Differences between revisions 2 and 31 (spanning 29 versions)

Stellar Day Orbits

Launch loop vehicle mass is limited; however, with high precision launch it is "relatively" easy to rendezvous vehicles to connect into larger assemblies. For example, launching rocket/heatshield packages, followed later by cargo packages to complete a vehicle for apogee plane change and subsequent aerobraking-assisted orbit changes.

The launch loop will rotate under the perigee of a previous launch once per stellar day, every 86164.0989 seconds, about 4 minutes shorter than a solar day. Launch times should be scheduled for exact integer multiples of these orbit times. Otherwise, the loop will not be exactly positioned at the perigee time.

The actual rendezvous time will be modified by the nonlinear gravity field of the Earth due to the equatorial bulge, and the gravity of the Sun, Moon, and Jupiter. These perturbations are not included here, but for a real system they can be predicted and computed to millimeter accuracy. There will also be less-predictable perturbations due to residual atmospheric gas drag (a function of solar UV activity), spacecraft out-gassing, light pressure, and other small effects. The following is for rough estimates only.

The Earth orbital period T (in seconds) is (to first order) a function of the standard gravitational parameter μ = 398600.44 km³/s² and the semimajor axis a . ~~~ T =~ \sqrt{ a^3 / \mu } ~ . ~ ~ The semimajor axis is half the sum of the perigee and apogee radius: 2 ~ a ~=~ r_a + r_p . So, we can rearrange to compute the sum, r_a + r_p = \left( 2 \mu T^2 / \pi^2 \right)^{1/3} . If D is the number of sidereal days, t = 86141.09 \times D . ombining terms, r_a + r_p = 84313.33 \times D^{2/3} km .

Here's a table of r_a + r_p in kilometers versus D in sidereal days :

D	r_a + r_p	D	r_a + r_p	D	r_a + r_p	D	r_a + r_p	D	r_a + r_p
S. days	kilometers	S. days	kilometers	S. days	kilometers	S. days	kilometers	S. days	kilometers
1	84334.87	6	278467.60	12	442039.77	25	721052.99	50	1144600.27
2	133873.26	7	308607.07	14	489883.19	30	814244.21	55	1219688.61
3	175423.60	8	337339.48	16	535493.04	35	902372.54	60	1292532.12
4	212510.55	9	364895.79	18	579235.96	40	986386.61	65	1363377.37
5	246596.65	10	391447.79	20	621384.63	45	1066961.75	70	1432427.12

Choose a perigee r_p to avoid existing satellite orbits. Subtract that from the sum to compute the apogee radius.

The average radius of the Moon is around 385,000 km, and orbits to those large distances will be strongly influenced by lunar gravity. Beyond 1.5 million kilometers, solar tidal forces are stronger than the Earth's gravity. These very slow orbits cannot be estimated simply.

While difficult to compute, and requiring continual accurate measurement and tiny correction thrusts, the multi-body tidal effects can be exploited to pump energy into or out of long period orbits; a subject for more adept orbit calculators than I am. Belbruno's Fly Me to the Moon describes these "chaotic" effects for lay audiences.

-  ⇤ ← Revision 2 as of 2018-03-26 18:06:24 → 
  Size: 2558
  Editor: KeithLofstrom
  Comment:
+   ← Revision 31 as of 2018-03-29 06:29:52 → ⇥
  Size: 3766
  Editor: KeithLofstrom
  Comment:
-Deletions are marked like this.
+Additions are marked like this.
 Line 1:
+## page was renamed from SiderealOrbits
-Line 2:
+Line 3:
-= Sidereal Orbits =
+= Stellar Day Orbits =
-Line 6:
+Line 7:
-The launch loop will rotate under the perigee of a previous launch once per sidereal day, every 86164.089 seconds, about 4 minutes shorter than a solar day.  Launch times should be scheduled for exact integer multiples of these orbit times.  Otherwise, the loop will not be exactly positioned at the perigee time.
+The launch loop will rotate under the perigee of a previous launch once per stellar day, every 86164.0989 seconds, about 4 minutes shorter than a solar day.  Launch times should be scheduled for exact integer multiples of these orbit times.  Otherwise, the loop will not be exactly positioned at the perigee time.
-Line 10:
+Line 11:
-------
+----
-Line 12:
+Line 13:
-The Earth orbital period is (to first order) a function of the standard gravitational parameter μ = 398600.44 km³/s² and the semimajor axis a. 

