Tuesday, January 21, 2014

Proposed "fixes" ; MeanMo(n);

7 . Cycler Paths & Kepler's Laws
Link lost, find material; perhaps delete.

1. HABITATS FOR HUMANITY.
add Sidebar: Dr. Gerald O'Neill

3. HABITAT GOES TO MARS.
add sidebar: Hohlman and his transfer.

4.  CYCLERS FOR HUMANITY.
Reformat text following:May the Force be with you.

5. SPEED DETERMINES POSITION:
Narrow table

Earth OrbitAp
Widen table starts with Step 7)


6. MARSONANCE:
Might be good as is, closely check.

8. LINE OF NODES 
Replace following with bio of guy who discovered "line of nodes"

Sidebar: Johannes Kepler

9. Transponder on NEA
Rename and sync up with TabContents
Fix missing pix..

10. AI: Virtual presence via Robots and Nanobots
Rewrite.

11. Robotic observatory.
"Dark side of the moon" on steroids best vantage point to observe the Universe.

12. Terraformation notes To become Earth-like, habitats must transform via export of Earth's soil with both minerals and microbes.

13. Centripetal: Rotation is Essential Human habitats, orbiting cylinders of huge dimensions, will rotate around their longitudinal axis. They will orbit the Earth and throughout the Solar System.

14. Impart Spin with Mass Driver Moving a huge habitat from zero spin to g-force spin will take a significant amount of power.

15. Mining Operations Asteroids are a plentiful source of materials required to construct and maintain habitats in space.

16. Habitats: Alpha and Omega Orbiting Sol at 60° ahead or behind Earth might have certain advantages.

17. Launching Alpha The most convenient way would leverage Luna and its vast resources.

2. KEPLER AND HIS LAWS


good for now.
 Mean Anomaly (MA) Summary
MA area value = value of TA area.


 For this orbit, we now know this particular True Anomaly(70°) takes about 10% of orbit period.
 
 
Determine MA Ratio
MA Area = TA Area
Determine TA Ratio
Angle Ratio is Area Ratio
AMA =πa²×∠MA / 360° = ATA
TTA : TOrb = ATA : AOrb
∠MA


∠AC-Total
=
E-e×sin(E)


360°
 
=
37.5°


360°
=
.104
AMA = 3.14×(1.5AU)²×10.4% = ATA
AMA = .734 AU2 = ATA
Mean Anomaly (MA) area is certain percentage of Auxiliary Circle (AC) area. Per Kepler, resulting area value is same area encompassed by True Anomaly (TA) inside elliptical orbit.
TTA
=
∠MA


360°
×
TOrb
For ν = 70°, MA Area is 10.4% of AC area.
TTA = 37.5°/360° = .104 ×TOrb


Mean Motion (n) (see Wikipedia)
Mean motion, n, is the average angular velocity
of a satellite progressing through one cycle of an orbit.
Most orbits are elliptical with angular velocities
which vary according to their position vector.  
Thus, mean motion is an average value
which coincides with only one midrange angular velocity.
In the special case of circular orbits, object maintains a constant angular velocity.
Thus, if an object orbits about an auxiliary circle (AC), 
orbital speed would be a constant (mean motion) for all orbital positions. 
Following equation leverages Kepler's Third Law.
n = √(G×MSol)

√(a3
=√(μSol)

 √(a3)
DEFINE DIMENSIONS
Gravitational constant:G = 6.67384×10-11m3kg-1sec-2
Sun's mass: MSol = 1.98855 × 1030 kg
Standard gravitational parameter:μSol = G × MSol
= 13.271265×1019
m3/sec2
semi-major axis a in meters
mean motion n in radians per sec.
Standard Mean Motion
Determine n for Earth's orbit.a = 1.0 AU = 149,597,870,700 m = 1.49598×1011 m
n =√(μSol)

 √(a3)
=(13.27×1019m3s-2)

(1.4×1011m)3
=√(3.96×10-14) 

(sec2)
= 1.991×10-7 rad

sec
Convert rad/sec to a more convenient degrees per day:
n =1.991×10-7 rad

second
×86,400 sec

day
=.0172 rad

day
×180°

π rad
=.9856°

day
To obtain n as deg/day and/or rad/day, input a as AU in following: 
.0172 rad/day

√(a3)
 = n=.9856°/day

 √(a3)

EXAMPLE-1: Determine mean motion, n, for Apollo.
Semimajor axis: aApollo = 1.47 AUs = 2.199 x 1011 m
nApollo=Sol)

(a3)
=(13.27126×1019m3/s2)

((2.199×1011m)3)
=(13.27126×1019) rad

(10.6335×1033) sec
nApollo = (1.248×10-14) rad/sec = 1.11717×10-7  rad/sec
nApollo=1.11717×10-7 rad

second
×180°

π rad
×86,400 sec

day
xxxx
Standard Mean Motion
Determine n for Earth's orbit.a = 1.0 AU = 149,597,870,700 m = 1.49598×1011 m
n =√(μSol)

 √(a3)
=(13.27×1019m3s-2)

(1.4×1011m)3
=(3.96×10-14) 

(sec2)
= 1.991×10-7 rad

sec
Convert rad/sec to a more convenient degrees per day:



n =1.991×10-7 rad

second
×86,400 sec

day
=.0172 rad

day
×180°

π rad
=.9856°

day
To obtain n as deg/day and/or rad/day, input a as AU in following: 
.0172 rad/day

√(a3)
 = n=.9856°/day

 √(a3)


EXAMPLE-1: Determine mean motion, n, for Apollo.
Semimajor axis: aApollo = 1.47 AUs = 2.199 x 1011 m


nApollo=Sol)

