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Assuming that your spacesuit does not impede your movement at all, and you use the same length sling.. exactly the same speed as you could slingshot on Earth. The speed of a whirled sling is not limited by gravity, or air resistance, but by inefficiency in you arm's muscular movement. Top speed is about 160km/h = 44.5 m/s On Earth the length is partially ...


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I'm not sure what altitude they are aiming for at periapsis, but the $\Delta V$ to go from an 800x800km to 800x100km orbit is 194 m/s, and the $\Delta V$ to go from a 900x900km orbit to a 900x50 km orbit is around 234 m/s, so that's around the range that would be necessary. If the arm in their schematic travels 1 meter (which is probably very generous), then ...


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Orbit Guardians - bs, right? Scary? Yes! But no, not necessarily 100% bs. Answers to How hard do you have to throw something off the ISS to make it deorbit? are in the "ballpark" of 90 m/s (fastest thrown Cricket ball is 45 m/s). However, at 800 km you'd need to throw it harder. Reusing my stuff from there: $$v^2 = GM \left(\frac{2}{r} - \frac{1}{...


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I don't have a derivation readily available, but here are some hints on solving this. The B-Plane vectors are defined by cross products of the eccentricity and orbital momentum vectors: https://nyxspace.com/MathSpec/celestial/orbital_elements/#b-plane-b_plane. The inclination is the arc cosine of the Z component of the orbital momentum vector over the ...


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To compute the arrival asymptotes, convert the heliocentric position and velocity into the planetocentric frame (the Mars J2000 frame in your case). Then, compute the B-Plane parameters as derived here , and a decent visualization is here. Once you are in that frame, you may compute the declination $\delta$ and right ascension $\alpha$ at arrival as follows: ...


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I see two main things throwing off your calculations: You can not simply subtract the velocity of one planet from another to get interplanetary transfer costs. An optimal transfer consists of an elliptic orbit touching the orbit of the inner planet at perihelion, and the outer planet at aphelion. Thus, the numbers you should try to obtain are the ...


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If you have the arrival hyperbolic excess velocity vector $\vec{v_{\infty}}$ and its magnitude $\big|\vec{v_{\infty}}\big|$, then you can calculate the unit vector $\hat{s}$ parallel to it by doing the following: $$\hat{s} = \frac{\vec{v_{\infty}}}{\big|\vec{v_{\infty}}\big|}$$ You can then use the algorithm to calculate the right-ascension and the ...


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In practical terms, these hypothetical single-element manoeuvrers are not useful, as one cares very little for what intermediate path one takes through empty space, while the propellant consumption is an absolute bottleneck. Argument of periapsis can be changed the way you describe, by applying zenith thrust at apoapsis or nadir thrust at periapsis: $$...


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https://sci.esa.int/documents/33960/35865/1567260128466-JUICE_Red_Book_i1.0.pdf ESA JUICE Red book, chapter 5.1.4 page 89 The inclination will be increased by several Callisto flybys. Interesting that the JUICE team had a tradeoff - higher inclination but less Callisto mapping (becase the flybys would be over the same region), or less inclination but flybys ...


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I think you're correct that it's using repeated gravitational fly-bys in a synchronized orbit to do it. Starting at 4:57 in the video, where the "high inclination orbits" portion of the mission begins, it's clear that each inclination change coincides with the orbiter meeting one of the moons. It then does the same trick in reverse to get back to ...


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I just got this announcement from the JPL mail list ssd-announce@list.jpl.nasa.gov: During the week of April 12, the Horizons ephemeris system will be updated to replace the DE430/431 planetary ephemeris, used since 2013, with the new DE440/441 solution and sixteen most massive small-body perturbers. The new DE440/441 general-purpose planetary solution ...


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In Vallado's Fundamentals of Astrodynamics and Applications, Section 8.6.1: Gravity Field of a Central Body, he derives the equations to calculate perturbations due to aspherical bodies. In Section 8.7.1: Application: Simplified Acceleration Model, he walks through an example calculating the acceleration due to J2 through J6 (as well as from atmospheric drag)...


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My interpretation of your question is this: At 0 horizontal velocity, an aircraft needs to provide 100% of its weight in upward force to maintain its altitude. At orbital velocity, with the surface of the Earth curving away exactly as fast as the aircraft falls, it needs to provide 0% of is weight in upward force to maintain altitude. The question: For an ...


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The ECI x-axis is defined as pointing towards the intersection between Earth's equatorial plane and the ecliptic plane, which is shown in the screenshot here. The red line across the sky is the ecliptic and the blue grid is the celestial sphere (also sometimes called equatorial grid). So you can see here the red arrow labeled "J2000: X" is pointing ...


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I've got a set of Keplerian orbital elements $e_0$, $a_0$, $i_0$, $\omega_0$, $\Omega_0$, and $\theta_0$, and I'd like to get to a different orbit with orbital elements $e$, $a$, $i$, $\omega$, $\Omega$, and $\theta$. How do I calculate (a) the amount of delta-v I'll need for this maneuver or set of maneuvers, and (b) which maneuver or set of maneuvers I ...


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What you're looking for is Lambert's problem, which is used both for trajectory design and orbit determination, and to produce porkchop plots. Your hunch that this is not a simple problem is correct. pykep has a solver for Lambert's problem that supports multiple revolutions as well as solvers for various related problems such as low-thrust trajectories.


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The Oberth effect exists because imparting a fixed change in momentum increases kinetic energy more when traveling at a faster speed than at a slower speed. This is due to the fact that kinetic energy is proportional to the square of velocity. Let's look at an example. Assume your spacecraft has a mass of 1 kg, and you have a rocket engine that can provide a ...


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If your field of view defined as a typical solid angle, as shown on Wikipedia, then you'd define the half angles as the half of 100x100: 50x50. STK (and probably other modeling tools) accounts for the visibility of a sensor using a dual-conic model similar to an eclipsing model when assuming spherical celestial objects. Am I missing something?


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I originally posted this answer here. This is a figure that I have from a class assignment from a few years back. While definitely not a practical trajectory, it shows the characteristics of how to transfer from an elliptical to a circular orbit. This solution was computed using indirect optimization. This problem assumed constant thrust magnitude (so the ...


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I don't have an exact GTO-to-GEO transfer on hand, but I have this figure from a class assignment from a few years back. While definitely not a practical trajectory, it shows the characteristics of how to transfer from an elliptical to a circular orbit. This solution was computed using indirect optimization. This problem assumed constant thrust magnitude (...


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