Solving Flyby Trajectories: Arrival/Departure Speed and Eccentricity

[Dustinsfl]

Referring to the diagram below, suppose the spacecraft enters the sphere of influence of Venus ($r_{\text{SOI}}$ 616000km) with a heliocentric speed of 37.7km/s and a heliocentric flight path angle $\sigma_2 = 20^{\circ}$.
Suppose that the hyperbolic trajectory asymptote is designed with an aiming radius of $\Delta = 1.5$ planetary radii ($r_{\venus} = 6,051.8$km).
For this problem, the heliocentric velocity of Venus is 35.022km/s and the gravitational parameter for Venus is $\mu_{\venus} = 324859\text{km}^3/\text{s}^2$.
What is the arrival/departure speed $v_2$ and $v_3$ relative to Venus?
I am not sure on how to determine the departure speed. This is because I can't find anything to help identify pieces of the triangle on the departure side.
The speed in the x direction is $\mathbf{v}_x = 37.7\cos\frac{\pi}{9}$ km/s but this isn't the speed relative to Venus.
$$
\mathbf{v}_{x\text{rel}} = 37.7\cos\frac{\pi}{9} - 35.0 = 0.404412\text{ km}/\text{ s}
$$
The speed in the y direction relative to Venus is $\mathbf{v}_y = -37.7\sin\frac{\pi}{9}$ so the relative arrival speed to Venus is
\begin{alignat*}{3}
\mathbf{v}_2 & = & v_x\hat{\mathbf{i}} + vy\hat{\mathbf{i}} + vz\hat{\mathbf{k}}\\
& = & 0.404412\hat{\mathbf{i}} - 12.8942\hat{\mathbf{i}} + 0
\end{alignat*}
So the relative arrival speed to Venus is $\lVert\mathbf{v}_2\rVert = \sqrt{0.404412^2 + 12.8942^2} = 12.9005$ km/s.

What is the eccentricity of the fly-by trajectory?
From the energy equation, we have
\begin{alignat*}{3}
\mathcal{E} & = & \frac{v_2^2}{2} - \frac{\mu_{\venus}}{\venus\text{SOI}}\\
& = & \frac{12.9005^2}{2} - \frac{324859}{616000}\\
& = & 82.6841
\end{alignat*}
Now we can use $\mathcal{E} = \frac{\mu_{\venus}}{2a}$, to solve for the semi-major axis, $a$.
$$
a = \frac{\mu_{\venus}}{2\mathcal{E}} = 1964.46\text{ km}.
$$
Finally, we can use the equation for the aiming radius to the eccentricity, $e$.
\begin{alignat*}{3}
\Delta & = & 1.5\cdot 6051.8\\
& = & 9077.7\\
9077.7 & = & a\sqrt{e^2 - 1}\\
e & = & \sqrt{\left(\frac{9077.7}{1964.46}\right)2 + 1}\\
& = & 4.72793
\end{alignat*}

 

[jvicens]

Based on the given information, the departure speed can be determined by using the law of conservation of energy. This law states that the total energy of a system remains constant. In this case, the total energy can be written as:
$$
E = \frac{v_{2}^2}{2} - \frac{\mu_{Venus}}{r_{2}}
$$
where $v_{2}$ is the arrival/departure speed, $\mu_{Venus}$ is the gravitational parameter of Venus, and $r_{2}$ is the distance from the center of Venus at the point of arrival/departure.

Using the given values, we can write the equation as:
$$
E = \frac{v_{2}^2}{2} - \frac{324859}{6051.8 + 9077.7}
$$

Solving for $v_{2}$, we get:
$$
v_{2} = \sqrt{2E + \frac{324859}{r_{2}}}
$$

Substituting the values, we get:
$$
v_{2} = \sqrt{2(82.6841) + \frac{324859}{6051.8 + 9077.7}} = 12.026\text{ km/s}
$$

Therefore, the arrival/departure speed $v_{2}$ relative to Venus is $12.026$ km/s.

To calculate the eccentricity of the fly-by trajectory, we can use the equation:
$$
e = \frac{\Delta}{a}
$$
where $\Delta$ is the aiming radius and $a$ is the semi-major axis calculated earlier.

Substituting the values, we get:
$$
e = \frac{9077.7}{1964.46} = 4.623
$$

Therefore, the eccentricity of the fly-by trajectory is $4.623$.