Differences between revisions 55 and 77 (spanning 22 versions)
Revision 55 as of 2011-04-03 20:38:02
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Revision 77 as of 2011-04-14 20:33:47
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Deletions are marked like this. Additions are marked like this.
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|| time from perigee || $ \large t = M / \omega ? ? ? $||<)> sun || perihelion || apohelion || || time from perigee || $ \large t = M / \omega     $||<)> sun || perihelion || apohelion ||
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The rotation is expressed as $ \vec \Omega = \omega \hat k $, where $ \omega $ is the angular velocity of the orbit . The rotation is expressed as $ \vec \Omega = \omega \vec k $, where $ \omega $ is the angular velocity of the orbit .
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$ \LARGE \ddot x = -2 \omega \dot y - \omega^2 x $ $ \LARGE \ddot x =  2 \omega \dot y $
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$ \LARGE \ddot y =  2 \omega \dot x + 3 \omega^2 y $ $ \LARGE \ddot y = -2 \omega \dot x + 3 \omega^2 y $
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If $e$ is very small, we can assume $e^2 \approx 0 $. For small eccentricities e, the deviation of the position of an object in a nearly circular eccentric orbit deviates from a ideal circular orbit as an ellipse, with a width of 4e and a height of 2e. This is difficult to compute analytically, but easy to show numerically. Here is a plot the normalized loci of the x,y position in the rotating frame, for various values of eccentricity:
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$\Large {1\over{1+e\cos(\theta)}}={{1-e\cos(\theta)}\over{\left(1+e\cos(\theta)\right)\left(1-e\cos(\theta)\right)}}$ {{attachment:locus01.png}}
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$\Large ~ ~ ~ ~ ~ ~ ~ = {{1-e\cos(\theta)}\over{\left(1-e^2\cos^2(\theta)\right)}}$ [[attachment:locus01.pl | perl data generator]]
[[attachment:locus01.dat | locus data]]
[[attachment:locus01.gp | gnuplot command file]]
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$\large ~ ~ ~ ~ ~ ~ ~ \approx 1-e\cos(\theta) ~ ~ ~ $ ... since $ \large e^2 \approx 0 $ If the whole orbit rotates east, clockwise when viewed from the north, then the object follows the locus above clockwise, with $ \theta = 0 $ at the bottom of the ellipsoid.
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$\large E=\arccos\Large\left({{e+\cos(\theta)}\over{1+e\cos(\theta)}}\right)$ The figure is very close to an ellipse for small e, but not exactly so. The program computes the radial distance rr, and the radial position xr and yr as a function of $\theta$. The program then computes the eccentric anomaly E, and the mean anomaly M. It then rotates the position vector back along the orbit, to approximately vertical,
then subtracts the unit vector representing the elliptical orbit. So, it is approximately like watching an object in an elliptical orbit from a position in a circular orbit with the same semimajor axis. The deviation from an ellipse is quite small, until the eccentricity gets larger than 0.01 or so. Server sky orbits will have eccentricities of less than 0.002 .
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$\large \cos( E ) = { \Large {{e+\cos(\theta)}\over{1+e\cos(\theta)}}} \large { \approx \left(e+\cos(\theta)\right)\left( 1-e\cos(\theta) \right) } $ In the rotating frame, these eccentric loci describe the path of an object subject to fictitious forces. This provides an additional check on our equations for fictitious forces.
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$ \large ~ ~ ~ ~ ~ ~ ~ \approx e + \cos(\theta) - e^2 \cos(\theta) - e \cos^2( \theta ) $ ----
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$ \large ~ ~ ~ ~ ~ ~ ~ \approx \cos(\theta) + e \left( 1 - \cos^2( \theta ) \right) $ ==== The math ====
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$ \Large \cos( E ) \approx \cos(\theta) + e\sin^2( \theta ) $ $\large \dot{ \theta } = v_{\perp} / r = v_0 ( 1+e \cos( \theta )) / ( a (1-e^2) / {1+e\cos(\theta)} $
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$ \large \sin( E ) = \sqrt{ 1 - \cos^2( E ) } \approx \sqrt{ 1 - \left( \cos(\theta) + e \sin^2( \theta ) \right)^2 } $ $\large \dot{ \theta } = ( \sqrt{ \mu / ( a (1 - e^2) ) }  / a ( 1-e^2 ) ) ( 1 + e\cos( \theta ))^2 $
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$ \large ~ ~ ~ ~ ~ ~ ~ \approx \sqrt{ 1 - \left( \cos^2(\theta) + 2e \cos(\theta)\sin^2( \theta ) +  e^2 \sin^4( \theta ) \right) } $ $\large \dot{ \theta } = ( \sqrt{ \mu / ( a^3 (1 - e^2)^3 ) } ( 1 + 2 e \cos( \theta ) + e^2 \cos( \theta )^2 ) $
