Timeline for To what extent could a single Triton flyby slow down a direct Hohmann transfer to Neptune for NOI?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 26, 2020 at 18:44 | comment | added | Star Man | @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 $v_\infty$? | |
| Feb 13, 2020 at 15:18 | comment | added | Mark Adler | 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. | |
| Feb 12, 2020 at 17:18 | comment | added | DarkRunner | Hi mark, how do you get the eccentricity in terms of v_infinity? | |
| Jan 17, 2020 at 4:11 | comment | added | Mark Adler | mu is the GM (Newton’s gravitational constant times the mass) of the body. Triton in this case. | |
| Jan 17, 2020 at 4:09 | comment | added | Mark Adler | 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. | |
| Jan 15, 2020 at 22:41 | comment | added | DarkRunner | 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 $ | |
| Jun 6, 2015 at 12:55 | vote | accept | TildalWave | ||
| Dec 20, 2014 at 17:18 | history | answered | Mark Adler | CC BY-SA 3.0 |