Thursday, June 27, 2013

HABITATS CAN TRANSPORT perhaps to Mars

Habitats could comfortably transport large human populations
via orbits throughout the Solar System.
Simulate earth gravity via continuous spin around longitudinal axis.
Large external mirrors could reflect sunlight into habitat
to provide power for production and life support requirements.
Island-3 habitats (r = 3,600 m, est. l= 7.2 km) has about 40,000 acres
of surface area for living quarters, agriculture and other needs.

Habitats can transfer payloads to Mars via transfer orbits. 

Once there, habitats could house Martian humans in orbits about Mars.
Martian habitats could even share same Martian orbits as Martian moons.

HUMANS CAN ORBIT MARS IN HABITATS
Deimos and Phobos
Two moons of Mars could serve as operational bases
and likely provide considerable construction materials.

Many agricultural items produced in habitat can transfer to Martian surface; such items can help terraform Mars over many years. Eventually, Mars (both surface and subsurface) will prove to be an expansive source of agriculture and raw materials.  However, Mars is unlikely to ever reproduce Earth like gravity; longitudinal spin of the cylindrical habitats can produce 1-g via centrifugal force.
Martian humans will always prefer to live in the large orbiting habitats; conditions will be more "Earth like" and access to interplanetary travel will be much easier than for the Martian humans who choose to reside on Mars itself.

In contrast with living submerged under Martian surface,
orbiting habitats provide a ready made,
comfortable and controlled environment.

Agriculture
flora would produce oxygen and food

fauna for petting and food
Simulated Earth Gravity
from longitudinal spin of cylindrical habitat.
Energy Source
sunlight from large exterior mirrors
which never tarnish.



Before habitat orbits Mars,
habitat has to get there.
USE HOHMANN TRANSFER (HT) FROM EARTH TO MARS.
PROBLEM: During lengthy space voyages, humans require plentiful supplies: food, water, oxygen as well as Earth like gravity.  None of those are now available in current space vehicles.
Infrequent, interplanetary flights are now accomplished by Artificial Intelligence (AI) devices inside small capsules. Sending humans in such vessels seems pointless; not only would travelers have to endure months/years of 0-g conditions which would likely permanently disable them; but providing required consumables would be challenging at best.

SOLUTION: On the other hand, interplanetary flight conditions could greatly improve if humans traveled on habitats during these lengthy missions. Habitats can be very comfortable "homes" during lengthy flights.
Earth
Pos.
Hab.
Pos.
Hab
Cumm.
Time
T
Mars
Pos.
0 dy57.3°
49.9°53°50 dy83.6°
99.9°99°100 dy109.9°
149.9°126°150 dy136.2°
199.8°151°200 dy162.5°
258.6°180°259 dy180.0°
T×.999°

day
Est.Σt57.3°+T×.526°

day
Power supply via sunlight from large external mirrors.  
Onboard water stores and agriculture provide oxygen and food.
Centrifugal force from longitudinal spin provides 1-g gravity.
Large human population provides essential social interaction.
Constant communication with "home planet" reinforces purpose.



As a matter of fact, habitats might be necessary for the several years required for trips to other planetary systems: Jupiter and moons, Saturn and rings and moons, Uranus, Neptune, perhaps even the planetoids such as: Pluto, Ceres, etc.
GETTING THERE
Transfer Orbit (TO) is a highly eccentric orbit between two planetary orbits. A non-powered object must use a transfer orbit to travel between planetary orbits. The most efficient transfer orbit is a Hohmann Transfer (HT), which uses least possible fuel  for "burns" to enter/exit TOs between orbits.

For HT from Earth to Mars, the perihelion (nearest point to Sol) is the exit point from Earth's orbit and its aphelion (farthest point from Sol) is where it intercepts Mars.
HT APHELION (QHT) AND HT PERIHELION (qHT)
HT Aphelion (QHT) is the semi-major axis of Mars, a.
QHT = a
HT Perihelion (qHT) is semi-major axis of Earth, a.
qHT = a = 1.0 AU
After 259 days of unpowered flight, habitat reaches destination aphelion. Vehicle emits a controlled thrust to orbit destination planet. At departure perihelion, vehicle emits controlled thrust to exit Earth's orbit and enter transfer orbit. Transfer must begin when properly planned.
SEMI-ORBIT GIVES TRANSFER TIME



From QHT and qHT, compute semi-major axis of transfer orbit, aHT
Semi-major axis:
aHT=QHT + qHT

2
=a + 1 

2
= 1.26 AU
From Kepler's Third Law, compute PHT , period of transfer orbit,
Period of Orbit
PHT[aHT]3=1.4143 year
Axiomatic: Transfer time is the duration, THT,
of the semi-orbit from qHT to QHT.
Transfer Time:
THT = PHT 