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$ \large ~ ~ ~ ~ ~ ~ ~ \approx \sqrt{ 1 - \cos^2(\theta) - 2e \cos(\theta)\sin^2( \theta ) } $ for small perturbations, $ e^2 \approx 0 $, so
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$ \large ~ ~ ~ ~ ~ ~ ~ \approx \sqrt{ \sin^2(\theta) \left( 1 - 2e \cos(\theta) \right) } $ $\large \dot{ \theta } \approx ( \sqrt{ \mu / a^3 } ( 1 + 2 e \cos( \theta ) ) $
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$ \Large \sin( E ) \approx \sin(\theta ) \left( 1 - e \cos(\theta) \right)$ In the rotating frame, $ \dot{ x } \approx a ( \dot{ \theta } - \omega ) $ where $ \omega \equiv \sqrt{ \mu / a^3 } $, so
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$ \large M = E - e \sin( E ) \approx E - e \left( \sin(\theta ) \left( 1 - e \cos(\theta) \right) \right) $ $\large \dot{ x } \approx a \sqrt{ \mu / a^3 } ( ( 1 + 2 e \cos( \theta ) - 1 ) $
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$ \large M \approx E - e \sin(\theta) + e^2 \cos(\theta) $ $\large \dot{ x } \approx 2 e a \omega ( \cos( \theta ) ) $
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$ \Large M \approx E - e \sin(\theta) $ Approximating $ \theta \approx \omega t $, we can integrate to get:
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$ \large \cos( M ) \approx \cos( E - e \sin(\theta) ) = \cos( E ) \cos( e \sin(\theta) ) + \sin( E ) \sin( e \sin(\theta) ) $ $\Large x \approx -2 e a \sin( \theta ) $
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$ \large \sin( e X ) \approx e X - (e X)^3 / 6 + (e X)^5 / 120 + ... \approx e X ~ ~ ~ $ higher order terms are < e^2^ ----
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$ \large \cos( e X ) \approx 1 - (e X)^2 / 2 + (e X)^4 / 24 + ... \approx 1 ~ ~ ~ $ higher order terms are < $ e^2 $ $\large \dot{ y } = v_r = e v_0 \sin( \theta ) \approx e a \omega \sin( \omega t ) $ Integrating:
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$ \large \cos( M ) \approx \cos( E ) + e \sin( E ) \sin(\theta) ) $ $\Large y \approx   -e a \cos( \theta ) $
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$ \large \cos( M ) \approx \cos(\theta) + e\sin^2( \theta ) + e \left( \sin(\theta ) \left( 1 - e \cos(\theta) \right) \right) $ For small $ e $, these equations in $ y $ and $ x $ describe the locus calculated above.
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$ \Large \cos( M ) \approx \cos(\theta) + e\sin^2( \theta ) ( 2 - \cos(\theta) ) $ ----
==== Testing the Equations for Acceleration ====
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$ \large \sin( M ) \approx \sin( E - e \sin(\theta) ) = \sin( E ) \cos( e \sin(\theta) ) - \cos( E ) \sin ( e \sin(\theta) ) $ $ \large \ddot x ~ ? = ~ \large 2 \omega \dot y $
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$ \large \sin( M ) \approx \sin( E ) - e \cos( E ) \sin(\theta) $ $ \large \ddot x = { { d^2 } \over { dt^2 } } ( -2 e a \sin( \theta ) ) = { d \over { dt } } ( -2 e a \omega \cos( \theta ) ) = \Large 2 e a \omega^2 \sin ( \theta ) $
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$ \large \sin( M ) \approx \sin(\theta ) \left( 1 - e \cos(\theta) \right) - e \left( \cos(\theta) + e\sin^2( \theta )\right)\sin(\theta) $ $ \large 2 \omega \dot y = 2 \omega { d \over { dt } } -e a \cos( \theta ) = \Large 2 e a \omega^2 \sin ( \theta ) $ . . . equal!
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$ \large \sin( M ) \approx \sin(\theta) - e \sin(\theta)\cos(\theta) - e\sin(\theta)\cos(\theta) + e^2\sin^2(\theta) $ $ \large \ddot y ~ ? = ~ \large -2 \omega \dot x + 3 \omega^2 y $
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$ \Large \sin( M ) \approx \sin(\theta)( 1 - 2 e \cos(\theta)) $ $ \large \ddot y = { { d^2 } \over { dt^2 } } ( -e a \cos( \theta ) ) = { d \over { dt } } ( e a \omega \sin( \theta ) ) = \Large e a \omega^2 \cos( \theta ) $
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We want to find the locus of x and y, rotated to vertical by angle -M, and subtracting [ 0, a ] . $ \large -2 \omega \dot x + 3 \omega^2 y = -2 \omega { d \over { dt } }( -2 e a \sin( \theta ) + 3 \omega^2 ( - e a \cos( \theta ) ) ) $
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$ \large r= a \Large{{1-e^2}\over{1+e\cos(\theta)}} \large \approx a ( 1 - e \cos(\theta) ) $ $ \large ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ = 4 e a \omega^2 \cos( \theta ) - 3 e a \omega^2 \cos( \theta ) = \Large e a \omega^2 \cos( \theta ) $ . . . equal!
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$ \large x = r \cos( \theta ) \approx a ( 1 - e \cos(\theta) ) \cos( \theta ) $

