4234
Comment:

← Revision 11 as of 20120327 17:05:27 ⇥
8048

Deletions are marked like this.  Additions are marked like this. 
Line 21:  Line 21: 
The moon's orbit takes it above and below the equatorial plane, which causes outofplane tidal forces. The Moon's orbital plane is inclined 5.145 degrees from the ecliptic plane (the plane of the earth's orbit), and precesses over a ''nodal period'' of 18.6 years. The earth's axial tilt is 23°26'21".4119 or 23.439°, so the moon's orbital plane is inclined between 18.294° and 28.584° with respect to the equatorial plane, varying sinusoidally between one and the other over the nodal period.  The moon's orbit takes it above and below the equatorial plane, which causes outofplane tidal forces. The Moon's orbital plane is inclined 5.145 degrees from the ecliptic plane (the plane of the earth's orbit), and precesses over a ''nodal period'' of 18.6 years. The earth's axial tilt is 23°26'21" or 23.439°, so the moon's orbital plane is inclined between 18.294° and 28.584° with respect to the equatorial plane, varying sinusoidally between one and the other over the nodal period. 
Line 23:  Line 23: 
According to Boden in Larsen and Wertz) , counteracting these tidal forces on a geostationary object requires a $ \Delta V $ budget of as much as $ 102.67 ~ cos ~ \alpha ~ sin ~ \alpha ~ $ m/s per year ( $ ~ \equiv ~ 51.335 ~ sin ~ 2 \alpha $ ), where $ \alpha $ is the angle between the equatorial (geostationary orbit) plane and the and the Moon's orbital plane. Boden claims the worst case is 36.93 m/s per year, which implies an $ \alpha $ of 23.00°. Actually, the worst case $ \alpha $ is 28.584°, so using assuming the first "102.67..." formula and its doubleangle "51.3345 ..." equivalent, the worst case annual $ \Delta V $ budget is more like 43.13 m/s per year. Let's see if we can derive that 102.67 m/s/y number.  According to Boden (in Larsen and Wertz) , counteracting these tidal forces on a geostationary object requires a $ \Delta V $ budget of as much as $ 102.67 \cos{ \alpha } \sin{ \alpha } $ m/s per year ( $ \equiv ~ 51.335 \sin { 2 \alpha } $ ), where $ \alpha $ is the angle between the equatorial (geostationary orbit) plane and the and the Moon's orbital plane. Boden claims the worst case is 36.93 m/s per year, which implies an $ \alpha $ of 23.00°. Actually, the worst case $ \alpha $ is 28.584°, so using assuming the first "102.67..." formula and its doubleangle "51.3345 ..." equivalent, the worst case annual $ \Delta V $ budget is more like 43.13 m/s per year. Let's see if we can derive that 102.67 m/s/y number. 
Line 25:  Line 25: 
Over the course of a month, the moon moves north of the equatorial plane, through it, then south of the equatorial plane, and back. When the moon is on the equatorial plane, there is no perturbation north or south, though there are tidal forces radially on an object in geostationary orbit. The inplane tidal acceleration is approximately $ 2 ~ R_G ~ \mu_M ~ ~ sin ~ \theta ~ / {R_M}^3 $, where $ \theta $ is the orbital angle difference between the orbiting object and the moon.  Over the course of a month, the moon moves north of the equatorial plane, through it, then south of the equatorial plane, and back. When the moon is on the equatorial plane, there is no perturbation north or south, though there are tidal forces radially on an object in geostationary orbit. The inplane tidal acceleration is approximately $ 2 ~ R_G ~ \mu_M \sin{ \theta } / {R_M}^3 $, where $ \theta $ is the orbital angle difference between the orbiting object and the moon. 
