CRITICAL ISSUES'
A) Gravitational attraction force and centrifugal force
An issue we encountered in rating the attraction
gravitational link between the Sun and the ellipsoid itself lies in the fact that, considering that the Sun
it would be located exactly in the center of the structure and at its center
mass and therefore their distance would be 0 m, due to the law of attraction
gravitational force of NewtonF we should divide by
g = G
mSun mellissoid
d
Sun−ellipsoid
2
0, which is not possible in this case. Mathematically speaking, we could
affirm that the result tends to positive infinity ( . It is
x 0
+
lim
→
f(x) =+ ∞
demonstration the graph of the function drawn with the graphing calculator inserted
just below. We used rather small masses as data, relatively i
planets of the Solar System, to avoid an overload of the calculation tools),
30
but physically we could not say the same, as we are not in the presence of
a singularity. There are two possible ways:
1) We consider the ellipsoid as a unitary body, as a result we get
that the Sun is located exactly at the point that corresponds to its center of
mass and since we are not dealing with a singularity we can
state that it is an equilibrium system.
2) We consider the ellipsoid as the set of many small parts (plates of
metal having a certain surface area in m each) which are attracted - in 2
since, individually, they would have a decidedly lower mass than that
of the Sun - from the Sun, each with a strength inversely proportional to
own distance distance from it (it follows that those of the North "poles" e
South of the ellipsoid, i.e. the points intersecting the axis of solar rotation,
are attracted more than those further away)
The second consideration therefore implies that the structure would tend to collapse
due to the force of gravity. How to avoid this? We imagined two strategies
possible. The first is to evaluate a gas pressure value sufficient for
generate a force on each of the metal plates equal and opposite to that
generated by the gravitational pull of the Sun. However, this could lead to
problems listening to the concert, for the reasons explained in SECTION 3. A
second solution makes things slightly more complicated: to exercise a
force that has the same magnitude and opposite direction to that of gravitational attraction
we could also put the entire structure into rotation, or in other words, insert into
solar orbit every single part of the structure (which would mean,
overall, put the structure into rotation). So there would be a force
centrifugal pointing in the opposite direction to the centripetal one (represented by the force
gravitational attraction of the Sun) which could guarantee the stability of the
31
structure. The only criticality of this solution lies in the fact that the ellipsoid is not
being a solid of rotation, it would risk colliding with Pluto. The only way
to solve this problem would be, given the fact that Pluto makes a
revolution around the Sun every 248 Earth years (so it has a low speed
orbital), pressurize the facility and perform the concert well in advance
before Pluto reaches aphelion, so as not to risk colliding with the
walls of the ellipsoid once it is rotated. Clearly, finished the
concert, it would be necessary to bring the ellipsoid back to its initial position, otherwise yes
would, in fact, risk the collision.
B) Survival
Musicians and listeners may have some difficulty breathing
an atmosphere consisting of only hydrogen or only helium, even if they had to do so for
only a few minutes: neither of the two gases is toxic to humans, but they cause asphyxiation
quickly in case of oxygen deficiency (in our case we are talking about
pure hydrogen). Therefore, cylinders filled with breathing gas mixtures would be needed
(such as those used by divers, which are filled with Nitrox, a mixture of
nitrogen and oxygen, Trinix, composed of nitrogen, oxygen and helium or even Eliox, a
mixture of helium and oxygen) primarily for workers, but also for musicians and
spectators. If we provided even a single gas cylinder to each worker (then ne
they would also be useful for musicians, who we would not want to play in apnea, and for
spectators) you should get 3, 887180925 · 10 gas cylinders.