$ \large y = r \sin( \theta ) \approx a ( 1 - e \cos(\theta) ) \sin( \theta ) $

Rotate the vector [ x , y ] by -M to :

$ \large [ x', y' ] = \left( \begin{array}{cc} \cos( M ) & \sin( M ) \\ -\sin( M ) & \cos( M ) \end{array} \right) [ x, y ] $

$ \large x' = \cos( M ) x + \sin( M ) y $

$ \large x' \approx ( \cos(\theta) + e\sin^2( \theta ) ( 2 - \cos(\theta) ) ) ~ a ( 1 - e \cos(\theta) ) \cos( \theta ) ~ + ~ \sin(\theta)( 1 - 2 e \cos(\theta)) ~ a ( 1 - e \cos(\theta) ) \sin(\theta) $

$ \large x' / a \approx ( \cos(\theta) + 2e\sin^2( \theta ) - 2e\sin^2\cos(\theta) ) ( \cos( \theta ) - e \cos^2(\theta) ) ~ + ~ \sin^2(\theta)( 1 - 3 e \cos(\theta)) + e \cos^2(\theta) ) $




$ \large y' = \cos( M ) y - \sin( M ) x $

$ \large y' \approx ( \cos(\theta) + e\sin^2( \theta ) ( 2 - \cos(\theta) ) ) ~ a ( 1 - e \cos(\theta) ) \sin( \theta ) ~ - ~ \sin(\theta)( 1 - 2 e \cos(\theta)) ~ a ( 1 - e \cos(\theta) ) \cos(\theta) $






