of the most important parameters of an spatial mission. To raise to LEO a mass of

1 Kg costs, in approximation: about 6000 — 12000e. To reach higher orbits or inter-planetary

trajectories is obviously more expansive.

...and they orbit with the angular frequency of the center ofmass.

For a tethered system, the trajectory of the system center ofmassG plays a central role; perhaps more

significant than in the case of a conventional (non tethered)satellite.


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String fundamentals...

The internal forces of a rigid body can be re-

duced to

one resultant~F

shearing force~Fs

normal force~N

one torque~M

twisting torque~Mt

bending torque~Mb

~F~F S

~N

~M~M b

~M t

A flexible body is considered astringif:

• the traction stiffness is very high

• the bending stiffness is zero

• the twisting stiffness is zero

• it does not support compression stress

As a consequence:

~M = ~M t = ~M b = ~0

~F s = ~0, ~N ≡ ~F

Thus, forstrings, there is only one internal

force thetensionwhich is atraction forceandnormal to the section


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Governing equations... little by little

Equilibrium equations forinextensiblestrings


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Equations of motion forinextensiblestrings


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Equilibrium equations forextensiblestrings


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Equilibrium equations forextensiblestrings

Equations of motion forextensiblestrings


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A hanging cable

x

y

z

Tmax = ρLgL0

Tmin = 0

Equilibrium

Equations

x = y = u = v = 0, w = 1 +T

AE

dz

ds= 1 +

T

AE

dT

ds+ ρLg = 0

⇒

T = ρLgL01 −s

L0

z = s1 +ρLgL0

AE(1 −

s

2L0

)

Tmax = ρLgL0 ⇒ L∗

0 =σ∗

ρvg

Here σ∗ is the breaking stress of the material (perhaps with some

safety coefficient). The lengthL∗

0does not depend on the cross-

section of the cable. For steelL∗

0≈ 26 Km. Problems: 1) there

is a maximum length, 2) the material is wasted


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A constant stress cable

x

y

z

Tmax = σ0Amax

Tmin = σ0Amin

Equilibrium

Equations for constant stress (σ = σ0)

x = y = u = v = 0, T = σ0A(s), w = 1 +σ0

E

dz

ds= 1 +

σ0

E⇒ z = s1 +

σ0

E

σodA(s)

ds+ ρvA(s)g = 0

A(s) = Amin exp

σ∗

σ0

·L0 − s

L∗

0

There is no limit for the length of the cable. The tether material is

optimized.Problem: the cable mass grows exponentially withL0

mt =σ0

gAmin

[

exp

σ∗

σ0

·L0

L∗

0

− 1

]


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Forces on tethers

Gravitational forces: A unit mass at a distance

R from the Earth centre is acted by the gravita-

tional force which is composed of the main term

~f = −µ

R2~uR

and perturbations. The more important pertur-

bation is of the order of

~f p ≈ −µ

R2J2 ⇒

‖ ~f p‖

‖ ~f ‖∼ J2 ∼ 10−3

The other perturbation terms associated with the

Earth gravitation are small compared with~f

∼ 10−6‖ ~f ‖

and usually they are neglected.

Lunisolar perturbation : the forces associated

with the attraction from the Sun and the Moon

are of the order

‖ ~f M‖

‖ ~f ‖∼ 10−8,

‖ ~f S‖

‖ ~f ‖∼ 10−8

for a LEO tethered system.

Aerodynamic forces:on a tether element are

‖d ~f a‖

‖d ~f ‖∼

ρa

ρ

R

dT

∼ 10−5

ρa = 2.5 · 10−12kg/m3 R = 6.8 · 106 m

ρ = 1.45 · 103kg/m3 dT = 10−3 m

Usually, the tether length is large and this force

become important below∼ 400 Km heightTether Fundamentals – p.25/48

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Relative motion in circular orbit

E

G

P ,m

~F i

~F g

x

y

z~r

~F g = −mµ

ζ3

~ζ, ~ζ =−→EP

To study the motion relative to the orbital frame

Gxyz of any particleP of massm we have to

consider the gravity~F g and the centrifugal and

Coriolis forces:

~F i1 = mω2R(1 +x

R, 0,

z

R)

~F i2 = 2mω(z, 0,−x)

The gravity gradient~Fgg = ~Fg + ~F i1

~Fgg ≈ mω2(3x~i− y~j)

The resultant force acting onP is m ~f where

~f ≃ ω2(3x~i− y~j) + 2ω(z~i− x~k)


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Stable equilibrium along the local vertical

θ

Local Vertical

3mΩ2

0d

d

G

x

z

Any deviationθ from the local vertical gives

place to a torque. In effect, the gravity gradi-

ent force breaks down in

• one component along the tether, which

is basically balanced by the tether

tension

• one component orthogonal to the tether,

which provides the restoring torque.