$ T ~=~ { \large \sqrt{ a^3 / \mu } } $

The semimajor axis is half the sum of the perigee and apogee radius:

$ 2 a ~=~ r_a + r_p $

So, we can rearrange to compute the sum $ r_a + r_p $:

$ r_a + r_p = { \large \left( 2 \mu T^2 / \pi^2 \right)^{1/3} } $

If $ D $ is the number of sidereal days, $ t = 86141.09 D $.  Filling in the numbers,

$ r_a + r_p = 84313.33 D^{2/3} $ km
+The Earth orbital period $ T $ (in seconds) is (to first order) a function of the standard gravitational parameter μ = 398600.44 km³/s² and the semimajor axis $ a $.  $ ~~~ T =~ \sqrt{ a^3 / \mu } ~ . ~ ~ $  The semimajor axis is half the sum of the perigee and apogee radius:  $ 2 ~ a ~=~ r_a + r_p $.  So, we can rearrange to compute the sum,
$ r_a + r_p = \left( 2 \mu T^2 / \pi^2 \right)^{1/3} $ .  If $ D $ is the number of sidereal days, $ t = 86141.09 \times D $.  ombining terms, $ r_a + r_p = 84313.33 \times D^{2/3} $ km .
-Line 30:
+Line 18:
-||<)> $ D $ ||<)> $ r_a + r_p $ ||
||<)>S. days||<)> kilometers    ||
||<)>     1 ||<)>      84313.33 ||
||<)>     2 ||<)>     133839.07 ||
||<)>     3 ||<)>     175378.79 ||
||<)>     4 ||<)>     212456.28 ||
||<)>     5 ||<)>     246533.67 ||
||<)>     6 ||<)>     278396.48 ||
||<)>     8 ||<)>     337253.32 ||
||<)>    10 ||<)>     391347.81 ||
||<)>    15 ||<)>     512810.70 ||
||<)>    20 ||<)>     621225.93 ||
||<)>    25 ||<)>     720868.83 ||
||<)>    30 ||<)>     814036.25 ||
||<)>    40 ||<)>     986134.69 ||
||<)>    50 ||<)>    1144307.94 ||
||<)>    60 ||<)>    1292202.00 ||
||<)>    70 ||<)>    1432061.28 ||
||<)>    80 ||<)>    1565391.24 ||
+||<)>$ D $ ||<)>$ r_a + r_p $|| ||<)>$ D $ ||<)>$ r_a + r_p $|| ||<)>$ D $ ||<)>$ r_a + r_p $|| ||<)>$ D $ ||<)>$ r_a + r_p $|| ||<)>$ D $ ||<)>$ r_a + r_p $||
||<)>S. days||<)> kilometers || ||<)>S. days||<)> kilometers || ||<)>S. days||<)> kilometers || ||<)>S. days||<)> kilometers || ||<)>S. days||<)> kilometers ||
||<)>  1 ||<)>   84334.87 || ||<)>  6 ||<)>  278467.60 || ||<)>  12 ||<)>  442039.77 || ||<)>  25 ||<)>   721052.99 || ||<)>  50 ||<)>  1144600.27 ||
||<)>  2 ||<)>  133873.26 || ||<)>  7 ||<)>  308607.07 || ||<)>  14 ||<)>  489883.19 || ||<)>  30 ||<)>   814244.21 || ||<)>  55 ||<)>  1219688.61 ||
||<)>  3 ||<)>  175423.60 || ||<)>  8 ||<)>  337339.48 || ||<)>  16 ||<)>  535493.04 || ||<)>  35 ||<)>   902372.54 || ||<)>  60 ||<)>  1292532.12 ||
||<)>  4 ||<)>  212510.55 || ||<)>  9 ||<)>  364895.79 || ||<)>  18 ||<)>  579235.96 || ||<)>  40 ||<)>   986386.61 || ||<)>  65 ||<)>  1363377.37 ||
||<)>  5 ||<)>  246596.65 || ||<)> 10 ||<)>  391447.79 || ||<)>  20 ||<)>  621384.63 || ||<)>  45 ||<)>  1066961.75 || ||<)>  70 ||<)>  1432427.12 ||
-Line 50:
+Line 26:
+Choose a perigee $ r_p $ to avoid existing satellite orbits.  Subtract that from the sum to compute the apogee radius.
-Line 51:
+Line 28:
+The average radius of the Moon is around 385,000 km, and orbits to those large distances will be strongly influenced by lunar gravity.   Beyond 1.5 million kilometers, solar tidal forces are stronger than the Earth's gravity.  These very slow orbits cannot be estimated simply.
-Line 52:
+Line 30:
-MoreLater
+While difficult to compute, and requiring continual accurate measurement and tiny correction thrusts, the multi-body tidal effects can be exploited to pump energy into or out of long period orbits; a subject for more adept orbit calculators than I am.  Belbruno's [[ https://press.princeton.edu/titles/8375.html | Fly Me to the Moon ]] describes these "chaotic" effects for lay audiences.

Diff for "StellarDayOrbits"

Stellar Day Orbits