35
Considering that a cylinder having a volume of 5 L filled with breathing gas
with a pressure of 150 bar it has an autonomy of about 75 minutes with a consumption of 10
L/min and since the workers don't need 75 minutes to build such a
megagalactic structure, you should get this number of cylinders:
no
workers
: t
autonomy 1 cylinder = n
total, cylinders
: t
ellipsoid construction
no
total, cylinders =
(No
workers
t
ellipsoid construction)
t
autonomy 1 tank
=
(3,887180925 10
35
·3.1536·10
9
)
4500 = 2.72414 10
41
cylinders
for a total of
V of gas needed,
gas = no
total, cylinders
c
capacity 1 cylinder = 2.72414 10
44
L
in addition to those needed to fill the structure. The volume occupied by all of these
oxygen cylinders would amount to
V ,
tot, cylinders, m
3 = v
1 cylinder, m
3 · no
cylinders =
5
1000
2, 72414 10
41 = 1.36207 10
39
m
3
equal to times the volume of the Sun.
v
tot, cylinders, m
3
v
Sun
=
1.36207 10
39
1.4122 10
27 = 9.65 10
11
32
Certainly a warehouse would have to be built to contain them, to compress the gas
even more to reduce their volume, or simply stack them outside
from the ellipsoid to introduce them little by little inside it as you go along
they run out.
In addition to the breathing problem there would be that of radiation to which
musicians and workers would be exposed: especially for those who were in the
vicinity of the Sun, the quantity of these would be very high and dangerous.
The radiation (per unit of time, i.e. per second) received by Mercury is worth
9083, about 7 times greater than that received on Earth outside W
m
2
of the atmosphere.
The musicians at the center of the Solar System would receive one more deal
higher radiation and would be in serious danger in the event of solar storms e
flares, so it would be necessary to carefully evaluate the right moment for
the performance of the concert, in accordance with the predictions of observers of the activity
solar like NASA's Solar Dynamics Observatory, so avoid
over toasting the musicians.
The solar radiation values would vary taking into account the presence of
gas between the Sun and the various observers, but certainly in the areas closest to the Sun
they would be quite high. At most the man in adulthood can receive without
report damage 20 mSv of radiation in one year. In case of higher values
spacesuits would be needed (that leave the ears outside, to be able to listen
the enchanting music of Holst) protectors made of materials that are able to shield the
radiation received from the Sun. The risk of gamma rays from the cosmos would be
lower as the hydrogen or helium in the atmosphere would be able to shield them
a good part (in fact their density, as highlighted in SECTION 3 is equal to
0.46 and 0.23 times that of the earth's atmosphere at sea level, so it would provide
good protection. Helium reduces the level of radiation received by exposed individuals
even more than hydrogen).
The designers of the radiation protection suits from the Sun, equipped with
breathing tanks, they should also take into account the high
temperatures to which individuals - musicians in particular - would be subjected e
use insulating materials as well as cooling systems.
C) Boiler problems
However there is a problem: before the indefatigable workers set about
building the ellipsoid the heat was transmitted by radiation and a good part of
it was lost in deep space, now the heat emitted by the Sun remains
trapped inside the structure. Indeed, since it is a closed system that does not
exchanges matter or energy with the outside, the temperature of the gas inside the
structure would tend to increase unabated. This fact could be
first of all annoying for those who particularly suffer from the heat, but would go to
gradually change the speed of sound propagation, which would increase
33
hand in hand with the temperature, the density would decrease and the density would increase
more and more pressure (similar effect to a pressure cooker). Beyond a certain
limit (although it is necessary, considering the size of the structure, a rather long time
long) the structure would risk exploding. Possible solutions could be
valves placed on the surface of the structure to expel a part of the gas, which
it should then be replenished with gas with a lower temperature, or systems of
cooling comprising radiators which allow the dispersion of part of the
heat outside the structure (similar to those used for satellites and spacecraft
space that need, in particular situations, to keep their own low
temperature, called ACTS, or Active Thermal Control System).
D) Hot areas and cold areas
The temperature established by us for the gas to be inserted inside the structure
(87.15 K) would undergo changes internally due to the activity of
Sun. The areas closest to it would be heated and would tend to be more
hotter than those further away. This would also affect the density of the gas,
lower near the Sun and higher as we move away and the speed of
sound propagation (in the areas of Mercury and Venus the sound would propagate
more rapidly, while going towards Jupiter, Saturn and the outermost planets the most
low gas temperature would slow down the sound waves). The
heat emitted by the Sun inside the mega structure would propagate by
convection: the Sun in fact would transmit heat (and kinetic energy) to the particles of
adjacent gases causing displacement, with the generation of convective motions. Not
it would be radiation as between the bodies within the Solar System not
there would no longer be a vacuum but a gas, and one could not speak of conduction since
the gas particles are free to move and transmit energy and heat to them
that they meet.