(a3)
=(13.27126×1019m3/s2)

((2.199×1011m)3)
=(13.27126×1019) rad

(10.6335×1033) sec
nApollo = (1.248×10-14) rad/sec = 1.11717×10-7  rad/sec
nApollo=1.11717×10-7 rad

second
×180°

π rad
×86,400 sec

day
 = .553°/day


For convenience, restate mean motion formula
to input a as AU and output n as deg/day.
REDEFINE DIMENSIONS
Restate G with AUs.GAU = 1.99346×10-44 AU3kg-1sec-2
Recompute μSol.μSol = GAU × MSol = 3.964×10-14AU3/sec2
Convert secs to days.(86,400sec/day)2×μSol = 296.102×106AU3/day2
Square Root√(μSol) = 17.208×10-3 AU √(AU) /day
Convert n to deg/day.
n = (180°/π rad)× √(μSol/a3)
√(μSol/a3) = n = .0172 rad/day ÷ a3/2   = .9856°/day ÷ a3/2
EXAMPLE-2: Apollo, let a = 1.47AU
n = .9856°/day ÷ (1.47)3/2
n = .9856°/day ÷ 1.78228   =  0.553°/day 
Mean Anomaly (see Astronomical Glossary)
Mean Anomaly, M, has a second meaning; it relates mean motion to the auxiliary circle.

For any solar orbit, object's velocity varies
from minimum at aphelion (Q) to maximum at perihelion (q).
Between min and max is an average speed; i.e., "Mean" Motion (n).
If object could enter orbit's AC, velocity would become constant n.
Let n be an angular velocity: radians/sec, radians/day, degrees/day.
Mean motion times time provides the angle traveled for a given duration.
∠ = n × t
Example-1: Recall Apollo's mean motion is n = 0.553°/day.
Thus, for duration, t = 30 days,
= 0.553°/day × 30 days = 16.59° on Apollo's AC.

On the other hand, for any AC angle,
above equation enables us to determine associated duration.
Since Mean Anomaly (M) is an AC angle; determine duration for any given M.
Example-2: If a given M for Apollo happens to be 16.59°;
 determine associated duration:
t = M ÷ n
t = 16.59° ÷ 0.553°/day = 30 days

Recalling that mean motion, n, is defined as n = √(μ/a3); then above expression for M can be rewritten as:
Mt = t × (μ/a3)
Mean Anomaly(M)
was previously interpreted as the circular sector angle
which yielded same area as a specified True Anomaly (TA).

MTA= E - e × Sin(E)
where M and E are in radians





Kepler's Equation

t × (μ/a3) = M = E - e × Sin(E)

Rearrange to solve for t:
t = (E - e × Sin(E)) × a × (a/μ)
t = MTA × a × (a/μ)
where e is the eccentricity and t is time counted from a pericentre passage.
True Anomaly, ν, is object's angular distance from orbit's perihelion.
Any given ν maps to a specific Eccentric Anomaly, E.

EXAMPLE: Let eccentricity, e = 1/3.
True Anomaly(ν)Eccentric Anomaly (E)
45°32.65°
Given
E =Cos-1(Cos(ν) + e

1 + e Cos(ν)
)
For any given ν, there is a Mean Anomaly, M,
a circular sector with same area as
ν, an ecliptic section.
νEMean Anomaly (M) 
45°32.65°22.35°
90°70.53°52.52°
GivenPrev. Tab.E - e ×  Sin(E)
Mean Motion, n, can be easily computed for any orbit.
EXAMPLE: Let semimajor axis, a = 1.5.
νEMMean Motion (n) 
45°32.65°22.35°.5365°/day
90°70.53°52.52°.5365°/day
135°119.28°102.62°.5365°/day
GivenPrevious Table .9856°/day ÷  √(a3)
For any orbit, any given ν requires t amount of travel time.
νEMntravel time (t)
45°32.65°22.35°.5365°/day41.7 days
90°70.53°52.52°.5365°/day97.9 days
135°119.28°102.62°.5365°/day191.3 days
180°180.00°180.00°.5365°/day335.5 days
GivenPrevious Table MTA ÷ n


SUMMARY
For any orbit, a straight forward procedure can determine travel time for any True Anomaly (ν).
This is elapsed time since asteroid last passed perihelion (q), closest point of orbit to Sun. 

(ν) True
Anomaly 
Elapsed
Time (tν) 
 (ν) True
Anomaly
Elapsed
Time (tν) 
(ν)  True
Anomaly
Elapsed
Time (tν) 
15°13.26 dy135°191.28 dy255°547.46 dy
30°26.97 dy150°234.56 dy270°573.12 dy
45°41.65 dy165°283.38 dy285°594.68 dy
60°57.88 dy180°335.51 dy300°613.15 dy
75°76.34 dy195°387.64 dy315°629.37 dy
90°97.90 dy210°436.47 dy330°644.05 dy
105°123.56 dy225°479.74 dy345°657.77 dy
120°154.38 dy240°516.64 dy360°671.02 dy
E =Cos-1(Cos(ν) + e

1 + e Cos(ν)
)
MTA = E-e×Sin(E)
n =√(μSol)

 √(a3)
=.9856°/day

(a)3
tν =MTA

 n
CONCLUSION: Kepler's equation provides a way to predict times of flight
After Kepler inherited Tycho's post as Imperial Mathematician, he spent the rest of his life working with Brahe’s data and publishing many works. Johannes Kepler died in Regensburg in 1630, while on a journey from his home in Sagan. Two years later, action from the Thirty Years War demolished his gravesite. Though Kepler lived and died in turbulent times, his scientific works had a far greater effect on humanity than all the historical events during his time.

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