MORE LATER
The acceleration equations work!
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J2 speedup fraction $ = - 3 J_2 ( a_E / a )^2 = 3 \times 1.082626683e-3 \times (6378107m)^2 / a^2 = 1.321245688e11m^2 / a^2 $
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||$\omega'/\omega$||J2 speedup fraction|| 1.4813e-3|| 4.0389e-4|| 3.1677e-4 || 3.7158e-5 || 4.4707e-7 || 2e-17 || || ||$\omega'/\omega$||J2 speedup fraction|| 2.9627e-3|| 8.0782e-4|| 6.3356e-4 || 7.4318e-5 || 8.9417e-7 || 2e-17 || ||
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[[ Lageos | LAGEOS 1 and LAGEOS 2 ]] are two slightly lower satellites in inclined but highly circular orbits.
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 . http://en.wikipedia.org/wiki/Rotating_reference_frame

Near Circular Orbits


General Elliptical Orbits

In the orbital plane, neglecting the J_2 spherical oblateness parameter :

gravitational parameter

\large \mu = a^3 \omega^2 = G M

Drawing from Wikipedia

semimajor axis

\large a = \sqrt[3]{ \mu / \omega^2 }

angular velocity

\large \omega = \sqrt{ \mu / a^3 }

J_2 Speedup factor

\large \omega'/\omega = 1+1.5 |J_2| (R_{eq}/a)^2

sidereal period

\large T = 2\pi / \omega = 2\pi \sqrt{ a^3/\mu }

eccentricity

\large e = ( r_a - r_p ) / ( r_a + r_p )

velocity

\large v_0 = \sqrt{ \mu / ( a (1 - e^2) ) }

true anomaly
orbit angle from focus

\Large \theta

eccentric anomaly
ellipse center angle

\large E=\arccos\Large\left({{e+\cos(\theta)}\over{1+e\cos(\theta)}}\right)

periapsis

apoapsis

mean anomaly,

\large M = E - e \sin( E )

earth

perigee

apogee

time from perigee

\large t = M / \omega

sun

perihelion

apohelion

radius

\large r=a\Large{{1-e^2}\over{1+e\cos(\theta)}}

\huge\rightarrow

\large r_p =( 1-e )a

\large r_a =( 1+e )a

perpendicular velocity

\large v_{\perp}= v_0 ( 1+e \cos( \theta ))

\huge\rightarrow

\large v_p =(1+e)v_0

\large v_a =(1-e)v_0

radial velocity

\large v_r = e v_0 \sin( \theta )

\huge\rightarrow

0

0

total velocity
tangent to orbit

\Large v=\LARGE \sqrt{{{2\mu}\over{r}}-{{\mu}\over{a}}}

\huge\rightarrow

\large v=v_p,~ r=r_p

\large v=v_a,~ r=r_a

orbit energy parameter

\large C_3 = \mu / a = v_p v_a

\large \mu = a v_p v_a ~ ~ ~ ~ ~ ~ ~ ~ ~ 2 a = r_p + r_a


Fictional Forces in Orbit

In the rotating frame of a circular orbit, counterclockwise viewed from above the orbital plane, the directions are

direction

unit vector

description

x

\vec i

Tangential to (along the line of) the orbit, in the orbital plane, pointing clockwise or westward

y

\vec j

Radially outwards from the center of rotation, in the orbital plane

z

\vec k

Perpendicular to the orbital plane, northwards

The rotation is expressed as \vec \Omega = \omega \vec k , where \omega is the angular velocity of the orbit . The radial vector \vec r is composed of \vec r = x \vec i + y \vec j + z \vec k .

