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Economics of Space Exploration

Bill Gibson

UVM Fall 2012

Bill Gibson University of Vermont
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Hohmann transfer

10/14/11 6:03 PMHohmann transfer orbit.svg - illustration of a Hohmann transfer orbit

Page 1 of 2http://upload.wikimedia.org/wikipedia/commons/d/df/Hohmann_transfer_orbit.svg

1

23

O

RR'

!v

!v'

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Hohmann transfer

Velocity changes are are tangent to the initial and final orbits(changes magnitude of velocity but not direction).

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Hohmann transfer


Burns are short (2-5 min) compared with length of orbit.Treat them as if they were instantaneous.


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Hohmann transfer



Coapsidal orbits: two elliptical orbits have their major axes

aligned.


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Hohmann transfer




aligned.

Whenever we add or subtract V, we change the orbitsspecific mechanical energy and hence its semi-major axis.

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Hohmann transfer




aligned.


ExampleHow do we calculate the burns?


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Hohmann transfer




aligned.


ExampleHow do we calculate the burns?

Answer: It is a two-step procedure!


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Hohmann transfer

Conservation of energy states that the sum of the kineticenergy and the potential energy remains constant

0 =V2

2 GM

r

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Hohmann transfer


0 =V2

2 GM

r

From Keplers Law (this is derived from conservation of

angular momentum).

rpvp = rava

where ra = radius at apoapsisand rp at periapsis

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Hohmann transfer


0 =V2

2 GM

r


angular momentum).

rpvp = rava

where ra = radius at apoapsisand rp at periapsisAt the two points, apogee and perigee of the orbit, we have

conservation of energy

PE1 + KE1 = PE2 + KE2

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Hohmann transfer


0 =V2

2 GM

r


angular momentum).

rpvp = rava

where ra = radius at apoapsisand rp at periapsisAt the two points, apogee and perigee of the orbit, we have

conservation of energy

PE1 + KE1 = PE2 + KE2

This is all we need to calculate the required burnBill Gibson University of Vermont
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Hohmann transfer

Substituting

mv2p

2 GMm

rp=

mv2a2

GMmra

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Hohmann transfer

Substituting

mv2p

2 GMm

rp=

mv2a2

GMmra

Next cancel out the mass and rearrange the terms:

v2p v2a = 2GM(1

rp 1

ra)


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Hohmann transfer

Substituting

mv2p

2 GMm

rp=

mv2a2

GMmra

Next cancel out the mass and rearrange the terms:

v2p v2a = 2GM(1

rp 1

ra)

Substitute Keplers second law: vp =ravarp

r2a v2a

r2p v2a = 2GM(

1

rp 1

ra)

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The math

Simplify to find

v2a =2GM(

1rp

1ra )

( r2a

r2p 1)

= 2rpGM(ra + rp) ra

=2GMrp(1 rpra )

r2

a (1 r2p

r2a )

=2GMrp(1 rpra )

r2

a

(1

rp

ra)(1 +

rp

ra)

=2GMrp

ra(ra + rp)

v2p =2GMra

rp(rp+ ra)

v2a = v2p

2rpGM

(ra + rp) ra/

2GMra

rp(rp+ ra)

=r2p

r2a

va =rp

ravp


Th h
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The math

Simplify to find

v2a =2GM(

1

rp 1

ra )( r

2a

r2p 1)

= 2rpGM(ra + rp) ra

=2GMrp(1 rpra )

r2

a (1 r2p

r2a )

=2GMrp(1 rpra )

r2

a

(1

rp

ra)(1 +

rp

ra)

=2GMrp

ra(ra + rp)

v2p =2GMra

rp(rp+ ra)

v2a = v2p

2rpGM

(ra + rp) ra/

2GMra

rp(rp+ ra)

=r2p

r2a

va =rp

ravp

Done!Bill Gibson University of Vermont

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Hohmann transfer example

Example

A spacecraft is in a circular earth orbit with an altitude of 150statute miles. Calculate the vs required to change to a circularorbit with an altitude of 250 statute miles.GM= 3.99221 1014Nm2/kg and Re = 6.378 106m

Use the basic equation for circular orbits v = GM/r.


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Example


Use the basic equation for circular orbits v = GM/r.Given that the perigee has to be the 150mile orbit

rp = r1 = (6.378106m+ (150mi)1609.3m

mi) = 6.619 4106m


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Example


Use the basic equation for circular orbits v = GM/r.Given that the perigee has to be the 150mile orbit

rp = r1 = (6.378106m+ (150mi)1609.3m

mi) = 6.619 4106m

Apogee has to be the 250mi orbit

ra = r2 = (6.378106m+ (250mi)1609.3m

mi) = 6.7803 106m


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For the lower orbit the velocity is:

vlow = [(3.99221 1014)m3

s2/(6.6194 106m)]0.5 = 7766m

s


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vlow = [(3.99221 1014)m3

s2/(6.6194 106m)]0.5 = 7766m

s

Now do the second higher orbit

vhigh = [(3.99221 1014m3

s2/(6.7803 106m)]0.5 = 7673.3m

s


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vlow = [(3.99221 1014)m3

s2/(6.6194 106m)]0.5 = 7766m

s

Now do the second higher orbit

vhigh = [(3.99221 1014m3

s2/(6.7803 106m)]0.5 = 7673.3m

s

Note that higher orbit is slower.


Solution
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Solution

Now apply the equations above for the elliptical transfer orbit:

va =

2GMrp

ra (ra + rp)

=

2(3.99221 1014 m3s2 )(6.6194 106m)

(6.7803 106m)(6.6194 106m + 6.7803 106m)= 7627.1

m

s


Solution
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Solution

Now apply the equations above for the elliptical transfer orbit:

vp =

2GMra

rp(ra + rp)

=

2(3.99221 1014 m3s2 )(6.7803 106m)

(6.6194 106m)(6.6194 106m + 6.7803 106m)= 7812.5

m

s


Solution
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Solution

v (1st burn) = vpv1 = 7812.5

ms

7766.0m

s = 46.5ms

.


Solution
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Solution

v (1st burn) = vpv1 = 7812.5

ms

7766.0m

s = 46.5ms

.

v (2nd burn) = v2 va = 7673. 3ms 7627.1ms = 46.2ms


Time of Flight
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Time of Flight

Hohmann transfer fuel efficient but not economically efficient!


Time of Flight
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Time of Flight


It can take a long time...things can go wrong


Time of Flight
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Time of Flight



Radiation, solar storms, psychological problems, food, water...


Time of Flight
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Time of Flight




T=

P/2 =

a3GM


Time of Flight
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g




T=

P/2=

a3GM

a is the semi-major axis


Time of Flight
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g




T = P/2 = a3GM


Example

What is the time of flight in this problem?


Time of Flight
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g




T = P/2 = a3GM


Example

What is the time of flight in this problem?

Answer: calculate!

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20f12 Hohmann