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4

The potential of fly-by ping pong is pretty much unlimited, provided you have enough time to your disposal. Given an initial transfer with a perihelion slightly lower than the orbit of Venus, a Venus flyby can increase the aphelion to a bit further out than the Earth's orbit. On the following Earth flyby, the perihelion can be lowered, and you can just ...


10

The eccentricity is 1.0. The eccentricity $e$ of an orbit can be found from the radius of apoapse and periapse as: $$e=\frac{r_a-r_p}{r_a+r_p}$$ and the semimajor axis $a$ can as well, from: $$a=\frac{r_a+r_p}{2}$$ If you throw an object horizontally (velocity perpendicular to position vector) you will end up in a closed orbit if you throw at slower ...


7

The eccentricity of a radial orbit is $1$, regardless of its energy. This is a class of orbits where the type of orbit cannot be inferred from the eccentricity alone. With a "traditional" parabolic orbit of $e=1$, the angular momentum $L$ has a well defined value, but the semi-major axis $a$ is not defined. In the case of a vertical bounded free-fall orbit, ...


2

Brilliant question. I'm surprised nobody referred to the amazing JSWT User Documentation available at STScl-JWST. (This question was active again, somehow, so adding an answer.) Antzi's answer was only regarding momentum management, stationkeeping is about orbit maintenance. But let's address both here. From the JWST user documentation, it is clear that ...


7

No, it's a textbook case of conservation of the linear momentum vector in the absence of any external forces. Linear momentum of a system Sum(mv) is a conserved quantity even if individual parts are allowed to change their momentum vectors. Actually reaction wheels also conserve angular momentum of the total system (ship + wheel) as well! But that's Okay ...


40

Previously posted comments are correct: in free space (assumed free of any other bodies' gravity fields) there is no way to convert the reaction wheels' angular motion to translational motion. There is one tongue-in-cheek way: throw a reaction wheel off the spacecraft in the direction opposite the direction of the desired delta-V! ;-) If you abandon the ...


5

In this answer, I'm going to assume the tunnel is a perfect vacuum and there's no intense heat or pressure. This is a thought experiment. If you and John fall at the same time, you guys will meet approximately 20 minutes after you jump into the tunnel. Depending on how wide the tunnel is, you guys will most likely collide at a velocity of around 11 km/s. ...


2

In terms of gravity, there will be no force (ish). This is a consequence of the shell theorem, which states that inside a sphere of constant density the shell of the sphere exerts no gravitational force. Now, some caveats: Earth is not a perfect sphere, but it's so close to one for our purposes that it doesn't matter Earth is not constant density. ...


6

Exactly at the center-of-mass of the Earth there is no net gravitational force. Gravity comes from mass, and all the mass of the Earth is evenly distributed around the center-of-mass (by definition), so all the forces cancel out. As you get deeper and deeper, the net gravitational force falls -- linearly if density is constant. In practice, of course, you ...


1

Welcome to the site! Nope. The deepest the human race has gotten is the now-defunct Kola Borehole, which didn't even make it through the crust. Though the crust is thinner at the ocean floor, that presents its own challenges. Pressure and heat, mostly. You've got ~6*10^24 kilos of (mostly) iron above you at around 5000 degrees C and ~3e6 atmospheres. If ...


3

The article assumes that the Earth's shape is an oblate spheroid, i.e., a "squashed" sphere. If we "unsquash" everything back, including satellites' positions (which means multiplying the $z$-coordinates of the satellites by $1/\sqrt{1-\epsilon_e^2}$), the line segment connecting the new positions would intersect the sphere if and only if the line segment ...


4

This is another example of a perpetual motion machine. Although you may believe it will produce net motion without violation of conservation of momentum: Any net motion produced would be a violation of conservation of momentum, so it cannot occur You've not fully accounted for the net center of mass motion throughout your machine's cycle, or how momentum is ...


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