Resonant cyclers would have orbits in exact multiples of years. Thus, they return near Earth at same date during those particular years. Thus, a resonant cycler could conveniently rendezvous with Earth at same date every two years. Thus, every two years, returning passengers could disembark; departing passengers could board for a two year voyage.
PROBLEM: Unfortunately, this doesn't necessarily mean a subsequent rendezvous with Mars. Since orbit Mars is not resonant with the orbit of Earth, relative position of Mars will inconveniently shift everytime the cycler returns. Thus, Mars bound passengers might wait longer than expected, or they might not even have an opportunity to go to Mars.

Upon cycler's closest point to Sol (perihelion),
it could ideally lag the Earth by 41°.
In this example, it would rendezvous with the Earth 50 days later. 
Resonant Cycle: Year 1
Resonance indicates a multiple.
(In music, resonant wavelengths are multiples of a base wavelength.)
An Earth resonant cycler would have an orbital period
with an exact multiple of Earth's period, one year.
For a cycler orbit of exactly two years, compute semimajor axis:
a = ∛(T^{2}) = ∛(2^{2}) = 1.5874 AU
For convenience, arbitrarily choose semilatis rectum: l = 1.000 AU
While a 2 year orbit must have a semimajor axis of 1.588 AU, the semilatis rectum can be selected from an infinity of values from a reasonable range. For example, there is an infinity of values between .9 AU and 1.1 AU. An l of 1 AU leads to following orbital elements:
Focus: c = √(a^{2}  al ) = .96563AU
Perihelion: q = a  c = .622 AU
Aphelion: Q = a + c = 2.553 AU


After exactly 2 years, cycler would again be at perihelion,
As an precise mutliple of Earth's period, cycler would once again lag the Earth by 41°.
It would once again rendezvous with the Earth 50 days later (see below). 
Resonant Cycle: Year 2
ASSUME COPLANAR CYCLER. For convenience, assume cycler orbit is human engineered to be coplanar with the Ecliptic. Thus, angle of inclination , i, is about zero, and inclination distances are disregarded for this example, but would be considered in real cases. Even for human engineered orbits such that i < 1.0°, recall there is an infinity of values between 0° and 1°.
TWO INTERSECTIONS  ONE RENDEZVOUS. Cycler's orbit first intersects Earth's orbit in 50.6 days for a successful rendezvous. After 1 year + 315 days, cycler again intersects Earth's orbit in a different position. However, the Earth is nowhere near, and a second rendezvous is not feasible in this example.

Resonant Cycle:To Mars
ASSUME INITIAL RENDEZVOUS TO BE PERFECT. Every year (about March 21), Earth crosses exactly opposite Line of Aries at same time/point in annual orbit (date changes between Mar 20 and 21 due to Leap Year phenomena). This example assumes a cycler period of exactly 2 years; further assume cycler's orbit has been adjusted to rendezvous with Earth exactly at March 21 on alternate years (every other March 21).
We further assume a fortuitoous circumstance such that on Year 0, Mars leads Earth by 5 degrees as shown in diagram. If the orbits of Earth and Mars are properly positioned; then, cycler could accomplish a subsequent rendezvous with Mars 63 days laters on May 23.
Earth  Cycler  Mars 
1.0 year  0.5 orbit 
.532 orbit

2.0 year  1.0 orbit  1.064 orbit 
For every Cycler period of two years, Earth does two complete orbits about Sol while Mars does 1.064 orbits, one complete orbit plus 23°. 
Repeat Rendezvous is Rare
Problem: Orbit of Mars does not resonate with orbit of Earth.
Two years later (Year Two), Cycler again passes near Earth on March 21 and proceeds to orbit of Mars on May 23. Unfortunately, Mars is not there; insteand, it is 23° down track (about .61 AU or 91,000,000 km), a considerable distance and impractical for a rendezvous.
Earth  Cycler  Mars 
2 year  1 orbit 
1 orb + 23°

4 year  2 orbit 
2 orb + 46°

. . .  . . .  . . . 
30 yrs  15 orb 
15 orb+345°

32 yrs  16 orb 
17 orb + 8°

. . .  . . .  . . . 
94 yrs  47 orb 
50 orb+3°

Subsequently, Mars further increases the angular difference as shown in the diagram. Eventually, this angular difference approaches 360° (comes full circle to approach the Year 0 position). Unfortunately, 23° does not divide evenly into 360°; thus, after 15 orbits, May 23 position of Mars has progressed 345° around its orbit and the 16th May 23 position (not shown) winds up 8° (31.75 million km) beyond the desired rendezvous position. After 47 Cycler orbits (94 Earth years), Cycler's May 23 position finally intersects Mars's orbit near Mars itself (2.9° away, about 11 million km) and might prove practical for a payload exchange.
RESERVED for later discusion:
Analyze ways to modify cycler orbit: gravity assists from Earth; intraorbit thrusts (fuel burns).
Analyze retrograde rendezvous possibilities where cycler intersects oribt of Mars on return portion of orbit.

