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
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"
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 Earthlike, 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 gforce 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(70Â°) takes about 10% of orbit period.
Standard Mean Motion
Determine n for Earth's orbit.a = 1.0 AU = 149,597,870,700 m = 1.49598×10^{11 }m
Convert rad/sec to a more convenient degrees per day:
To obtain n as deg/day and/or rad/day, input a as AU in following:
EXAMPLE2: 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
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.
For any orbit, any given ν requires t amount of travel time.
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.
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.
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 Orbit  Ap 

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 NEARename 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 Earthlike, 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 gforce 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(70Â°) takes about 10% of orbit period.
Determine MA Ratio

MA Area = TA Area

Determine TA Ratio


Angle Ratio is Area Ratio

A_{MA} =Ï€aÂ²Ã—âˆ _{MA} / 360Â° = A_{TA}

T_{TA} : T_{Orb} = A_{TA} : A_{Orb}


âˆ _{MA}
âˆ _{ACTotal}

=

EeÃ—sin(E)
360Â°

=

37.5Â°
360Â°

=

.104

A_{MA} = 3.14Ã—(1.5AU)Â²Ã—10.4% = A_{TA}
A_{MA }= .734 AU^{2} = A_{TA}
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.

T_{TA}

=

âˆ _{MA}
360Â°

Ã—

T_{Orb}


For Î½ = 70Â°, MA Area is 10.4% of AC area.

T_{TA }= 37.5Â°/360Â° = .104 Ã—T_{Orb}

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.

Standard Mean Motion
Determine n for Earth's orbit.a = 1.0 AU = 149,597,870,700 m = 1.49598×10^{11 }m
EXAMPLE1: Determine mean motion, n, for Apollo.
Semimajor axis: a_{Apollo} = 1.47 AUs = 2.199 x 10^{11} m
 
xx  xx 
Determine n for Earth's orbit.a = 1.0 AU = 149,597,870,700 m = 1.49598×10^{11 }m
n =  √(μ_{Sol}) √(a^{3})  =  √(13.27×10^{19}m^{3}s^{2}) √(1.4×10^{11}m)^{3}  =  √(3.96×10^{14})^{ } √(sec^{2})  =  1.991×10^{7 }rad sec 

n =  1.991×10^{7} rad second  ×  86,400 sec day  =  .0172 rad day  ×  180° π rad  =  .9856° day 

.0172 rad/day √(a^{3})  =  n  =  .9856°/day √(a^{3}) 

EXAMPLE1: Determine mean motion, n, for Apollo.
Semimajor axis: a_{Apollo} = 1.47 AUs = 2.199 x 10^{11} m
 

n_{Apollo} = √(1.248×10^{14}) rad/sec = 1.11717×10^{7} rad/sec  

For convenience, restate mean motion formula
to input a as AU and output n as deg/day.
to input a as AU and output n as deg/day.
Restate G with AUs.  G_{AU} = 1.99346×10^{44 }AU^{3}kg^{1}sec^{2}  
Recompute μ_{Sol}.  μ_{Sol} = G_{AU} × M_{Sol }= 3.964×10^{14}AU^{3}/sec^{2}  
Convert secs to days.  (86,400sec/day)^{2}×μ_{Sol} = 296.102×10^{6}AU^{3}/day^{2}  
Square Root  √(μ_{Sol}) = 17.208×10^{3 }AU √(AU) /day  
Convert n to deg/day. 
n = (180°/π rad)× √(μ_{Sol}/a^{3})

√(μ_{Sol}/a^{3}) = n  =  .0172 rad/day ÷ a^{3/2 }= .9856°/day ÷ a^{3/2} 

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
Example1: 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.
Example2: If a given M for Apollo happens to be 16.59°;
determine associated duration:
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.
Example2: 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/a^{3}); then above expression for M can be rewritten as:
M_{t} = t × √(μ/a^{3})
Mean Anomaly(M)
was previously interpreted as the circular sector angle
which yielded same area as a specified True Anomaly (TA).
t = M_{TA }× a × √(a/μ)
Mean Anomaly(M)
was previously interpreted as the circular sector angle
which yielded same area as a specified True Anomaly (TA).
M_{TA}= E  e × Sin(E)
where M and E are in radians
where M and E are in radians
Kepler's Equation
t × √(μ/a^{3}) = M = E  e × Sin(E)

Rearrange to solve for t:
t = (E  e × Sin(E)) × a × √(a/μ)t = M_{TA }× a × √(a/μ)
where e is the eccentricity and t is time counted from a pericentre passage.
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 

For any given ν, there is a Mean Anomaly, M,
a circular sector with same area as ν, an ecliptic section.
a circular sector with same area as ν, an ecliptic section.
ν  E  Mean Anomaly (M) 

45°  32.65°  22.35° 
90°  70.53°  52.52° 
Given  Prev. Tab.  E  e × Sin(E) 
Mean Motion, n, can be easily computed for any orbit.
EXAMPLE: Let semimajor axis, a = 1.5.
ν  E  M  Mean Motion (n) 

45°  32.65°  22.35°  .5365°/day 
90°  70.53°  52.52°  .5365°/day 
135°  119.28°  102.62°  .5365°/day 
Given  Previous Table  .9856°/day ÷ √(a^{3}) 
ν  E  M  n  travel time (t) 

45°  32.65°  22.35°  .5365°/day  41.7 days 
90°  70.53°  52.52°  .5365°/day  97.9 days 
135°  119.28°  102.62°  .5365°/day  191.3 days 
180°  180.00°  180.00°  .5365°/day  335.5 days 
Given  Previous Table  M_{TA }÷ 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 (t_{ν})  (ν) True Anomaly  Elapsed Time (t_{ν})  (ν) True Anomaly  Elapsed Time (t_{ν})  

15°  13.26 dy  135°  191.28 dy  255°  547.46 dy  
30°  26.97 dy  150°  234.56 dy  270°  573.12 dy  
45°  41.65 dy  165°  283.38 dy  285°  594.68 dy  
60°  57.88 dy  180°  335.51 dy  300°  613.15 dy  
75°  76.34 dy  195°  387.64 dy  315°  629.37 dy  
90°  97.90 dy  210°  436.47 dy  330°  644.05 dy  
105°  123.56 dy  225°  479.74 dy  345°  657.77 dy  
120°  154.38 dy  240°  516.64 dy  360°  671.02 dy  

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.
0 Comments:
Post a Comment
Links to this post:
Create a Link
<< Home