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| 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? | 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? |
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| The destination orbit has a radius of $ r_d $ and a velocity of $ v_d = \sqrt{ \mu / r_d } $ were $ \mu = $ 398600.4418 km^3^ / s^2^ . | The destination orbit has a radius of $ r_d $ and a velocity of $ v_d = \sqrt{ \mu / r_d } $ were $ \mu = $ 398600.4418 km^3^ / s^2^. ---- === 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 } } $ |
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 } }