1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A challenging transfer orbit question

  1. Dec 27, 2015 #1
    Assume that these orbital elements remain good for Vesta and for Ceres in the year 2080.

    Vesta.
    a = 2.3617272 AU
    e = 0.08889958
    i = 7.14°
    Ω = 103.8467°
    ω = 151.1467°
    T = 2456923.543

    Ceres.
    a = 2.7680973 AU
    e = 0.07575951
    i = 10.5917°
    Ω = 80.3229°
    ω = 72.7187°
    T = 2456552.887

    Part A.

    Prove that a boost-and-coast elliptical transfer orbit having its perihelion at Vesta at the moment of departure—0h UT on 1 June 2080—exists.

    Part B.

    Determine

    1. The position of the Ceres intercept in heliocentric ecliptic coordinates.

    2. The time in transit.

    3. The arc of true anomaly from departure to arrival.

    4. The Keplerian elements of the transfer orbit.

    5. The delta-vees needed for transfer orbit insertion at departure and for velocity matching at arrival.
     
  2. jcsd
  3. Dec 27, 2015 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Sounds like a HW exercise.

    Please use the Homework Template when posting such threads and post them in the relevant HW forum. You must show an attempt at solving the problem in order to receive help.
     
  4. Dec 27, 2015 #3
    This isn't homework. I finished school in 1982 and am presently retired. I posted this just to see whether anyone could do the math. But, okay, I'll make a start to solving the problem.

    Part A.

    This problem is actually a bit more than a transfer orbit problem, as it involves an interception with an object in the destination orbit. Proving that the transfer orbit results in an intercept of Ceres at the arrival position requires that one demonstrate the equality of two differences in time:

    1. The time required for the spaceship to travel from Vesta's position at the moment of departure to the position at which Ceres will be found at the moment of arrival. We call this time difference the calculated transit time Δt.

    2. The time required for Ceres to travel from where it is at the moment the spaceship departs from Vesta to the position at which the spaceship crosses Ceres' orbit. We call this time difference the required transit time: t₂−t₁.

    Finding a position of arrival for which Δt=t₂−t₁ isn't trivial. Generally the student, playing the part of the spaceship's navigator, must search for a point on Ceres orbit such that the time differences match each other. The search could presumably be made easier with an appropriate algorithm, but the hunt-and-peck method will suffice.

    In practice, many of the transfer orbits proved to exist by a match in the transit times will be retrograde with respect to the general angular motion of the solar system, and one might additionally require that the transfer orbit have a maximum inclination to the ecliptic.

    I assert that a suitable transfer orbit from Vesta to Ceres exists having the aforementioned departure instant, intercepting Ceres on 21 May 2081 at 1h 13m 13s UT. I will develop the math needed to prove my assertion. I'll assume that my readers already know how to reduce the Keplerian elements of an orbit and a specified time to a position and velocity in heliocentric ecliptic coordinates.

    Departure time, heliocentric position and sun-relative velocity of Vesta in ecliptic coordinates.

    t₁ = 2480916.5 = 0h UT on 1 June 2080
    x₁ = +0.990674250328 AU
    y₁ = −1.960739031050 AU
    z₁ = −0.061709732998 AU
    Vx₁ = +18868.91718568642 m/s
    Vy₁ = +8318.62962562096 m/s
    Vz₁ = −2544.32403043785 m/s

    The heliocentric distance and longitude of departure:

    r₁ = 2.197667197119 AU
    λ₁ = 296.8054404448836°


    Arrival time, heliocentric position and sun-relative velocity of Ceres in ecliptic coordinates.

    t₂ = 2481270.5508449073 = 1h 13m 13s UT on 21 May 2081
    x₂ = +2.304282544207 AU
    y₂ = +1.641923128791 AU
    z₂ = −0.373147469306 AU
    Vx₂ = −10690.43095525598 m/s
    Vy₂ = +13460.76837587118 m/s
    Vz₂ = +2393.72410187866 m/s

    The heliocentric distance and longitude of arrival:

    r₂ = 2.853921624405 AU
    λ₂ = 35.471881159415354°


    The required transit time,
    t₂−t₁ = 354.05084490729496 days


    Since r₁<r₂, if there is an apside of the transfer orbit at r₁, then it must be its perihelion.

    The line-of-sight distance between departure and arrival,
    d = 3.847302282288 AU


    I define the integer variable β and permit it to have only the values 1 and 2.

    If β=1, an apside (perihelion or aphelion) of the transfer orbit occurs at departure.
    If β=2, an apside (perihelion or aphelion) of the transfer orbit occurs at arrival.

    β = either 1 or 2
    φ = 3 − β
    N = (−1)^φ

    The variables β and φ will usually be subscripts. The variable N is a sign toggle factor.

