1.2 advantages of this position are the

1.2 Location and Orbit
1.2.1 Which Planet Will Our Space Settlement Orbit?
As The Iris will be the first space settlement built by humans, it would be wise to position
the spacecraft close to Earth. The main advantages of this position are the lowering of both
transportation and construction costs as much as possible, and the easier communication with
Earth. Moreover, if the settlement will orbit our planet, its inhabitants will be able to permanently
see the Earth, which will make them feel more secure and calmer. After considering all
of these benefits, we decided that The Iris will orbit the Earth.
1.2.2 The Lagrangian Points
In 1772, Joseph Louis Lagrange discovered five points in an orbital configuration of two large
bodies where can be placed a third body, of comparatively negligible mass, so as to maintain
its position relative to the two massive bodies. These are called the Lagrangian points and are
labeled L1 to L5. In each of these positions, the combined gravitational attractions of the large
bodies are balanced by the centrifugal force of the small body required to orbit with them.
We will refer to the Earth-Moon system, since we decided The Iris will orbit Earth. Hence, our
settlement will orbit the Earth with the same period as the Moon.
Figure 6: The Lagrangian points of the Earth-Moon system
Image Credit: space.stackexchange.com
Below, we will analyze and compare all five points:
The L1 lies on the line defined by the two massive bodies (M1 and M2) and between them. This
is the only point which appears in stationary systems. Explanation: the Earth’s gravitational
pull is canceled by the Moon’s gravitational attraction and the centrifugal force which acts on
the settlement.
The L2 is also situated on the line through the two large masses, but beyond the smaller one
(the Moon). Explanation: both of Earth’s and Moon’s attraction forces are counterbalanced
by the centrifugal force.
The L3 is placed on the line defined by the two large masses, beyond the larger one (the Earth).
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Explanation: Earth’s and Moon’s attraction forces are canceled by the centrifugal force exerted
on The Iris.
Figure 7: Gravitational accelerations at L4
Image Credit: en.wikipedia.org/wiki/Lagrangian point
The L4 and L5 are equidistantly situated from the Earth and Moon and are symmetrical about
the line defined by the larger bodies. Thus, each of the two Lagrangian points forms an equilateral
triangle with the Earth and Moon.
Further, we will demonstrate that our spacecraft would be in equilibrium in these two positions:
In the diagram above, point A represents the center of the Earth (M1), B is the position of the
Moon (M2), and C is The Iris (m). The sides of the ABC triangle are noted with a,b and c.
Our settlement is in equilibrium with Earth and Moon, so the distance between them remains
constant. Point D represents not only the center of mass between Moon and Earth but also the
frame’s center of rotation (barycenter). Hence, all the three bodies will orbit around it with
the same period T. R represents the space settlement’s radius of rotation and v its velocity. We
have also noted with vM the Moon’s velocity.
As D represents the barycenter of the frame, the Moon’s radius of rotation is:
BD = c
M1
M1 + M2
= r
Since distance is equal to velocity times period, it results:
2?R = vT and 2?r = vMT
Dividing both sides of the equations above by (RT) or (rT) we obtain:
2?
T
=
v
R
and 2?
T
=
vM
r
Therefore,
v
R
=
vM
r
(1.1)
The centrifugal force exerted on the Moon is:
FcfM =
M2v
2
M
r
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The gravitational attraction of the Earth which acts on the Moon is:
FG = ?
M1M2
c
2
,
where ? symbolizes Newton’s constant of gravitation.
The two forces balance each other:
M2v
2
M
r
= ?
M1M2
c
2
Dividing the expression by M2, we have:
v
2
M
r
= ?
M1
c
2
We substitute r in the above equation:
v
2
M(1 + M2
M1
) = ?
M1
c
(1.2)
Let Fcf be the centrifugal force that acts on the settlement:
Fcf =
mv2
R
Fcf is counterbalanced by the Earth’s and Moon’s pulls:
mv2
R
= cos?FM + cos?FE
where ? and ? represent the angles into which R divides the angle C. Since FE = ?
M1m
a
2
and
FM = ?
M2m
b
2
,after substituting and dividing by m, we have:
v
2
R
= ?
M2
a
2
cos? + ?
M1
b
2
cos? (1.3)
In order to be in equilibrium, the components of the attraction forces that pull The Iris in
directions perpendicular to R have to cancel each other, thus:
FEsin? = FMsin?
?
M1m
b
2
sin? = ?
M2m
a
2
sin?
Dividing by ?m, we obtain the equation:
M1
b
2
sin? =
M2
a
2
sin? (1.4)
By squaring (1.1), it results:
v
2
R2
=
v
2
M
r
2
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Considering that r = c
M1
M1 + M2
, we have:
v
2
R2
=
v
2
M
c
2
(1 + M2
M1
)
2
We divide both sides of the above equation by (1 + M2
M1
) and multiply it by c
2
:
v
2
(
c
2
R2
)
1 +
M2
M1
= v
2
M(1 + M2
M1
) (1.5)
Now we use (1.2) and (1.5) equations:
v
2
(
c
2
R2
)
1 +
M2
M1
= ?
M1
c
Multiplying by
(1 + M2
M1
)R
c
2
results:
v
2
R
= ?