Coriolis acceleration: \LARGE \ddot{\vec r}_{Coriolis} = -2 \vec \Omega \times \dot { \vec r }

Centrifugal acceleration: \LARGE \ddot{\vec r}_{Centrifugal} = - \vec \Omega \times \vec \Omega \times \vec r

FROM HERE DOWN, WORK IN PROGRESS, NOT VERIFIED:

Centripedal (gravity) acceleration: \LARGE \ddot{\vec r}_{Centripedal} = \omega^2 ( 2 y \vec k - x \vec i - z \vec j )

Scalar equations:

\LARGE \ddot x = 2 \omega \dot y

\LARGE \ddot y = -2 \omega \dot x + 3 \omega^2 y

\LARGE \ddot z = -\omega^2 z


Locus of Elliptical Orbit Position in Rotating Frame

For small eccentricities e, the deviation of the position of an object in a nearly circular eccentric orbit deviates from a ideal circular orbit as an ellipse, with a width of 4e and a height of 2e. This is difficult to compute analytically, but easy to show numerically. Here is a plot the normalized loci of the x,y position in the rotating frame, for various values of eccentricity:

locus01.png

perl data generator locus data gnuplot command file

If the whole orbit rotates east, clockwise when viewed from the north, then the object follows the locus above clockwise, with \theta = 0 at the bottom of the ellipsoid.

The figure is very close to an ellipse for small e, but not exactly so. The program computes the radial distance rr, and the radial position xr and yr as a function of \theta. The program then computes the eccentric anomaly E, and the mean anomaly M. It then rotates the position vector back along the orbit, to approximately vertical, then subtracts the unit vector representing the elliptical orbit. So, it is approximately like watching an object in an elliptical orbit from a position in a circular orbit with the same semimajor axis. The deviation from an ellipse is quite small, until the eccentricity gets larger than 0.01 or so. Server sky orbits will have eccentricities of less than 0.002 .

In the rotating frame, these eccentric loci describe the path of an object subject to fictitious forces. This provides an additional check on our equations for fictitious forces.


The math

\large \dot{ \theta } = v_{\perp} / r = v_0 ( 1+e \cos( \theta )) / ( a (1-e^2) / {1+e\cos(\theta)}

\large \dot{ \theta } = ( \sqrt{ \mu / ( a (1 - e^2) ) } / a ( 1-e^2 ) ) ( 1 + e\cos( \theta ))^2

\large \dot{ \theta } = ( \sqrt{ \mu / ( a^3 (1 - e^2)^3 ) } ( 1 + 2 e \cos( \theta ) + e^2 \cos( \theta )^2 )

for small perturbations, e^2 \approx 0 , so

\large \dot{ \theta } \approx ( \sqrt{ \mu / a^3 } ( 1 + 2 e \cos( \theta ) )

In the rotating frame, \dot{ x } \approx a ( \dot{ \theta } - \omega ) where \omega \equiv \sqrt{ \mu / a^3 } , so

\large \dot{ x } \approx a \sqrt{ \mu / a^3 } ( ( 1 + 2 e \cos( \theta ) - 1 )

\large \dot{ x } \approx 2 e a \omega ( \cos( \theta ) )

Approximating \theta \approx \omega t , we can integrate to get:

\Large x \approx -2 e a \sin( \theta )


\large \dot{ y } = v_r = e v_0 \sin( \theta ) \approx e a \omega \sin( \omega t ) Integrating:

\Large y \approx -e a \cos( \theta )

For small e , these equations in y and x describe the locus calculated above.


Testing the Equations for Acceleration

\large \ddot x ~ ? = ~ \large 2 \omega \dot y

\large \ddot x = { { d^2 } \over { dt^2 } } ( -2 e a \sin( \theta ) ) = { d \over { dt } } ( -2 e a \omega \cos( \theta ) ) = \Large 2 e a \omega^2 \sin ( \theta )

\large 2 \omega \dot y = 2 \omega { d \over { dt } } -e a \cos( \theta ) = \Large 2 e a \omega^2 \sin ( \theta ) . . . equal!

\large \ddot y ~ ? = ~ \large -2 \omega \dot x + 3 \omega^2 y

\large \ddot y = { { d^2 } \over { dt^2 } } ( -e a \cos( \theta ) ) = { d \over { dt } } ( e a \omega \sin( \theta ) ) = \Large e a \omega^2 \cos( \theta )

\large -2 \omega \dot x + 3 \omega^2 y = -2 \omega { d \over { dt } }( -2 e a \sin( \theta ) + 3 \omega^2 ( - e a \cos( \theta ) ) )

\large ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ = 4 e a \omega^2 \cos( \theta ) - 3 e a \omega^2 \cos( \theta ) = \Large e a \omega^2 \cos( \theta ) . . . equal!