Line 27:  Line 27: 
As the moon moves around its orbit, it moves north or south. This puts a northerly component of $ R_M ~ sin ~ \omega $ on the moon's position, where \omega is the argument of periapsis, the angle around the moon's orbit from the ascending node relative to the equatorial plane. We can approximate the tidal acceleration as $ 2 ~ R_G ~ \mu_M ~ ~ sin ~ \theta ~ sin ~ \omega ~ / {R_M}^3 $  As the moon moves around its orbit, it moves north or south. This puts a northerly component of $ R_M ~ sin ~ \omega $ on the moon's position, where $ \omega $ is the argument of periapsis, the angle around the moon's orbit from the ascending node relative to the equatorial plane. We can approximate the tidal acceleration as $ 2 ~ R_G ~ \mu_M \sin{ \theta } \sin{ \omega } \sin{ \alpha } / {R_M}^3 $. Perhaps a more accurate analysis would have a $ \sin{ \alpha } \cos{ \alpha } ~ \equiv ~ { 1 \over 2 } ~ \sin{ 2 \alpha } $ factor like Boden. Let's just see if we get in the ballpark. The thrust needed to counteract the acceleration is the absolute value of the acceleration, and that absolute value can be integrated over time: $ Integrated ~ thrust ~ = ~ \int{ \left 2 ~ R_G ~ \mu_M ~ \sin{ \theta } \sin{ \omega } \sin{ \alpha } / {R_M}^3 \right dt } $ $ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ = ~ 2 ~ R_G ~ \mu_M ~ / {R_M}^3 ~ \int{ \left \sin{ \theta } \sin{ \omega } \sin{ \alpha } \right dt } $ $ \sin{ \alpha } $ varies slowly, over the 18.6 year nodal period. We are more interested in its maximum, as that is what we must design the thrust our north and south correction thrusters and fuel supply for, assuming frequent resupply. For the monthly $ \sin{ \omega } $ and daily $ \sin { theta } $ terms, we can assume the maximum of each ( = 1 ) for the maximum correction thruster thrust, and the average of each ( = $ 2 / \pi $ ) for fuel supply. So the maximum thrust per second ( at maximum $ \alpha $ } is: Maximum thrust $ \approx ~ 2 ~ R_G ~ \mu_M ~ \sin { \alpha_{max} } / {R_M}^3 ~ = ~ $ 3.48e6 m/s^2^ And the average thrust per year ( at maximum $ \alpha $ } is: Average thrust $ \approx ~ \left( 8 \over \pi^2 \right) ~ year~ R_G ~ \mu_M ~ \sin { \alpha_{max} } / {R_M}^3 ~ = ~ $ 44.5 m/s/y That is 3% above the Boden value (after correction), so we are close. If we use the sin*cos correction from Boden, we come in 10% low, 39.1 m/s/y , so there is still something slightly wrong in my analysis. Use the Boden numbers. === Solar NorthSouth Perturbation === The perturbation from the sun is simpler; the inclination of the equatorial plane to the ecliptic plane ( $ \gamma = $ 23.439° ) does not change much over a few decades. By similar reasoning, the maximum solar thrust per second is: Maximum thrust $ \approx ~ 2 ~ R_G ~ \mu_S ~ \sin { \gamma } / {R_S}^3 ~ = ~ $ 1.33e6 m/s^2^ And the average solar thrust per year is: Average thrust $ \approx ~ \left( 8 \over \pi^2 \right) ~ year~ R_G ~ \mu_S ~ \sin { \gamma } / {R_S}^3 ~ = ~ $ 17.0 m/s/y The latter number is 18% lower than Boden, so there is something more than slightly wrong. Again, use the Boden numbers. == Correction Delta V for an offequator Launch Loop == For a launch loop exactly on the equator, the destination GEO station will be easy to rendezvous with, requiring at most a few meters per second $ \Delta V $ for rendezvous. Launch loops to GEO may be sited 5 degrees south of the equator to avoid equatorial weather. That means the minimum equatorial inclination is 5°. A geostationary transfer orbit to a geostationary CaptureRail has an apogee velocity of 1335 m/s, and we must have zero northsouth velocity for that to work right (unless the rail is spinning diagonally). So, the vehicle's northsouth velocity must be corrected by approximately 1335 m/s sin( 5° ) or 116 m/s in the 18500 seconds transfer fractional orbit time, an acceleration of 6.3e3 meters per second per second. Assuming a 5 ton vehicle, that is a thrust of 31 newtons. If that is made with a laser ablative thruster with an effective exhaust velocity of 5000 meters per second, that is a propellant rate of 6.3 grams per second, or about 120 kilograms of propellant expended, and a power delivery rate of 80 kW . A planar thrust panel will be inefficient, so both the power and the propellant expended will be higher than that. Offequator launch loops will require some asyetunimagined modification to capture rail, or expend a lot of propellant, to arrive. This is bothersome, and violates the "minimal propellant in orbit" goal. More invention needed. 