This torque leads the tether again to the

local vertical

Thus, the local vertical is a stable equilibrium

position for the tethered system


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DG VII Jornadas de Trabajo en Mecánica Celeste

Mass geometry considering the tether mass

To describe the system mass geometry taking

into account the mass of the tether, is bet-

ter to use the parameters(m, φ, Λ) instead of

(m1, m2, mt),

m = m1 + m2 + mt

Λ = mt/m

cos2 φ =1

mm1 +

1

2mt

sin2 φ =1

mm2 +

1

2mt

m1 = m

(

cos2 φ −1

2Λ

)

m2 = m

(

sin2 φ −1

2Λ

)

hG = L cos2 φ

m1

m2

G

dm

~f e

h

hG

~u

Masses of the tethered system

Is =1

12mL2(3 sin2 2φ − 2Λ)


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Influence of the tether mass

Equation for the tension (inextensible

tether):

dT

dx+ 3ρLΩ2

0x = 0

The solution is a parabolic distribution

T = Tmax −3

2ρLΩ2

0x2

with a maximum in the center of gravity

(≈ the center of massG).

A detailed analysis provides the values

T1 =3

4

mµ

r2

G

ǫ

sin2 2φ − 2Λ sin2 φ + o(ǫ)

T2 =3

4

mµ

r2

G

ǫ

sin2 2φ − 2Λ cos2 φ + o(ǫ)

Tmax =3

4

mµ

r2

G

ǫ

sin2 2φ (1 −Λ

2) + o(ǫ)

whereT1 andT2 are the tensions at the tether’s ends.

~um1 G m2

hG

dx

x~F

dx= 3ρlΩ

2

0x dx~u


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Influence of the tether mass

T1 > T2

m

m1 > m2

For light tethers the difference

∆T = Tmax − Tmin

is alwayssmall

∆T ≪ Tmax, Tmin

T

r1 r2r0 ≈ rG

T1 T2Tmax

Tension profile along the tether


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Preliminary design

The tensionTmax provides the length

Tmax =3

2ρLΩ2

0L2 ⇒ L =1

Ω

√

2Tmax

3ρL

It does not depends on thetether sectionA:

Tmax = σmaxA

ρL = ρvA

⇒ L =1

Ω

√

2σmax

3ρv

Taking the coordinateζ = x/L:

T = Tmax −3

2ρLΩ2

0x2 =3

2ρLΩ2

0L2(1 − ζ2)

T1 =3

2ρLΩ2

0L2(1 − ζ2

1 )

T2 =3

2ρLΩ2

0L2(1 − ζ2

2 )

l1, mt1

m1 G m2

l2, mt2

On the other hand, the gravity gradient provides

T1 = 3m1Ω2l1 ⇒ m1 =ρL(1 − ζ2

1)

2ζ1

T2 = 3m2Ω2l2 ⇒ m2 =ρL(1 − ζ2

2)

2ζ2

The mass of the segmentsl1 andl2 are

mt1 = ρLl1 = ρLLζ1, mt2 = ρLl2 = ρLLζ2

mt1

m1

=2ζ2

1

1 − ζ2

1

,mt2

m2

=2ζ2

2

1 − ζ2

2


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Exercise(taken from the book of Beletskii)

Exercise:A tether made of Kevlar, flying in LEO,

should place a mass of500 Kg, 100 Kmbelow the

center of mass. What section does the tether need?

