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| For transfer orbits below geosynchronous, the apogee of the orbit is the release radius $ r_a $, and the angular velocity is $ \omega_E = 2 \pi / P_{sidereal} $, where $ P_{sidereal} $ is the sidereal day, 86164.0905 seconds. We want to find $ r_a $ and the perigee velocity $ v_p $. | For transfer orbits below geosynchronous, the apogee of the orbit is the release radius $ r_a $, and the angular velocity is $ \omega_E = 2 \pi / P_{sidereal} $, where $ P_{sidereal} $ is the sidereal day, 86164.0905 seconds. We want to find $ r_a $ and the perigee velocity $ v_p $. We will have to work backwards. |
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| $ v_p = r_a v_a / r_d = \omega_E { r_a }^2 / r_d $ | $ {v_a}^2 / 2 \mu = 1 / r_a - 1 / ( r_a + r_p ) $ |
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| Orbital energy: | Solve for $ r_p $, perigee radius, the candidate $ r_d $ |
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| $ {v_a}^2 - 2 \mu / r_a = {v_p}^2 - 2 \mu / r_d $ Solve for $ r_a $ : $ ( \omega_E r_a )^2 - 2 \mu /r_a = ( \omega_E {r_a}^2 / r_d )^2 - 2 \mu / r_d $ $ {r_a}^3 - ( 2 \mu / {\omega_E}^2 ) = {r_a}^5 / {r_d}^2 - 2 \mu r_a / ( {\omega_E}^2 r_d ) $ Divide by $ {r_d}^3 $: $ ( r_a / r_d )^3 - ( 2 \mu / {\omega_E}^2 {r_d}^3 ) = ( r_a / r_d )^5 - ( 2 \mu / {\omega_E}^2 {r_d}^3 ) ( r_a / r_d ) $ Normalize for $ R = r_a / r_d $ and $ M = 2 \mu / ( {\omega_E}^2 {r_d}^3 ) $. $ R^3 - M = R^5 - M * R $ $ R^5 = R^3 + M * (R - 1) $ ... given M, solve iteratively for R |
$ r_p = \huge { r_a \over { ( 2 \mu / ( {r_a}^3 {\omega_E}^2 ) ) - 1 } } $ |
Orbit Circularization
What is the ΔV needed for apogee insertion into a circular equatorial orbit from a launch loop transfer orbit, and for perigee insertion from a space elevator transfer orbit?
The destination orbit has a radius of r_d and a velocity of v_d = \sqrt{ \mu / r_d } were \mu = 398600.4418 km3 / s2.
The launch loop calculation is fairly simple.
An 80 kilometer breech altitude launch loop defines a transfer orbit with a perigee r_p = 6378 + 80 km = 6458 km . The semimajor axis is a = 0.5 * ( r_p + r_d ) , the eccentricity e = ( r_d - r_p ) / ( r_d + r_p ) , the characteristic velocity is v_0 = \sqrt{ \mu / ( a * ( 1 - e^2 ) ) } , and the apogee velocity is v_a = ( 1 - e ) v_0 . Combining and simplifying:
{v_a}^2 = ( 1 - e )^2 {v_0}^2 = ( \mu / a ) ( 1 - e )^2 / ( 1 - e^2 )
{v_a}^2 = ( 2 \mu r_p ) / ( r_d ( r_d + r_p ) )
v_a = \sqrt{ ( 2 \mu r_p ) / ( r_d ( r_d + r_p ) ) }
\Delta V = v_d - v_a
The space elevator calculation is more complicated.
For transfer orbits below geosynchronous, the apogee of the orbit is the release radius r_a , and the angular velocity is \omega_E = 2 \pi / P_{sidereal} , where P_{sidereal} is the sidereal day, 86164.0905 seconds. We want to find r_a and the perigee velocity v_p . We will have to work backwards.
v_a = \omega_E r_a
{v_a}^2 / 2 \mu = 1 / r_a - 1 / ( r_a + r_p )
Solve for r_p , perigee radius, the candidate r_d
r_p = \huge { r_a \over { ( 2 \mu / ( {r_a}^3 {\omega_E}^2 ) ) - 1 } }