Problems on Intertrajectory orbits for spacecrafts

[reachshijo]

Hey guys, I am a Masters student in Space Engineering and my hands have been tied up with this question that I have encountered. Hope some of you guys can help me sort out this question.

Q. A spacecraft reaches Mercury while travelling on a heliocentric orbit, co-planar with the Mercury orbit. The apocenter velocity on this heliocentric orbit is 30.818 km/s while on the same orbit the maximum radial velocity is 8.877 km/s. The encounter with the planet occurs after the spacecraft has already experienced the passage at the transfer orbit apocenter.
The transfer orbit eccentricity vector is 180 (deg) away from the Mercury eccentricity vector. Mercury orbit is elliptical.
The proposed close approach to Mercury, if exploited to close the spacecraft with a maneuver at the pericenter on a circular orbit around the planet, asks for a braking maneuver of -7.5145 km/s.
1. Identify and depict the gravity assist experienced by the spacecraft as it approaches Mercury considering the maneuver aims to reduce the energy content with respect to the heliocentric reference systems. Also calculate the maximum height of the spacecraft above the planet, the eccentricity of the gravity assist, and the focal axis orientation in the Mercury orbital reference plane.
2. Compute the minimum distance the spacecraft will reach with respect to the sun.

 

[indy91]

It's pretty late for me, so I'll only post about the first part of the problem.

The first step would be calculating the parameters of the heliocentric orbit. You only have those two velocities, but you can get the eccentricity and semi-major axis from this. Because the radial velocity is at its maximum you can write the more general equation:

[math]v_r = \frac{\mu}{h} e sin(\theta)[/math]

as

[math]v_r = \frac{\mu}{h} e[/math]

with the true anomaly (theta) = 90°.

Let's keep this in mind. For the velocity at apoapsis you can write:

[math]h = r_A v_A[/math]

We don't know the apoapsis radius, but we can replace it with:

[math]r=\frac{h^2}{\mu} \frac{1}{1+e cos(\theta)}[/math]

For the apoapsis theta becomes 180°, therefore we can write:

[math]r_A=\frac{h^2}{\mu} \frac{1}{1-e}[/math]

We know the two velocities and the gravitational parameter for the sun and have three unknowns in equations 2,3 and 5. That should be easy to solve.

For the other questions, the part about the eccentricity vectors means, that the transfer orbit and the orbit of Mercury have lined up apsides. So where the transfer orbit has a pericenter, the orbit of Mercury has an apocenter.

Then you need to do some hyperbolic calculations in the vicinity of Mercury. I will try to solve this part tomorrow

EDIT:

First the numbers I got for the parameters of the heliocentric transfer orbit:

[MATH]h = 3.3433\cdot10^{15} m^2/s^2[/MATH]
[MATH]e = 0.2236[/MATH]
[MATH]r_A = 1.0849\cdot10^{11} m = 0.7252 AU[/MATH]
[MATH]r_P = 6.8832\cdot10^{10} m = 0.4601 AU[/MATH]

Now we need to calculate the encounter. It is supposed to happen after the apocenter, if I understand the problem correctly. We know the orbital parameters of Mercury from literature and can see, that the encounter is near the pericenter of the transfer orbit and the apocenter of the Mercury orbit respectively. Near, but not exactly, so that could make some things more difficult. We now should find out the true anomaly of the encounter. I'm not 100% sure this is the correct way, but I used my 4th equation for both orbits and set the two equations to be equal:

[MATH] \frac{h^2}{\mu_S}\cdot\frac{1}{1+e\cdot cos(\theta)}=\frac{h_M^2}{\mu_S}\cdot\frac{1}{1+e_M\cdot cos(\theta+\pi)}[/MATH]

with:
h = specific relative angular momentum of the heliocentric transfer orbit
mu_S = standard gravitational parameter of the sun
e = eccentricity of the heliocentric transfer orbit
theta = true anomaly of the encounter
h_M = specific relative angular momentum of the Mercury orbit
e_M = eccentricity of the Mercury orbit

the +pi part comes from the condition, that the eccentricity vectors are 180° away from each other.

Solving for theta we get two solutions, but the first solution after the apocenter is 345.4178°, which is about 15° away from pericenter of the transfer orbit.

To be continued...

EDIT2:

Now to the encounter. I have never calculated that with the planet being on an elliptical orbit around the sun, so I'm not 100% sure how to calculate [MATH]V_{\infty}[/MATH]. In vector notation it is usually written as:

[MATH]\mathbf{v}_{\infty,1}=\mathbf{V}-\mathbf{V}^{(v)}_{1}[/MATH]

with two V's being the velocity vector of Mercury and the velocity vector of the arriving transfer orbit relative to the sun, respectively. But as you'll find out, this doesn't produce good numbers...

Anyway, with the number for the DV to get into a circular orbit (which we wont do, because it's supposed to be a gravity assist flyby) we can get all desired parameters. The velocity difference at the Mercury pericenter is:

[MATH]\Delta v = v_{p,circ} - v_{p,hyp}[/MATH]

We can replace this with the definitions of pericenter velocity for a circular and a hyperbolic orbit:

[MATH]\Delta v = \sqrt{\frac{\mu_{M}}{r_p}}-\sqrt{v_{\infty}^2+\frac{2 \mu_{M}}{r_p}}[/MATH]

We know everything but r_p, so have fun rewriting the equation.

The next step would be calculating a few defining parameters of the flyby orbit, e.g. the semi major axis from the vis-viva equation and the eccentricity with:

[MATH]e=1+\frac{r_p v_{\infty}^2}{\mu_M}[/MATH]

 

[Linguofreak]

This looks very suspiciously like homework.