(factor of safety 4)

Material data:

ρv = 1.45 · 103 Kg/m3

σ∗

= 2.8 · 109 N/m2

σmax = σ∗

/4 = 0.7 · 109 N/m2

Orbital data:

Ω0 ≈ 1.164 · 10−3 s−1

m1 = 500 Kg

l1 = 100 Km

Procedure

1. DetermineL =1

Ω

√

2σmax

3ρv

≈ 488 Km

2. Determineζ1 =l1

L≈ 0.205

3. Determine the ratiomt1

m1

≈ 0.0878

4. DetermineA

A =m1

ρvl1

(

mt1

m1

)

≈ 0.303 mm2

For this tether

ρL ≈ 0.44 Kg/Km

mt1 ≈ 43.9 Kg


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Relation between(Λ, φ) and (ζ1, ζ2)

0

0.2

0.4

0.6

0.8

1

π/8 π/4 3π/8 π/2

0.01

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

0.01

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

Λ

ζ1 = cte ζ2 = cte

φ

There is a bijective map-

ping between the parame-

ters (ζ1, ζ2) and (φ, Λ)

given by the equations:

ζ1 =

√

Λ

2 − Λtan φ

ζ2 =

√

Λ

2 − Λcot φ

Λ =2ζ1ζ2

1 + ζ1ζ2

tan φ =

√

ζ1

ζ2


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Tether motion

The motion of the tether breaks down in

The motion of the center of mass

• is basically Keplerian

• the tether introduces a gravitational perturba-

tion of the order ofǫ2

• usually there are more perturbations associ-

ated with the tether (aerodynamic or electrody-

namic drag, for example). Some of them could

become important

• it is coupled with the motion relative to the

center of mass and both motions should be de-

termined simultaneously

The motion relative to the center of mass

It has two basic components:

• the motion of the whole system as a rigid body

(pendular motion)

• vibrations around equilibrium positions or

pendular motions

• usually there is a strong coupling between both

kind of motions

• the energy associated with the pendular mo-

tion is usually much more important that the en-

ergy associated with the tether vibrations (due to

the large difference in masses)


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Pendular motion of the tether

Consider the motion relative to the orbital frame of an inextensible tether

∂

∂s(T (s, t)~t(s, t)) + ρL

~f = ρL

∂2~x

∂t2

~f ≃ ω2(3x~i − y~j) + 2ω(z~i − x~k)

When the tether is light (ρL → 0) we have

∂

∂s(T~t) = ~0 ⇒

∂T

∂s

∂~x

∂s+ T

∂2 ~x

∂s2= ~0

Since both vectors are independent

∂T

∂s= 0 ⇒ T = T0

∂2~x

∂s2= ~0 ⇒ ~x = ~a(t) + s~u(t)

Boundary conditions:

at s = 0, ~x = ~0 ⇒ ~a(t) = ~0

Motion of the end massm2

m2l2 ~u = m2~f

2− T0 ~u

Dot product yieldsT0:

T0 = −m2(l2 ~u − ~f 2) · ~u

Cross product yields~u(t):

~u× (l2 ~u − ~f 2) = ~0

This is an exact solution even for heavy tethers

(largeρL)


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Pendular motion of the tether

x

y

z

θ

Gϕ

~u

m1

m2

~u = cos ϕ cos θ~i− sin ϕ~j +cos ϕ sin θ~k

Equations for the inextensible tether

θ − 2 (1 + θ)ϕ tan ϕ +3

2sin 2θ = 0

ϕ +1

2sin 2ϕ

(

(1 + θ)2 + 3 cos2 θ)

= 0

T0 =m1 m2

mΩ2

0Lϕ2+cos2 ϕ((1+θ)2+3 cos2 θ)−1


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Vibrations in a tether

Usually tethers are long. In a first approximation, they can be considered unlimited. In such a case some

classical results about vibrations in strings turns out to be of significan interest.

Transversal Vibrations

Let η be the small orthogonal deflection from a

straight tether. Its evolution is given by

∂2η

∂ξ2=

1

v2t

∂2η

∂t2

Here the propagation velocity is

vt =

√

T

ρL

Longitudinal Vibrations

Let u(ξ, t) be the longitudinal displacement of

the sectionξ in a straight tether. Its evolution is

given by

∂2u

∂ξ2=

1

v2

l

∂2u

∂t2

Here the propagation velocity is

vl =

√

AE

ρL

=

√

E

ρvl ≫ vt

High values ofvl leads to stiff problemsTether Fundamentals – p.37/48

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Tethers are thin strings...

T =3

4

µm

r2

G

· ǫ · sin2 2φ 1 − ǫ cos 2φ + o(ǫ2) (1)

It is interesting to compare the tension with the totalweight (mg) of the system

T = mg · k · ǫ, where k =3

4sen2 2φ(

RE

rG

)2 1 − ǫ cos 2φ + o(ǫ2) < 3/4 (2)

andRE is the radius of the Earth. Sinceǫ ≪ 1 ⇒ T ≪ mg . Thus,tethers are thin.

