SteamKing said:
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.
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?