Lewis: "Mining the Sky" (p. 115). Lewis speculates about a system of resonant cyclers, each cyclers has a two year orbit but at a different position along Earth's orbit. For example, a system of 6 carefully orchestrated cyclers might have arrival times sequenced for every four months.
Thus, first cycler might have timetable: Jan 2020, Jan 2022, Jan 2024, and so on.
2nd cycler: Mar 2020, Mar 2022, Mar 2024, and so on.
3rd and remaining cyclers would sequence throughout remaining portion of two year cycler 
Orbital Elements
Mars Solar Orbit
Aphelion  Q  =  1.6659 AU  =  a + c 
Perihelion  q  =  1.3815 AU  =  a  c 
Semimajor axis  a  =  1.5237 AU  
Given

Semiminor axis  b  =  1.5170 AU  =  (a^{2}  c^{2})^{1/2} 
Focus  c  =  0.1422 AU  =  a × e 
Eccentricity  e  =  0.0933  
Given

Semilatus Rectum  l  =  1.5104 AU  =  b^{2} /a 
Sidereal Period  T  =  1.88 Yr  =  (a^{3})^{1/2} 
Angular Velocity  ω_{M}  =  0.524 °/day  =  360°/T 
Orbital Elements
EarthMars Resonant Cycler
Aphelion  Q  =  2.5530 AU  =  a + c 
Perihelion  q  =  0.6218 AU  =  a  c 
Semimajor axis  a  =  1.5874 AU  =  (T^{2})^{1/3} 
Semiminor axis  b  =  1.2599 AU  =  (l × a)^{1/2} 
Focus  c  =  0.9656 AU  =  (a^{2}  b^{2})^{1/2} 
Eccentricity  e  =  0.6053  =  c / a 
Semilatus Rectum  l  =  1.0000 AU  
Given

Period  T  =  2.000 Yr  
Given

Terran Orbital Elements
Assume Circular Solar Orbit
Aphelion  Q  =  1.0 AU  =  a + c 
Perihelion  q  =  1.0 AU  =  a  c 
Semimajor axis  a  =  1.0 AU  
Given

Semiminor axis  b  =  1.0 AU  
Given

Focus  c  =  0.0 AU  =  (a^{2}  b^{2})^{1/2} 
Eccentricity  e  =  0.0  = 

Semilatus Rectum  l  =  1.0 AU  =  b^{2} /a 
Sidereal Period  T  =  1.0 Yr  =  (a^{3})^{1/2} 
Angular Velocity  ω_{E}  =  0.986 °/day  =  360°/Txxx 
Cycler Table
R_{ν}  =  l_{C}
1 + e_{C} × cos(ν) 
V_{ν}  =√(  2μ_{Sol}
R_{ν}  +  μ_{Sol}
a_{C}  ) 
Cycler Relevant Positions
Pos.  Time  Range  Cycler
Velocity 
ν

Days
 AU 
km/sec

0°  0  0.614 AU  48.6 
90°  50  1.000 AU  35.2 
127°  112  1.611 AU  23.8 
180°  392  2.703 AU  11.02 
Given

Deg
 AU 
km/sec



x Remarks
At Perihelion
Nearest Earth
Nearest Mars
At Aphelion
x
2010 KK37 is a fortuitous find! With a two year period, its orbit is resonant with Earth's 1 year orbit; thus, every two years, it intersect's Earth's orbit at same relative position to Earth.
It has a comparatively large inclination angle (about 9°); however, this asteroid's point of ascension (asteroid transits Earth's orbital plane, Ecliptic, from south to north) happens to fall very close to Earth's actual position. Thus, a "perfect storm" of circumstances make 2010 KK37 an excellent cycler candidate.
 Asteroid's point of ascension falls very near Earth's orbit.
 Asteriod's two year orbital period is an exact multiple of Earth's orbit (1 year); thus, it is a "resonant" orbit and asteroid's position with relation to Earth recurs every 2 years.
 Every two years, Earth's position is very near point of ascension, a safe but close distance from Earth, .0065 AU, about twice the distance to the Moon.

0 Comments:
Post a Comment
<< Home