Rendezvous with Geostationary Destinations
Geostationary orbits are nontrivial to maintain. They are perturbed by nonroundearth forces represented by the J_{22} term in the spherical harmonics of the gravity field, forming attractors at 75 degrees east (above a point slightly east of the Maldives and south of India) and 105 west (west of the Galapagos, directly south of Denver, Colorado). Evading these attractors can take as much as 1.715 m/s delta V per year.
The NorthSouth Perturbation
The north/south perturbation by the moon and sun are much larger.
For the following analysis, I will approximate both the orbit of the Earth and moon as circles. I also assume that R_S \gg R_M \gg R_G and approximately equal tidal forces at near and far sides of the orbit. A more exact analysis suitable for precision rendezvous would numerically compute the actual multibody elliptical orbits, the spherical harmonic gravity terms for Earth and Moon, gravitational contributions from Jupiter and Venus, and optical perturbations. For now, we are looking for an approximation of maximum \Delta V , but do not forget that the launch, position, and velocity for vehicle in a mature launch loop system will be timed and optically measured to nanoseconds and micrometers, and continuously measured and controlled to this precision throughout the transfer orbit.
R_G 
4.216 e4 km 
geostationary orbit diameter 
\mu_M 
4.903 e3 km^{3}/s^{2} 
moon's standard gravitational parameter 
R_M 
3.844 e5 km 
moon/earth semimajor orbit radius 
\mu_S 
1.327e11 km^{3}/s^{2} 
sun standard gravitational parameter 
R_S 
1.496 e8 km 
earth/sun semimajor orbit radius 
Lunar NorthSouth Perturbation
The moon's orbit takes it above and below the equatorial plane, which causes outofplane tidal forces. The Moon's orbital plane is inclined 5.145 degrees from the ecliptic plane (the plane of the earth's orbit), and precesses over a nodal period of 18.6 years. The earth's axial tilt is 23°26'21" or 23.439°, so the moon's orbital plane is inclined between 18.294° and 28.584° with respect to the equatorial plane, varying sinusoidally between one and the other over the nodal period.
According to Boden (in Larsen and Wertz) , counteracting these tidal forces on a geostationary object requires a \Delta V budget of as much as 102.67 \cos{ \alpha } \sin{ \alpha } m/s per year ( \equiv ~ 51.335 \sin { 2 \alpha } ), where \alpha is the angle between the equatorial (geostationary orbit) plane and the and the Moon's orbital plane. Boden claims the worst case is 36.93 m/s per year, which implies an \alpha of 23.00°. Actually, the worst case \alpha is 28.584°, so using assuming the first "102.67..." formula and its doubleangle "51.3345 ..." equivalent, the worst case annual \Delta V budget is more like 43.13 m/s per year. Let's see if we can derive that 102.67 m/s/y number.
Over the course of a month, the moon moves north of the equatorial plane, through it, then south of the equatorial plane, and back. When the moon is on the equatorial plane, there is no perturbation north or south, though there are tidal forces radially on an object in geostationary orbit. The inplane tidal acceleration is approximately 2 ~ R_G ~ \mu_M \sin{ \theta } / {R_M}^3 , where \theta is the orbital angle difference between the orbiting object and the moon.
As the moon moves around its orbit, it moves north or south. This puts a northerly component of R_M ~ sin ~ \omega on the moon's position, where \omega is the argument of periapsis, the angle around the moon's orbit from the ascending node relative to the equatorial plane. We can approximate the tidal acceleration as 2 ~ R_G ~ \mu_M \sin{ \theta } \sin{ \omega } \sin{ \alpha } / {R_M}^3 . Perhaps a more accurate analysis would have a \sin{ \alpha } \cos{ \alpha } ~ \equiv ~ { 1 \over 2 } ~ \sin{ 2 \alpha } factor like Boden. Let's just see if we get in the ballpark.
