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5

In constrast to the other answers here, I would like to point out that the Oberth Effect does allow you to gain kinetic energy from a gravity well without needing to rob it of momentum ... if you fire your rocket motor at periapsis. Atomic Rockets has an excellent discussion of what the Oberth effect can do. So by burning 6 km/s of Δv, you get an actual ...


11

I think the question is based on a misconception about how gravity assists work. If you just let yourself get pulled to a distant object then continue out the other side, the same gravity that attracted you to it will then begin pulling you back again. You'll just oscillate around it like a bouncing ball. Gravity assists work because the target itself (e.g....


23

It doesn't really work that way. To begin with, the closest stars (apart from the Sun) are not close. If we were somehow to reach escape velocity from the Solar System (which this method won't do, see below), we would still be moving at only a small fraction of the speed of light unless we develop a propulsion system that generates energy internally or ...


20

The "gravitational" (slingshot) maneuvers space probes are performing are actually not so much about gravity. The gravity is method to "tie" temporarily these two bodies, but you could (purely hypothetically of course) use something else, some superstrong tether or so ... "Slingshot maneuver" is in fact much better name in this regard. What actually happens ...


1

This is a supplemental answer for now because while we know that a two body orbit can be reduced to a one body orbit around a central potential, doing that here will be a little distracting and I think the result for the one body in central potential looks cleaner. See also answers to Can the radial oscillations of an elliptical orbit be solved using a ...


2

The distance from the focus of attraction of an orbit can be expressed as a function of the true anomaly (angle) given by $r(\theta)=a\frac{1-e^2}{1+ecos(\theta)}$, where $a$ is the semi-major axis and $e$ is the eccentricity.


3

Let's go through this one thing at a time. First off, your value for $\delta$ is clearly in degrees. With an eccentricity of $200$, the reciprocal $1/e$ is so small it's nearly equal to it's own inverse sine -- in radians. $\delta$ is basically $2/200$ radians $=0.573°$. Next, realize that geometrically, the eccentricity of a hyperbola equals the center ...


5

There are two major issues that I can see. Whatever you're using to calculate the arcsin is indeed giving you a value in degrees. The parameters you've provided are basically that of a satellite racing by an object ad a moderate distance, moving far above escape velocity all along its trajectory. It neither gets close enough, nor hangs around long enough ...


23

I found it. It was NOSS 3-1, a satellite pair. Found it through heavens above. The pair also explains the apparent tumbling. However, it seemed much brighter than 4.1, but it‘s definitely it.


2

In-the-sky.org doesn't show anything that quite matches your specification. There was an H-2A rocket body (i.e. upper stage) overhead at 7:46 going from NE to ESE; there was an SL-16 (Zenit) rocket body a couple of minutes later going ESE to N. Tumbling rocket bodies can vary considerably in brightness.


5

The rotating reference frame is usually preferable over an inertial reference frame when analyzing three-body motion. Two main reasons for this: The motion is determined by the two primary bodies, so it makes sense to use these to define the reference system The typical three-body dynamics for which you would use such a reference system, such as Libration ...


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