    Definitions.
    m : mean anomaly
    u : eccentric anomaly
    θ : true anomaly

    If the apside at the apsidal endpoint of the intended trajectory is the perihelion, then
    mᵦ = uᵦ = θᵦ = 0

    If the apside at the apsidal endpoint of the intended trajectory is the aphelion, then
    mᵦ = uᵦ = θᵦ = π radians

    In the example problem, we have a perihelion at departure (from Vesta) so that

    β = 1
    φ = 2
    N = 1
    m₁ = u₁ = θ₁ = 0


    The eccentricity of a conic section, having the sun at a focus, which includes the point of departure and the point of arrival, is found by solving, simultaneously:

    (a) The equation which relates the heliocentric distance with the true anomaly,
    cos θ₂ − cos θ₁ = { [a (1−e²) / r₂ − 1] − [a (1−e²) / r₁ − 1] } / e

    (b) The law of cosines,
    d² = r₁² + r₂² − 2 r₁ r₂ cos(θ₂−θ₁)

    The semimajor axis can always be eliminated because one or the other endpoints of the intended trajectory occurs at one or the other apside of the transfer orbit. That is, rᵦ=a(1±e). This is why you don't need a third point on the transfer orbit to determine its elements.

    After some algebra, we get
    e = 2 (cos θᵦ) rᵦ (rᵦ−rᵩ) / (rᵩ² − rᵦ² − d²)

    In the example problem,

    e = 2 (cos θ₁) r₁ (r₁−r₂) / (r₂² − r₁² − d²)
    e = 0.2511148483338123


    The semimajor axis of the hypothetical transfer orbit is found from
    a = rᵦ / (1 − e cos θᵦ)

    In the example problem,

    a = r₁ / (1 − e cos θ₁)
    a = 2.934585085883742 AU


    The true anomaly in the hypothetical transfer orbit at the non-apsidal endpoint of the intended trajectory is found as follows:

    θᵩ = θᵦ + N arccos{(rᵦ² + rᵩ² − d²) / (2rᵦrᵩ)}

    In the example problem,

    θ₂ = θ₁ + N arccos{(r₁² + r₂² − d²) / (2r₁r₂)}
    θ₂ = 1.7169743527182846 radians


    The eccentric anomaly in the hypothetical transfer orbit at the non-apsidal endpoint of the intended trajectory is found as follows (observe the two-dimensional arctangent function):

    sin uᵩ = (rᵩ/a) sin θᵩ / √(1−e²)
    cos uᵩ = (rᵩ/a) cos θᵩ + e
    uᵩ = arctan(sin uᵩ , cos uᵩ)

    In the example problem,
    u₂ = 1.461115971266753 radians


    The mean anomaly in the hypothetical transfer orbit at the non-apsidal endpoint of the intended trajectory is found from Kepler's equation.

    mᵩ = uᵩ − e sin uᵩ

    In the example problem,
    m₂ = 1.2115100376022254 radians


    The period of the hypothetical transfer orbit is
    P = (365.256898326 days) a^1.5

    In the example problem,
    P = 1836.1936701754814 days


    The mean motion in the hypothetical transfer orbit is
    μ = 2π/P

    In the example problem,
    μ = 0.0034218532659352403 radians/day


    For short path trajectories (for which the arc of true anomaly going from departure to arrival is less than π radians), the calculated transit time in the hypothetical transfer orbit is

    Δt = (N/μ) [mᵩ − π sin(θᵦ/2)]

    For long path trajectories (for which the arc of true anomaly going from departure to arrival is between π and 2π radians), the calculated transit time in the hypothetical transfer orbit is

    Δt = P − (N/μ) [mᵩ − π sin(θᵦ/2)]

    We can infer by the fact that the arc of heliocentric longitude from the departure position to the arrival position (λ₂−λ₁), adjusted to the interval [0,2π), is 1.722054251692 radians, which is less than π radians, that the transfer orbit follows the short path, and hence, in the example problem,

    Δt = (N/μ) [m₂ − π sin(θ₁/2)]
    Δt = 354.05084422025993 days


    Here's the test that determines whether the choices of departure time and arrival time result in a valid transfer orbit. The following condition MUST be true, or else the spaceship's navigator will have to change either the time of departure, or the time of arrival, or both.

    Δt ≈ t₂ − t₁

    In general, that condition will not be met. The spaceship's navigator must be clever about finding moments for departure and for arrival that do result in a very near equality between the required and the calculated transit times.

    Fortunately, in the example problem,
    t₂ − t₁ − Δt = −59.36 milliseconds


    That's close enough to prove that a boost-and-coast transfer orbit exists between r₁ departing at t₁ and r₂ arriving at t₂.

    This completes Part A and the first three sub-questions of Part B, and it includes parts of the answer to Part B sub-question #4. Does anyone want to finish Part B?
     
    Last edited: Dec 27, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: A challenging transfer orbit question
  1. Orbit Question (Replies: 2)

  2. Lunar orbit transfer (Replies: 0)

Loading...