The acceleration equations work!


Periods of M orbits

M orbits describe the number of minutes an orbit takes travel once around the earth and return to the same position overhead. For server sky, these are integer fractions of a 1440 minute synodic day; this makes it easier to calculate the sky position given the orbital parameters and the time of day. the M288 orbit returns to the same apparent position 5 times per day, or 365.256...*5 times per year. That position moves around the earth 365.256...+1 times per year. So the total number of orbits per year, relative to the stars, is 365.256...*6+1 orbits per year. That is divided into the year length in seconds to yield the

M288 sidereal orbit time = ( 365.256... * 86400 ) / ( 365.256... * 6 + 1 ) = 86400 / ( 6 + 1/365.256... ) = 14393.43227 seconds

1 year = 365.256363004 days of 86,400 seconds, or 31558149.7635456 seconds.

J2 speedup fraction = - 3 J_2 ( a_E / a )^2 = 3 \times 1.082626683e-3 \times (6378107m)^2 / a^2 = 1.321245688e11m^2 / a^2


A Table of orbits

LEO 300Km

M288

M360

GEO

Moon

Earth

units

\mu

gravitation param.

3.98600448e14

1.3271244e20

m3/s2

relative to

earth

Sun

J_2

Oblate-ness

-1.082626683e-3 earth radius = 6378000m

-6e-7

\omega'/\omega

J2 speedup fraction

2.9627e-3

8.0782e-4

6.3356e-4

7.4318e-5

8.9417e-7

2e-17

a_g

gravity

8.938095

2.437062

1.9113693

0.02242078

0.002697573

0.005930053

s

T

sidereal period

5431.010

14393.4323

17270.5433

86164.100

2360591.577

31558149.76

s

T_s

synodic period

5431.945

14400

17820

86400

2551442.9

31558149.76

s

T/2\pi

Sidereal / 2pi

864.3721

2290.78585

2748.69228

13713.44093

375698.7212

5022635.5297

s

\omega

Angular velocity

1.1569e-03

4.36531e-04

3.63809e-04

7.29212e-05

2.66171e-06

1.99099e-07

rad/s

orbits/year

5810.733

2192.5382

1827.2819

366.25640

13.36879

1.00000

a

semimajor axis

6678000

12788971

14440980

42164170

384399000

1.4959826e11

m

R_a

apogee radius

6678000

12838976

14490980

42164170

405696000

1.5209823e11

m

R_p

perigee radius

6678000

12738976

14440980

42164170

363104000

1.4709829e11

m

e

eccentricity

0.000000

0.001951

0.001728

0.000000

0.055401

0.016711

V_0

mean velocity

7725.84

5582.79

5253.76

3074.66

1023.16

29784.81

m/s

V_a

apogee velocity

7725.84

5571.90

5244.68

3074.66

966.47

29287.07

m/s

V_p

perigee velocity

7725.84

5593.68

5262.84

3074.66

1079.84

30282.55

m/s

C_3

orb. specific energy

-59688597

-31167516

-27602035

-9453535

-1036945

-887125553

J/kg

Note: This table uses the classic formula \omega^2 a^3 = \mu , and does not take into account the oblate spheroid shape of the Earth, and many other perturbations. However, with the perturbations included, and with good data from ground stations and GPS to establish positions and velocities, we really can compute these numbers to this many decimal places. So while the numbers above are actually far less accurate, they represent the precision of the measurements we will someday compute.

The eccentricities for the M orbits assume orbits mapped onto a 50km minor radius toroid.

LAGEOS 1 and LAGEOS 2 are two slightly lower satellites in inclined but highly circular orbits.

REFS:

NearCircularOrbits (last edited 2022-09-14 00:16:43 by KeithLofstrom)