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For the following analysis, I will approximate both the orbit of the Earth and moon as circles. I also assume that $ R_M >> 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 $,  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 $, 
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The north/south perturbation by the moon and sun are much larger. 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.  === Lunar NorthSouth Perturbation === 
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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 ( $ ~ == ~ 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.  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. 
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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.  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. 
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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 $  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 2 ~ R_G ~ \mu_M ~ ~ sin ~ \theta ~ sin ~ \omega ~ sin ~ \alpha ~ / {R_M}^3 dt $ 
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 2 ~ R_G ~ \mu_M ~ ~ sin ~ \theta ~ sin ~ \omega ~ sin ~ \alpha ~ / {R_M}^3 dt
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