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2-Year Earth-to-Earth Trajectory

By Frances Campbell,2014-03-22 07:34
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2-Year Earth-to-Earth Trajectoryto,Earth,2year,earth,year

    2-Year Earth-to-Earth Trajectory

    ASEN 5519

    VComputing the ’s for the Earth-to-Earth portion of the VEEGA trajectory is one of the

    trickier parts of the final project. At Venus, we can use Lambert’s problem to compute

    VVVVthe and . We can compute the for the first EGA and for the second inoutinout

    EGA using Lambert’s problem as well. However, we can not use Lambert’s problem to

    VVcompute the for the first EGA or for the second EGA. The only constraint outin

    that we have is that the spacecraft must return to the same position two years later (since we already know that Earth will essentially return to the same place in two years). Using this information, it is straightforward to compute the magnitude of the sun-centered velocity vector immediately following the first EGA, as follows:

    365.242189 days 86,400 secP 2 years * * 1 year1 day

    1/32?P??a ?(sun2??()?

    sun E -2a

    ??sun?V 2E s/c_sun?rEarth_EGA1??

    However, in order to compute the appropriate B-plane parameters, we need to know the vector components of the velocity. Figure 1 shows the geometry of the velocity vector immediately after the first EGA. This diagram shows that there are infinitely many possibilities that would place the spacecraft into a 2-year orbit, all of which lie on the illustrated circle.

     Vsc_sun

    V

    ;(;

    Locus of 2-

    Year Orbits V Earth

    Figure 1: Diagram illustrating the possible orientations of the spacecraft velocity

    vector immediately after the first EGA.

Using the law of cosines, the angle can be computed.

    222;?VVV2VVcosθ s/c_sunEarthEarth

    VVVwhere, inout

    VNow we can compute the outgoing hyperbolic excess velocity vector () as a out

    function of and (.

    ?Vcos-;?out(( VVsin-sin;?;?out_1out(

    (;?;?Vsin-cos(out)?

    This vector is expressed in a coordinate system in which the x-axis lies along the Earth’s velocity vector (as can be seen from figure 1). We’ll call this reference frame “frame 1”.

    The Earth’s velocity vector is not necessarily along the x-axis of the ecliptic frame, thus V needs to be rotated from frame 1 into the ecliptic coordinate frame. First, we out_1

    assume that the Earth’s velocity vector lies in the ecliptic x-y plane. Figure 2 shows the geometry of the Earth’s velocity vector and the outgoing hyperbolic excess velocity vector in the ecliptic reference frame. Note that Earth’s velocity vector is in the x-y

    frame, however the outgoing hyperbolic excess velocity vector is not.

     yecliptic z Vecliptic

    V Earth

    ;

    x ecliptic

    Figure 2: Diagram illustrating the geometry of the Earth’s velocity vector and the

    outgoing hyperbolic excess velocity vector in the ecliptic reference frame.

From figure 2, we can see that the rotation from frame 1 to the ecliptic frame is simply a

    rotation of about the z-axis.

    VRV out_ecl1~eclout_1

    cossin0;?;??

    (;?atan2Vy,Vxwhere Rsincos0 and ;?;?EarthEarth1~ecl(

    (001)?

    Algorithm

    VVGiven: , for EGA 1 inEarth

    V Compute P, a, E, and s/c_sun

    ;?VVV2VVcosθCompute using: s/c_sunEarthEarth

    ;?atan2Vy,VxCompute EarthEarth

    cossin0;?;??

    (Rsincos0Compute ;?;?1~ecl(

    (001)?

    Loop over 0 < ( < 2*pi

    ?Vcos-;?out((Compute VVsin-sin;?;?out_1out(

    (;?;?Vsin-cos(out)?

    VRVCompute out_ecl1~eclout_1

    VUse to make sure EGA 1 and EGA 2 do not go below the surface of Earth out_ecl

    If both EGAs work, save the vector.

    End loop

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