For example, for the TSS-1 tether (Atlantis, 1992) we have:

m1 ≈ 100000 Kg r1 ≈ 6804 Km L ≈ 20 Km

m2 ≈ 500 Kg r2 ≈ 6824 Km ǫ ≈ 2.94 · 10−3

φ ≈ 4 rG ≈ 6804.1 Km k ≈ 1.305 · 10−2

The tension required to

tie together both satellites

turns out to be

T ≈ 38 Nw

The massless tether assumption is plenty of sense....


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Les Johnson showing the Spectra 2000 fiber of the non conductive part of the ProSEDS tether

The picture also shows the cutter and the pulley barber used to break the tether in the last phase of the deployment process


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... and they orbit with the angular frequency of the center ofmass.

r0

∆

rG

m1

m2

SinceΩ2 ≤ Ω0 ≤ Ω1 for a certain point of the tether,the center of gravity, its free

angular frequency coincides with the system orbital frequencyΩ0. This point is given

by the condition

Ω2

0 =µ

r3

0

⇒ r0 = (µ

Ω0

)13

Ω0= ΩG(1 + ε23

16(1 − cos 4φ) + o(ε2)), Ω2

G =µ

r3

G

r0= rG(1 − ε21

4sen2 2φ + o(ε2))

Thus, the orbital frequencyΩ0 ≡ ΩG apart from terms of orderε2. The difference

∆ = rG − r0 is positive but small

∆ =m1m2

(m1 + m2)2ε L + o(ε)

and the center of massG is a bit higher than the center of gravity.


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Equilibrium equations for aninextensiblestring

• s is the arc length with a known origin

• ~F is the external force per unit length

~F ds

s + dss

−~t(s)

~T(s + ds)− ~T (s)

~t(s + ds)

ds

We consider the forces which are acting over an

element of string of lengthds:

• through the end sectionss, s + ds

• the volume forces

Unknowns:

• the equilibrium shape:~x = ~x(s)

• the tension distribution:~T = ~T (s)

Linear momentum equation:

d ~T

ds+ ~F = ~0

Angular momentum equation:

~t(s)× ~T (s) = ~0 ⇒ ~T (s) = T (s)~t(s)

Inextensible string:

|~t(s)| = |d~x

ds| = 1

z


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Equilibrium equations in Cartesian coordinates

d

ds(T (s)~t(s)) + ~F = ~0

|d~x

ds| = 1

~x = (x(s), y(s), z(s))

~t = (u(s), v(s), w(s))

~F = (X, Y, Z)

X = X(x, y, z, u, v, w, s). The same forY, Z.

The problem is closed: 7 equations and 7 un-

knownsx(s), y(s), z(s), u(s), v(s), w(s) and

T (s). The order of the system is SIX, because

one them is an algebraic equation. There are SIX

integration constant.

dx

ds= u

dy

ds= v

dz

ds= w

d

ds(Tu) + X = 0

d

ds(Tv) + Y = 0

d

ds(Tw) + Z = 0

u2 + v2 + w2 = 1

+ initial and/or boundary conditions


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Equilibrium equations in cylindrical coordinates

x

yz

z

~ur

~uθ

~uz

θ

r

~x = r~ur + z~uz

~t =dr

ds~ur + r

dθ

ds~uθ +

dz

ds~uz

~F = Fr ~ur + Fθ ~uθ + Fz ~uz

dr

ds= u

rdθ

ds= v

dz

ds= w

d

ds(Tu) −

1

rTv2 + Fr = 0

d

ds(Tv) +

1

rTuv + Fθ = 0

d

ds(Tw) + Fz = 0

u2 + v2 + w2 = 1



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The equations of motion for aninextensiblestring

• s is the arc length with a known origin

• ~F is the external force per unit length

~F ds

s + dss

−~t(s)

~T(s + ds)− ~T (s)

~t(s + ds)

ds

−ρL

∂2~x

∂t2ds

We consider the forces which are acting over an

element of string of lengthds:

• through the end sectionss, s + ds

• the volume forces +inertia force

Unknowns:

• the shape:~x = ~x(s, t)

• the tension distribution:~T = ~T (s, t)

Linear momentum equation:

∂

∂s(T (s, t)~t(s, t)) + ~F = ρL

∂2~x

∂t2

Inextensible string:

|~t(s, t)| = |∂~x

∂s| = 1

+ initial and boundary conditions

z


T


The equations of motion for aninextensiblestring

∂

∂s(T (s, t)~t(s, t)) + ~F = ρL

∂2 ~x

∂t2

~t =∂~x

∂s

~x = (x(s, t), y(s, t), z(s, t))

~t = (u(s, t), v(s, t), w(s, t))

~F = (X, Y, Z)

X = X(x, y, z, u, v, w, s, t). The same forY, Z.