M1R
c
3
(1 + M2
M1
)
We substitute (1.3) relation in the one above:
?
M2
a
2
cos? + ?
M1
b
2
cos? = ?
M1R
c
3
(1 + M2
M1
)
After dividing the relation by ?M1, we obtain:
1
a
2
(
M2
M1
)cos? +
1
b
2
cos? =
1
c
2
(
R
c
)(1 + M2
M1
) (1.6)
From (1.4) equation, we can infer that
a
2 =
M2sin?b2
sin?M1
(1.7)
Substituting (1.7) relation in (1.6), we have:
1
c
2
(
R
c
)(1 + M2
M1
) = 1
b
2
(sin?cos?
sin? ) + cos? (1.8)
Therefore,
1
c
2
(
R
r
) = 1
b
2
(
1
sin? )(sin?cos? + cos?sin?)
Since sin?cos? + cos?sin? = sin(? + ?) = sinC, it results:
1
c
2
(
R
r
) = sin?R
b
2
(
sinC
sin? ) =?
sin?R
c
2R1
=
sinC
b
2
(1.9)
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Furthermore, we apply the law of sines in the triangle BCD:
sin?
r
=
sinB
R
It means that:
sinB =
sin?R
r
(1.10)
Using (1.9) and (1.10) equations, we can conclude:
sinB
c
2
=
sinC
b
2
But if we apply the law of sines in the ABC triangle, we have:
sinB
b
=
sinC
c
Thus,
b
3 = c
3
b = c
In this way, we have demonstrated that The Iris would truly be in equilibrium in both L4 and
L5 points.
The most important criterion when choosing our location is stability. The L1, L2 and L3 are
unstable equilibrium points. It means that we will be compelled to correct the spacecraft’s
course in order not to wander off. This action would imply constant burning of large amounts
of fuel, which is a major disadvantage. However, the L4 and L5 points are far more stable due to
the exertion of the Coriolis force. Even though the settlement would experience a perturbation,
the Coriolis force will act so as to send The Iris into a stable orbit around the point.
Consequently, if we were to position our spacecraft in a Lagrangian point, L4 and L5 would
represent the most suitable option. However, the huge distance between L4 and L5 and Earth
would cause lots of disadvantages and renders them less acceptable.
1.2.3 Earth Orbits
Regarding the position of The Iris, we also have to reckon with the Earth orbits. There are
essentially three classes of orbit: Low Earth Orbit (LEO), Medium Earth Orbit (MEO) and
High Earth Orbit (HEO), which includes a special orbit called Geostationary Orbit (GEO).
Below, we will analyze and compare these orbits.
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Figure 8: Earth orbits
Image Credit: http://www.snsweather.com/spacexlandgeo.html
The Low Earth Orbit has the substantial advantage of being the closest orbit to the Earth,
ranging in altitude from 180 km to 2000 km. Thus, it would be the cheapest orbit to build
the settlement on. Furthermore, placing The Iris on LEO would provide a high transmission
capacity to Earth and lower the communication time lag. Although the part of LEO lower than
1000 km altitude is regularly safe with respect to radiations, the inner Van Allen Radiation Belt
may go down in case of intense solar activity, sometimes reaching 200 km height. Accordingly,
a radiation shield would be a must-have.
However, positioning the spacecraft there would imply some major disadvantages. As our space
settlement would orbit at such low altitudes, The Iris would travel through the uppermost layers
of the atmosphere, where air resistance is still strong enough to tug at it, pulling it closer to
the Earth.1 Hence, we will be compelled to artificially maintain the spacecraft’s orbit, which
necessarily means to always burn significant quantities of fuel. Another reason to correct its
orbit is the large amount of space debris that may collide with The Iris at very high speeds.
As we can see in Figure 9, space debris is very concentrated in LEO, definitely increasing the
collision danger.
The Medium Earth Orbit, despite having the drawback of being further than LEO, would still
involve acceptable costs of construction. Besides, placing The Iris on MEO would involve plenty
of important benefits. For instance, MEO is stable enough to naturally keep the spacecraft on
the orbit since it does not interfere with atmosphere. MEO is located between the outer and
inner Van Allen Radiation Belts, thus having the level of radiation considerably lower than the
other Earth orbits. Moreover, the space debris is very sparsely scattered over the MEO, making
the collision hazard insignificant.
Most of the High Earth orbits are too far from Earth compared to the other orbits and don’t
present any advantages to outweigh this major drawback. We’ve found that the best High orbit
is the Geostationary Orbit, which has the asset of orbiting at the same speed like Earth. However,
GEO is located in the center of the outer Van Allen Radiation Belt, making impossible to
us to entirely protect The Iris’ inhabitants from harmful radiations.
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Figure 9: Space debris orbiting Earth
Image Credit: https://earthobservatory.nasa.gov
1.2.4 Conclusion
After an thorough and attentive examination of every possibility, we’ve eventually accomplished
our elusive goal of establishing the most suitable location for The Iris. We’ve opted for MEO to
be our spacecraft’s position since it is the only choice whose advantages clearly overbalance its
disadvantages. Although L4 and L5 are the most stable options, they would imply exorbitant
construction costs and difficult communication with Earth.