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I can't seem to find any good sources online for this, all I get are documents on how Nice model was used to compute Neptune's gravity capture of Triton via a binary dissociation, possibly because of similar terminology. So to the question;

To what extent, in terms of % delta-v, could Triton's retrograde orbit around Neptune gravitationally assist in slowing down a direct Hohmann transfer from Earth to Neptune for Neptune Orbit Insertion (NOI) during a single close-shave flyby of Triton?

I realize that the transfer would take about good three decades (I calculated 30.61652 years at Hohmann semi-major axis of 15.53545 AU using Kepler's third law and semi-major axes of Earth and Neptune), transfer time is not a concern, but I'm stuck with calculating velocity relative to Neptune at Triton's altitude for this transfer (I'm assuming Triton to be at the right place at the right time) and I didn't yet get to calculate delta-v that a low altitude (100 km above surface) gravity assist flyby of Triton could shed, so I'm not sure I'd know how to do that either. Neptune NOI would be for a Neptune orbit with semi-major axis of roughly 5,000 km above its surface (at 1 atm pressure). There wouldn't be any plane change for the target orbit.

This is not a homework question. It's been about 20 years since I last had to do such calculations back in the college years and I'd appreciate a bit of help in brushing up on this. I could plug this in some software but, call me masochistic, I wanted to do it the old-fashioned way. Back-of-an-envelope calculations would do, preferably discussing any shortcuts there might be in getting first-order approximations faster. No need to be too academic either, but I would like to see some calculations here, if that's not terribly inconvenient, in which case I guess a good reference with a short-ish writeup would do, too.

If you'll opt for the latter option using external references, please also include optimal launch dates, delta-v and Hohmann transfer times.

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    $\begingroup$ What you really want to do for Neptune orbit insertion is aerobraking at Neptune. Also you should be concerned with the transfer time. Neptune mission studies typically go for transfer times of ten years or so, due to spacecraft lifetime qualification, and due to Principal Investigator and government sponsor patience. This results in much higher $v_\infty$'s at Neptune. $\endgroup$
    – Mark Adler
    Commented Dec 20, 2014 at 18:17
  • $\begingroup$ @MarkAdler The point of this exercise of mine is not to establish optimal transfers to NOI. Frankly, I just made something up for my exercise and didn't even put much thought to how much sense it makes. I'm basically just brushing up on the decades of rust that gathered on my orbital mechanics and looking for new ideas of how to approach such problems. Your answer is helpful and I'll continue with your help now. I might have additional questions, but I'll probably rather post them as new ones so we can wrap this one up. ;) $\endgroup$
    – TildalWave
    Commented Dec 20, 2014 at 18:29
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    $\begingroup$ Understood. Fun exercise. Note that you will need to take into account the tilt of Triton's orbit with respect to the solar orbit plane on which you will be approaching Neptune. That tilt is substantial, currently about 50°. $\endgroup$
    – Mark Adler
    Commented Dec 20, 2014 at 19:04

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I'll try to get you started anyway. From the frame of reference of the assisting body, the trajectory of the probe is hyperbolic, with the same $v_\infty$ going out as coming in, but in a different direction. The trajectory is simply bent. The bend angle is:

$$\delta=2\sin^{-1}\left(1\over e\right)$$

where $e$ is the eccentricity of the hyperbola. You can get $e$ from $v_\infty$, the closest approach radius $r$ (measured from the center of the body), and the $GM$ of the body, $\mu$:

$$e={r\,v^2_\infty\over\mu}+1$$

$e$ goes from $1$ to $\infty$, where $1$ is effectively an infinite gravity assist that reverses the direction of the velocity by 180°, and $\infty$ is no assist at all with a bend of 0°. The closer in you can get and the slower you can go, the greater the bend angle is.

With the vector addition of the velocity of the assisting body (Triton) in the frame of the body it is orbiting (Neptune), you can see how the velocity of the probe is changed by the flyby in the Neptune frame, which is what you care about. You will need to do the reverse to get the $v_\infty$ approaching Triton, i.e. subtract the velocity vector of Triton from the approach velocity vector in the frame of Neptune. To get that, you subtracted the velocity vector of Neptune from the velocity in the Sun frame of the Hohmann transfer. Lots of frame changes.

There is a range of geometries for the flyby you can try, depending on where in its orbit Triton is at the time, the altitude of the flyby, and whether the flyby closest approach is on the leading or trailing side of Triton in its orbit.

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  • $\begingroup$ Hi Mark; I'm a bit confused by your answer. First, I know that the eccentricity of a hyperbola is given by $e=\frac { \sqrt { { a }^{ 2 }+{ b }^{ 2 } } }{ a } $. So how did you incorporate $r$, ${ v }_{ \infty }^{ 2 }$ in that expression? Also, what's $\mu $ $\endgroup$
    – DarkRunner
    Commented Jan 15, 2020 at 22:41
  • $\begingroup$ You don’t need to know or compute your a and b. You just plug in the e from the second expression into the first. $\endgroup$
    – Mark Adler
    Commented Jan 17, 2020 at 4:09
  • $\begingroup$ mu is the GM (Newton’s gravitational constant times the mass) of the body. Triton in this case. $\endgroup$
    – Mark Adler
    Commented Jan 17, 2020 at 4:11
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    $\begingroup$ The formula is right there in the middle of the answer. However $e$ does not depend only on $v_\infty$. You choose the eccentricity independently of $v_\infty$ by aiming for a particular closest approach radius $r$. That $r$ is bounded on the low side by the body itself. $\endgroup$
    – Mark Adler
    Commented Feb 13, 2020 at 15:18
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    $\begingroup$ @MarkAdler Does $e={r\,v^2_\infty\over\mu}+1$ work if the initial trajectory is an ellipse. In other words, can I use the velocity at the apoapsis of an ellipse as $v_\infty$? $\endgroup$
    – Star Man
    Commented Mar 26, 2020 at 18:44

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