E) Pluto on a collision course.
Based on the calculations in sections 4 and 5 (related to listening time and
collision of the planets with the Sun), it is possible to infer that the listeners in orbit around
to the dwarf planet will collide with the Sun well before the sound of the concert
you get to touch the current orbit of the same, but this does not mean that it will not be possible
listen to the concert from the planet: it is evident that going in two opposite directions,
the sound waves and Pluto and traveling, the former, in all directions, in some
point in space our wandering planet will intercept the sound of the concert. The
the problem that manifests itself to us is therefore time: we need to fill the
Megastructure with gas and play the concert over a year (time for
collide with the Sun using hydrogen).
We consider that a cylinder with the capacity of 100 kg of gas is capable of
dispense about 2 kg/h. The quantities of gas that we have to take into account are
those calculated in section 3, therefore:
Helium = 4.9237 10
29 kgs
34
Hydrogen = 2.479 10
29 kgs
Let's assume we want to take things slowly instead of getting too long with the times e
therefore wanting to complete the pressurization work in 6 months (4380 h). We'll have to
then dispense:
Helium = 4.9237 10
29Kg/4380h
≈ 1, 124 10
26 Kg/h
Hydrogen = 2.479 10
29Kg/4380h
≈ 5, 660 10
25 Kg/h
Then dividing by the delivery capacity of the cylinders we get to be
needed:
Helium =
1.124 ·10
26 Kg/h
2 Kg/h ≈ 5, 62 · 10
25
cylinders
hydrogen =
5.660 ·10
25 Kg/h
2 Kg/h ≈ 2.83 10
25
cylinders
F) All united for Pluto.
As explained previously, fill the Solar System with gas
would cause the motion of the planets to slow down due to friction, and just in case
of poor Pluto would cause it to fall on the Sun in a rather short time.
Surely at the end of the concert all the spectators will be satisfied,
enthralled by the melodious music of Holst, but we are sure that none of these
would like to definitively say goodbye to Pluto, already a victim of discrimination from
when it was defined as a "dwarf planet" in 2006. So we imagined a
way to prevent the poor planet from having to end its existence
colliding with the Sun: we thought of counteracting the friction force exerted
from the air to the planet by installing geared rocket engines on Pluto's surface
so as to exert, once turned on, a force in the opposite direction to that of
friction.
The friction force exerted on Pluto holds
F = m
Pluto
a = 1.293 10
22
0.0001414438145 = 1.83 10
18 No
The most powerful rocket engine built in human history is the famous F-1
Pratt & Whitney Rocketdyne, designed by Wernher von Braun and used in the
Saturn V, the impressive multistage rocket that allowed man in 1969 to
reach the moon. A single F-1 engine is capable (in a vacuum) of generating a
force of 7,700,000 N. Therefore, to keep Pluto in orbit it would be necessary
turn on an even number
n F-1 engines =
f
friction
f
1 engine
=
1.83 10
18
7,7·10
6 = 2.38 10
11 engines
for about two years (the time needed for
pressurize the facility, perform the concert, ed
35
possibly depressurize it). Engines should be oriented so
exert (as explained above) a force in the opposite direction to that of friction. For
do this, considering that Pluto completes a full rotation on itself in about
154 hours, the engines should be moved periodically to correct
orientation or, tripling the number, reduce the ignition time to one
third (reducing to one third, consequently, also the maximum angle of
inclination of the engine relative to the direction of the frictional force, which would pass
therefore from 90° to 30° and considerably reducing the inclination error compared
to the friction force vector). So the number of engines would be triple that
previously indicated, i.e. 7, 14 · 10 . An F-1 engine consumes 11 per second
1565 liters of liquid oxygen and 976 liters of RP-1 (or more simply kerosene),
therefore to keep the engines running for two years it would be necessary
v
liquid oxygen, 2 years = V
fuel, 1s
t
seconds = 1565 63072000 = 9, 9 10
10
L
of liquid oxygen e
v
RP−1, 2 years = V
fuel, 1s
t
seconds = 976 63072000 = 6, 2 10
10
L
of RP-1.