Lecture 2 Conic Sections

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    Con ic sect ions

    r

    M

    m

    V

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    = Keplers constant = GM(m3s-2)

    h = angular momentum = = constant

    E = total energy = cons tant =

    e = eccentr ic i ty =

    2r

    r

    V 2

    2

    1

    12

    cos1

    2

    22

    Eh

    rh

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    ra r

    p

    semi minor

    axis b

    Semi major axis a

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    Solut ion

    21 1 coser h 1

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    Locat ion of per igee (closest app roach) can

    be found from 1 by set t ing to zero:

    2

    1

    p

    hr

    e

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    Locat ion of apogee (furthest departure) can be

    found f rom 1by sett ing to (no te that raonly

    has real values fo r e < 1, circ les and el lipses)

    2

    1a

    h

    r e

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    Combin ing perigee and apogee give a useful

    means of f inding eccentr ic i ty :

    pa

    pa

    ap

    rrrre

    ererh

    112

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    Semi major axis, a =

    2

    212

    1

    p ahr r

    e

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    Semi minor axis, b =

    21p ar r a e

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    fam i ly o f solut ions

    The value of eccentricity determines the

    category of orbit obtained

    e 1gives captive orbits

    e 1creates escape trajectories

    e = 0

    circular

    e > 0 1 hyperbola

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    example 2: Connect w ith

    el l ipt ical trans fer orb it

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    72.046500

    650040000

    pa

    pa

    rr

    rre

    10

    1 cos 1 1

    3.88 14*6.5 6*1.72 6.59*10

    p ah r e r e r e

    E E

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    smEr

    hV

    smEr

    h

    V

    r

    hV

    p

    p

    a

    a

    /130,1065.6

    10*59.6

    /650,1640

    10*59.6

    10

    10

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    sm

    sm

    sm

    v

    v

    v

    /3804

    /150816503158

    /2296783410130

    2

    1

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    Fuel requ irement for manouver

    For LH2/LOX rocket

    with c = 4500m/s

    429.04500

    3804exp

    0

    m

    mf

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    This was an example of a Hohmann transfer

    It was bel ieved to have the min imum energyrequirement for moving between co -planar

    c i rcu lar orb i ts , as no fuel is wasted onveloci ty vector rotation

    Disproved by US student who pos tulated thegrav ity ass is t manoeuvre for an ass ignment

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    Example 3: hyperbo l ic capture

    into circu lar orb i t

    Arr ival at mars w ith a relat ive app roach

    veloc i ty (V) o f 3km /sec, w ith cap tu re

    into a circular orb i t at an al t itude of

    4000km

    Findvrequired

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    V

    v

    V2

    Vcirc

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    2

    26

    2

    40008*10 /

    2

    VE

    r

    J kg

    Adjust hyperbolic approach so that its per igee is at

    the alti tude of the required circular orbit. The

    requi red velocity change wil l then be along theline of f l ight and at maximum velocity, giving the

    greatest change in E for a given v

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    smE

    EE

    rEV

    /611064

    1327.4*268*2

    2

    22

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    smE

    E

    rVcirc

    /327064

    1327.4

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    Combinations of these manoeuvres enable all

    bodies in the plane of the ecliptic to be visited.

    Some moons rotate out of plane, and require

    3D calculations of a more complex nature

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    LEO opt ions

    7834 m/s

    10130

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    GEO options

    1650

    3158

    Minimum v

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    Fuel requ irement for manouver

    For LH2/LOX rocket

    with c = 4500m/s

    28.04500

    5724exp

    0

    m

    mf