Interplanetary Earth to Mars

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    Interplanetary

    Transfer

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    A Simplified Example

    A transfer from an orbit around a inner planet tothe orbit around an outer planet.

    For example, to calculate the Dv required tolaunch a spacecraft from an orbit around the Earth

    (let us take LEO of 300 km altitude) to an orbit

    around the Mars.

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    Hohmann Transfer (from earth to Mars)

    seV

    sp eV t

    sp mV tsm

    V

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    Departure from Earth

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    Interplanetary Transfer (Departure)

    Here = The velocity of the space probe

    relative to the sun at the time of escaping earth (ie.)

    velocity of the space probe on the transfer ellipse at

    the departure point D (perihelion in the heliocentric

    orbit).

    = the circular orbital speed of the planet 1

    (here, the Earth) relative to the sun.

    Here | | always > | |.

    sp

    e

    V t

    seV

    sp eV tseV

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    Interplanetary Transfer (Departure)

    Thus, by subtracting the known value of velocity

    of the planet relative to the sun, one obtains the

    hyperbolic excess speed on the earth escape

    hyperbola.

    Therefore, the required Dv at D could be

    2

    1 2 1

    21

    sp se

    D e

    ep

    e

    Rv V t V

    R R R

    V t V

    D

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    Interplanetary Transfer (Departure)

    2.946 km/s --- Is it the launch speed of the space

    probe at the orbit around the Earth?

    8

    1 2

    11 3 2

    11

    1 8

    29.783 / 0.09891

    1 1.496 10 , 1.5241.327 10 /

    2 1.5241.327 10 11.496 10 1.524 1

    2.946 /

    sun

    km s

    R AU km R AUkm s

    AUvAU AU

    km s

    D

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    Interplanetary Transfer (Departure)

    NO! --- It is the velocity of the space probe after

    escape from Earth (after launching from the earth &

    achieving a great distance from the earth). This

    speed is called excess speed (not the escape speed).

    Therefore, we have to find out what could be the

    velocity of the space probe relative to earth at thetime of launch, , in order to achieve the

    excess speed of 2.946 km/s at the point of exiting

    the earths sphere of influence.

    2.946 /eV t km s

    ep LV t

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    Interplanetary Transfer (Departure)

    Conservation of specific energy to calculate the

    launch speed, required to achieve excess

    speed of 2.946 km/s. ep LV t

    2 2

    2

    2

    3 2

    1 1

    2 2

    10

    2

    2

    398600.4418 / , 6378 300

    11.3 /

    ep epe eL ep

    L

    e

    ep eL ep

    L

    ep

    e L

    ep

    L

    V t V t

    r t r

    Vr

    V t Vr t

    km s r t km

    V t km s

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    Interplanetary Transfer (Departure)

    Therefore, at this point, the Dv applied by the

    thrusters would be 11.3 7.726 = 3.574 km/s.

    11.3 /ep

    LV t km s

    6678 km

    7.726 /km s

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    Interplanetary Transfer (Departure)

    Thus after applying the velocity change of 3.574km/s to the original circular velocity of 7.726 km/s,

    the spacecraft (or space probe) is travelling with

    velocity 2.946 km/s relative to earth, or velocity of

    32.7 km/s relative to the Sun.

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    Arrival at Mars

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    Arrival at Mars

    Here = The velocity of the space probe

    relative to the sun at the time of approaching mars

    (ie.) velocity of the space probe on the transfer

    ellipse at the arrival point A (aphelion in the

    heliocentric orbit).

    = the circular orbital speed of the planet 2

    (here, the Mars) relative to the sun.

    Here | | > | |.

    sp mV t

    smV

    smV sp mV t

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    Arrival at Mars

    Therefore, the required Dv at A could be

    Plugging all the values in the above equation,

    2.650 km/s.

    1

    2 2 1

    21

    sm sp

    A m

    mpm

    Rv V V t

    R R R

    V t V

    D

    mV t

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    Arrival at Mars

    Now the space probe is approaching Mars on a

    hyperbolic trajectory from an infinite distance with

    velocity of 2.650 km/s relative to Mars.

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    Arrival at Mars

    Now, we have to find out what could be the

    velocity of the space probe relative to Mars at the

    time of reaching the orbit around the Mars ( ). mp mV t

    22mp marsm mmp

    m

    V t V t r t

    3 2

    42,823.3 /3397

    mars km sRadius of Mars km

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    Arrival at Mars

    Therefore, at this point, the Dv applied by the

    thrusters to slow down to orbit around Mars would

    be km/s.

    mp mV t

    2.650 /km s

    mp m oMV t V

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    Reference(s)

    Howard D. Curtis, Orbital Mechanics forEngineering Students, Elsevier Publications,

    2005.

    http://web.mit.edu/12.000/www/finalprese

    ntation/traj/hohman.html

    Youtube videos Design of Interplanetary

    Missions using Patched Conics (3 parts).

    http://web.mit.edu/12.000/www/finalpresentation/traj/hohman.htmlhttp://web.mit.edu/12.000/www/finalpresentation/traj/hohman.htmlhttp://web.mit.edu/12.000/www/finalpresentation/traj/hohman.htmlhttp://web.mit.edu/12.000/www/finalpresentation/traj/hohman.html