The problem is closed: 7 equations and 7 unknownsx(s, t),

y(s, t), z(s, t), u(s, t), v(s, t), w(s, t) and T (s, t).

Now there are 6 coupled partial differential equations of first

and one algebraic relation.

∂x

∂s= u

∂y

∂s= v

∂z

∂s= w

∂

∂s(Tu) + X = ρL

∂2x

∂t2

d

ds(Tv) + Y = ρL

∂2y

∂t2

d

ds(Tw) + Z = ρL

∂2z

∂t2√

u2 + v2 + w2 = 1



T


Introduction toextensiblestrings

It is usual to approach the theory on theextensi-

blestrings by introducing the following assump-

tions:

• the string is perfectly flexible

• elastic regime

• pure extension

• plane cross-sections remain plane

• there are two arc lengths

1. s → for theslackstring

2. s1 → for thetautstring

These assumptions lead to the relations

ε =ds1

ds− 1

ε =σ

E

σ =T

A

⇒ds1

ds= 1 +

T

AE

Here,AE is theextensional stiffnessof the

tether material. Usually is ahigh value. In

the limit AE → ∞ we recover the previ-

ous model for theinextensiblestrings.


T


Equilibrium equations for anextensiblestring

d

ds(T (s)~t(s)) + ~F = ~0

~t =d~x/ds

|d~x/ds|, ds1 = |

d~x

ds| ds

~t =d~x

ds·

1

1 + T

AE

~x = (x(s), y(s), z(s))

~t = (u(s), v(s), w(s))

~F = (X, Y, Z)

X = X(x, y, z, u, v, w, s). The same forY, Z.

The problem is closed: 8 equations and 8 unknownsx(s),

y(s), z(s), u(s), v(s), w(s), T (s) ands1(s). The order

of the system is SEVEN, because one of them is an algebraic

equation. There are SEVEN integration constant.

dx

ds= u

dy

ds= v

dz

ds= w

d

ds(T ·

u

1 + T

AE

) + X = 0

d

ds(T ·

v

1 + T

AE

) + Y = 0

d

ds(T ·

w

1 + T

AE

) + Z = 0

√

u2 + v2 + w2 − 1 =T

AE

ds1

ds= 1 +

T

AE



T


The equations of motion for anextensiblestring

∂

∂s(T (s, t)~t(s, t)) + ~F = ρL

∂2 ~x

∂t2

~t =∂~x/∂s

|∂~x/∂s|, ds1 = |

∂~x

∂s| ds

~t =∂~x

∂s·

1

1 + T

AE

~x = (x(s, t), y(s, t), z(s, t))

~t = (u(s, t), v(s, t), w(s, t))

~F = (X, Y, Z)

X = X(x, y, z, u, v, w, s, t). The same forY, Z.

The problem is closed: 8 equations and 8 unknownsx(s, t),

y(s, t), z(s, t), u(s, t), v(s, t), w(s, t), T (s, t) and

s1(s, t). Now there are 7 coupled partial differential equa-

tions and one algebraic relation.

∂x

∂s= u

∂y

∂s= v

∂z

∂s= w

∂

∂s(T ·

u

1 + T

AE

) + X = ρL

∂2x

∂t2

d

ds(T ·

v

1 + T

AE

) + Y = ρL

∂2y

∂t2

d

ds(T ·

w

1 + T

AE

) + Z = ρL

∂2z

∂t2

√

u2 + v2 + w2 − 1 =T

AE

∂s1

∂s= 1 +

T

AE



Tether Fundamentals J. Peláez - IEECT G D Advanced Topics In Astrodynamics – Tethered Systems...

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Transcript of Tether Fundamentals J. Peláez - IEECT G D Advanced Topics In Astrodynamics – Tethered Systems...