2
×365.25 days 

year 
= 258.3 days
From QHT and qHT , compute: and e:
Semi-latus rectum:
=2×Q×q

Q + q
= 1.20635 AU
Eccentricity:
e=Q - q

Q + q
= 0.20635
For reference (θ = 0°), consider a ray from Sol to perihelion, qHT.
As Habitat orbits from qHT  to QHT, angle, θ, goes from 0° to 180°.
Compute radius, distance from Sol to Habitat, for each angle, θ.
Rθ=

1 + e×Cos(θ)

In the plane of the Transfer Orbit, define a 2 dimensional (X,Y) coordinate for each θ.Use Sol as origin (0,0).
θRXY
DegAUAUAU
1.0000001.0000000.000000
1.000026 0.9998740.017453
Given

1+e×Cos(θ)
R×Cos(θ)R×Sin(θ)
APPROXIMATE INCREMENTALS
Use Pythagoras to approximate 
incremental distances
between each coordinate
[distance between (Xθ-1,Yθ-1) and (Xθ,Yθ) ]
Δxθ = xθ - xθ-1 Δyθ yθ - yθ-1
dθ = [(Δxθ)2 + (Δyθ)2]

Convert distance from AU to kilometers.
θXYdθ
DegAUAUAUkm
1.0000000.000000n/an/a
0.9998740.0174530.0174532,610,979
0.9994950.034903
0.017454
2,611,138
GivenR×Cos(θ)R×Sin(θ)AU =
149,597,870.7
km
Use point velocity to
approximate incremental travel times.
Velocity for any angle, θ.
μSol 887.123 AU-km2
Vθ= [ μSol × ( 2

Rθ
- 1

a
)]
Time between each coordinate
[between (Xθ-1,Yθ-1) and (Xθ,Yθ) ]
θXYdθVθtθ
DegAUAUkmkm/secday
1.0000000.000000n/an/an/a
0.9998740.017453 2,610,979 34.2390.8826
0.9994950.0349032,611,13834.2370.8827
0.9988640.0523482,611,45734.233 0.8829
GivenR×Cos(θ)R×Sin(θ) dθ

86,400×V
PROGRESS BY DEGREES
θRθVθ XθYθdθdkmtθ Tθ
Table shows total travel time
for selected angle, theta,(θ),
of Hohmann Transfer from Earth to Mars
BACKGROUND: Hohmann Transfers (HTs)
are the most energy efficient transfer orbits.
Hohmann transfer time is easily calculated.
THT=(1 + a)3/2

5.656 yr
A.E. Roy Orbital Motion pg. 354
For convenience, 
TE chooses to use HT time as a typical transfer orbit duration.
DegAUkm/secAUAUincr.
AU
incr.
km
incr.
days
cumm.
days
1.00034.239 kps1.00000.0000n/an/an/an/a
 0.9998 34.238 kps 0.99980.01740.01742,610,9790.920.92
0.999434.237 kps0.99940.03490.01742,611,1380.921.85
0.998834.233 kps0.99880.05230.01742,611,4570.922.77
. . .. . .. . .. . .. . .. . .. . .. . .. . .Cumulative Times
Approximate total travel time
for any angular position during transfer.
From perihelion(θ=0°) to 45°, 90° and 135°;
with respective durations of 43, 95 and 167 days .
45°1.052732.908  kps0.7440.7440.01852,767,3641.0243.03
90°1.206329.448 kps0.000
 1.206
0.0215
3,210,053
1.3495.46 
135°1.412025.523 kps-0.9990.9990.02493,736,3811.84166.97
. . .. . .. . .. . .. . .. . .. . .. . .. . .
 177°1.519523.710 kps-1.5170.0790.02652 3,967,1692.13251.96 Previous tables compute incremental distance/time
for each degree of habitat's transfer.
This table has examples for a few angles
of cumulative travel times as shown.
Source table has cumulative values for θ = 0° to 180°.

Kepler's Equation
uses an elegant method
for more accurate travel times.
178°1.59823.712 kps-1.5180.0530.02652 3,967,8662.13254.09
179°1.59923.715  kps-1.519
0.026
0.02652 
3,968,331
2.13256.22
180°1.520023.718  kps-1.5200.0000.026533,968,5642.13258.35
 Given

1+e×cos(θ)
[μSol(2

Rθ
-1

a
)]
Rθ ×
cos(θ)
Rθ ×
sin(θ)
[(Δx)2+(Δy2)]dθ × (km/AU)

1 AU =
149,597,871km
 
dkm÷VθΣtθ
Ideal transfer orbit goes directly from Earth to Mars.
However, it requires a precisely timed "launch window" of a few hours once every 758 days.

