10

No, such a calculation is not possible. The performance to different orbits is determined by the specific parameters of each launcher, e.g. specific impulse of each stage engines, fuel ratio, time of staging, flight profile... Have a look at this plot, taken from https://elvperf.ksc.nasa.gov As you can see the two launchers have the same performance to a ...


10

If we assume Keplerian/Newtonian mechanics, then we can see a way to rendering the same local curvature of the path at perigee and at apogee (terms for orbiting Earth, of course). At both points the motion is perpendicular to the applied gravitational force. So, the curvature is given from Newtonian mechanics as $f/(mv^2)$ where $f$ is the magnitude of the ...


9

See pp. 14 & 15 in this FAA guide for an explanation, including these great diagrams: I also recommend sketching it out on a roll of paper:


7

As someone who computes simple orbits for a living, I would say your question has meaning only in the original Keplerian model of projectile motion: gravitational fields of central bodies are point symmetric, barycenters are at the geometric centers, and other bodies are not present. Then the symmetry arises from the straightforward, simple math used to ...


6

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 ...


5

I understand I am (3 years) late to the party, but I was recently linked to this question and figured I could determine a pretty accurate way to answer question 2 of 3 (how far from Earth at separation). I know the question asks for "roughly" how far but I have seen many of OP's questions looking for far greater detail on other topics so I am ...


4

I think the premise is wrong: in physics "everything is similar" unless proven otherwise. It's one of the principles of physics. So you say that the orbit (shape? Or did you mean velocity, or something else) should't be "similar": but in what way should't it be similar? You should specify that, what did you expect? If your premise is: &...


4

There are two figures which matter here: one is the energy that ends up in the thing you are accelerating. This is just given by $$\frac{m_P \Delta v^2}{2}$$ Where $m_P$ is the mass of the thing, and $\Delta v$ is the change in velocity (I am tacitly assuming that this kinetic energy is relative to someone for whom the object is initially at rest). The ...


3

How much fuel would it take? And how large should be a booster / rocket ? That depends on the engine and propellant. Tsiolovksy rocket equation is $$\Delta V = \ln\frac{m_i}{m_f}I_{sp}g$$ where $\Delta V$ is change in velocity (in this case, 11000 $\frac{m}{s}$) $m_i$ is the initial mass before the burn (payload plus propellant plus dry mass for the ...


3

But the cause is air drag which is a force acting backwards, and because of F = ma that means an acceleration backwards at least initially. But for how long? An infinitesimal time? What makes it hard is the continuous action of the drag. Drag for the orbiting station is has the effect of a continuous rearward force acting on the station but not directly on ...


3

Totally by accident I've just happened to run across the 2014 open access paper A peculiar stable region around Pluto with the abstract below. The purpose of the paper was to see if stuff might already be in some long-lived orbits around the system, things that New Horizons might pass near and photograph, or perhaps even collide with. I don't know if New ...


3

The Help for Orbital Simulator gives the following for Launches: $\phi$ is the latitude of the launch site. $\lambda$ is the longitude of the launch site. Ω is the ascending node of the orbit. Those three items determine the inclination of the orbit.


2

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 ...


2

Sounds perfectly plausible. As the other answers have shown, you need roughly 200 m/s delta V to reliably and quickly deorbit such a droplet. For every gram of mass in a droplet, you need to deliver an energy of 20 Joules. An ordinary consumer-class electric jackhammer easily delivers 50-70 Joules in a single impact, so you could readily deorbit hundreds of ...


2

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}{...


2

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 ...


2

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: $$...


2

The acceleration due to the spherical harmonics is a vector in 3D space. How you choose to describe it (with an inertial or non inertial frame) doesn't change the vector itself. However when it comes to plugging in that acceleration to propagate the orbit (solve the differential equation), it matters which frame you describe it in because $F = \cfrac{dp}{dt}$...


2

Kinetic energy is $\frac {1}{2}mV^2$ so you're looking at $55$ GigaJoules of energy. That's enough to power ~86 U.S. homes for a year according to this U.S. EIA FAQ


2

A first-order method I used before is to exploit the total impulse capability of the launch vehicle. This is specified for nearly every variety of booster, whether space launcher or short-range ballistic missile. Knowing the velocity change required to establish a trajectory in orbit (nearly independent of payload mass), one divides this into the total ...


1

You're on the right track. Kepler's equation is an analytical solution for the position of an object in an orbit with respect to time assuming two body dynamics. And given some initial state vector (position and velocity) you can calculate all the orbital elements (like you said), from which you can get orbital period and mean motion. And if you know the ...


1

A skyhook is simply about providing a free boost to the spacecraft, it isn't a magical teleportation device. All the normal realities of interplanetary transit apply. Launch windows are simply the time when the energy requirements are at their lowest, the energy requirements go up substantially outside the optimal times and there's a period where there's ...


1

The way I've modeled that in the past is to using Python to generate a script with N spacecraft in it, each with their own initial states (chosen from the known distribution), and using a Propagate Synchronized statement. Each phase of the mission would have its own script: upper stage in one script, and use the final state of that propagation statement to ...


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