The thrust needed to counteract the acceleration is the absolute value of the acceleration, and that absolute value can be integrated over time:
Integrated ~ thrust ~ = ~ \int{ \left 2 ~ R_G ~ \mu_M ~ \sin{ \theta } \sin{ \omega } \sin{ \alpha } / {R_M}^3 \right dt }
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ = ~ 2 ~ R_G ~ \mu_M ~ / {R_M}^3 ~ \int{ \left \sin{ \theta } \sin{ \omega } \sin{ \alpha } \right dt }
\sin{ \alpha } varies slowly, over the 18.6 year nodal period. We are more interested in its maximum, as that is what we must design the thrust our north and south correction thrusters and fuel supply for, assuming frequent resupply. For the monthly \sin{ \omega } and daily \sin { theta } terms, we can assume the maximum of each ( = 1 ) for the maximum correction thruster thrust, and the average of each ( = 2 / \pi ) for fuel supply.
So the maximum thrust per second ( at maximum \alpha } is:
Maximum thrust \approx ~ 2 ~ R_G ~ \mu_M ~ \sin { \alpha_{max} } / {R_M}^3 ~ = ~ 3.48e6 m/s^{2}
And the average thrust per year ( at maximum \alpha } is:
Average thrust \approx ~ \left( 8 \over \pi^2 \right) ~ year~ R_G ~ \mu_M ~ \sin { \alpha_{max} } / {R_M}^3 ~ = ~ 44.5 m/s/y
That is 3% above the Boden value (after correction), so we are close. If we use the sin*cos correction from Boden, we come in 10% low, 39.1 m/s/y , so there is still something slightly wrong in my analysis. Use the Boden numbers.
Solar NorthSouth Perturbation
The perturbation from the sun is simpler; the inclination of the equatorial plane to the ecliptic plane ( \gamma = 23.439° ) does not change much over a few decades. By similar reasoning, the maximum solar thrust per second is:
Maximum thrust \approx ~ 2 ~ R_G ~ \mu_S ~ \sin { \gamma } / {R_S}^3 ~ = ~ 1.33e6 m/s^{2}
And the average solar thrust per year is:
Average thrust \approx ~ \left( 8 \over \pi^2 \right) ~ year~ R_G ~ \mu_S ~ \sin { \gamma } / {R_S}^3 ~ = ~ 17.0 m/s/y
The latter number is 18% lower than Boden, so there is something more than slightly wrong. Again, use the Boden numbers.
Correction Delta V for an offequator Launch Loop
For a launch loop exactly on the equator, the destination GEO station will be easy to rendezvous with, requiring at most a few meters per second \Delta V for rendezvous.
Launch loops to GEO may be sited 5 degrees south of the equator to avoid equatorial weather. That means the minimum equatorial inclination is 5°. A geostationary transfer orbit to a geostationary CaptureRail has an apogee velocity of 1335 m/s, and we must have zero northsouth velocity for that to work right (unless the rail is spinning diagonally). So, the vehicle's northsouth velocity must be corrected by approximately 1335 m/s sin( 5° ) or 116 m/s in the 18500 seconds transfer fractional orbit time, an acceleration of 6.3e3 meters per second per second. Assuming a 5 ton vehicle, that is a thrust of 31 newtons. If that is made with a laser ablative thruster with an effective exhaust velocity of 5000 meters per second, that is a propellant rate of 6.3 grams per second, or about 120 kilograms of propellant expended, and a power delivery rate of 80 kW . A planar thrust panel will be inefficient, so both the power and the propellant expended will be higher than that.
Offequator launch loops will require some asyetunimagined modification to capture rail, or expend a lot of propellant, to arrive. This is bothersome, and violates the "minimal propellant in orbit" goal. More invention needed.
MORE LATER
References
Larsen and Wertz (1999). Space Mission Analysis and Design, Kluwer, 3rd Ed., Page 157 (Chapter Author Daryl G. Boden, PhD, USNA). (?) Soop, E. M. (1994). Handbook of Geostationary Orbits. Springer