To do this, habitat must enter Hohmann Transfer (HT) exactly as Mars leads Earth by 52.7°.  
This precise position happens one time during each Earth-Mars Synodic orbit (about 758 days).
This is a great opportunity for that one day out of 758;  however, this requirement seems to prevent travel to Mars during the other 757 days.
TE considers this to be too restrictive.
To slightly expand launch window, consider a "Parking Orbit (PO)".
EXAMPLE: If habitat departs Earth two days late, it intercepts Mars orbit about one degree behind the planet Mars (about 4 million kilometers).  If habitat remains on Mars orbit, it forever lags Mars by 4 million km.  To catch up with Mars, habitat could instead intercept a slightly smaller, concentric orbit.  In this slightly quicker PO, habitat slowly decreases angular distance to Mars. PO is a good "last resort" if fuel is limited (which is why we bother to wait for the HT "launch window".)
1) Semimajor axis (a)
Orbit
aAU
akm
Earth
1.00 AU
149,597,871 km
PO
1.50 AU
224,396,806 km
Mars
1.52 AU
227,388,764 km
Given
CAU×149,597,871km/AU
a - aPO  = 2,991,958 km
For circular orbit, a is constant distant from Sol.
For Mars, a is 1.52 Astronomical Units (AU).
Let parking orbit (PO), aPO , be 1.50 AU.
Assume orbits (Mars and PO) to be concentric and about 3 million kilometers apart.
Object in inner orbit (PO) will travel slightly faster than Mars.
2) Circumference (C)
OrbitSemi-
Major
Axis
Circumference
1° arc
Dist.
aCAUCkmΔkm
Earth1.00 AU6.283 AU939,923,423 km2,610,898 km
PO1.50 AU9.424 AU1,409,926,718 km3,916,347 km
Mars1.52 AU9.550 AU1,428,725,740 km3,968,682 km
Given
2 π a
CAU×149,597,871km/AUCkm/360°

For circular orbits, approximate C:
C= 2 π  a
 
For PO, one degree arc distance:
Δkm = CPO ÷ 360
Δkm =  3,916,463 km
3) Period (P)
OrbitSemi-Major
Axis
Period
1° arc
Time
aPYrPdyΔdy
Earth1.00 AU1.000 Yr365.25 days1.015 dy
PO1.50 AU1.837 Yr671.0 days1.864 dy
Mars1.52 AU1.874 Yr684.5 days1.901 dy
Givena3PYe × 365.25 dy/YrPdy/360°

For circular orbits, approximate P:
P= a3
 For PO, one degree arc time:
Δdy = PPO ÷ 360
Δdy =  1.864 days
4) Orbital Velocity: Linear (V)
OrbitSemi-Major
Axis
Circumference
Period Linear
Velocity
a
CAU
Ckm
PYr
PdyVkm/dayVkm/sec
Earth1.00 AU6.283 AU   939,951,145 km1.000 Yr365.25 days2,573,456 kpd  29.79 kps
PO1.50 AU9.424 AU1,409,926,718 km1.837 Yr671.0 days2,101,232 kpd  24.31 kps
Mars1.52 AU9.550 AU1,428,725,740 km1.874 Yr684.5 days2,087,223 kpd24.16 kps
Given
2 π a
CAU×149,597,871km/AU
a3
PYr×365.25 dy/YrCkm÷Pdy29.785/√a
V= [μSol(2

Rθ
-1

a
)]
Adjust for circular orbit, Rθ = a; and assume:
μSol ≈  887 AU-km2/sec2
V = [887.13 AU-km2/sec2

a
] = 29.785 km/sec

√a
5) Orbital Velocity: Angular (ω)
Compute angular velocity
directly from a, semimajor axis.
Divide total circumference angle (360°) by orbital period (in days).
CDeg ÷ PPO = 360° ÷ 671.0 days = 0.537°/day
Reduce above tables to:
OrbitSemimajor
Axis
Angular
Velocity
aAUω
Earth
1.00 AU
0.999°/day
PO
1.50 AU
0.537°/day
Mars
1.52 AU
0.526°/day
Given
360°

 √a3 ×365.25 dys/yr
Differential Closure (δ)
With Habitat in the smaller, quicker orbit, it will daily decrement difference between Habitat in Parking Orbit and Mars in its orbit.
δ = ωPO - ω
δ = 0.537°/day - 0.526°/day = 0.011°/day

If Habitat is 1° behind Mars, Habitat will completely align with Mars in about 91 days.
TAlign = 1°/δ
TAlign = 1° ÷ .011°/dy= 91 days
Even when aligned with Mars, Habitat will still be in smaller orbit and displaced from Mars about 3 million kilometers.
Perhaps, flight planners could use following maneuver:
1) During the 91 day catch up period, Space Tug could gradually nudge Habitat  closer to Martian orbit, perhaps daily nudges are best.
2) For each daily nudge, the habit assumes a slightly larger orbit and thus closer to Mars orbit, unfortunate when Habitat's orbit gets slightly larger, it also gets slightly slower with an even smaller daily differential.  Thus, total transit time (from PO to MO) might take 100 days.

CONCLUSION: HT and subsequent PO might be a good "last resort" for limited fuel scenarios.
However, if fuel is available, compute direct transfer orbit instead of HT.

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