New answers tagged orbital-mechanics
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The PFS- satellites were inserted into a lunar orbit in order to map, amongst other things, its gravitational field. So you would expect their orbit to be accurately known and monitored. The Apollo 16 mission report gives the following orbital parameters in Sect. 5-5
PFS-1 : Perilune 105 km, Apolune 144 km, Inclination -28 deg
PFS-2: Perilune 97 km, Apolune ...
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Uhoh gave a good summary on the subtle differences and losses, but missed a big one: the mass of the central body. The same state vector applied to a craft in orbit around a different body will result in vastly different TLE.
This is rarely of significance because what body is being orbited (and by extension, it's mass) is hardly ever an unknown parameter - ...
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You should be able to convert a TLE (element set) to a state vector (Cartesian XYZ and velocity) and back to TLE without loosing possitions information, provided you also provide bstar (drag coefficient).
Going forward is easy/standard, going back a little bit less common/might require running an optimiser if you can't find a tool.
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I'll take a partial stab at this. But first, I believe this rocket is not on a GTO trajectory (judging by the (relatively) high latitude view of the CONUS + Mexico, indicating inclined orbit). This is probably GPS III-05 (17 June 2021 16:09 UTC launch, image was tweeted ~4 hours later), TLE history for reference:
Not knowing the specs of the camera used, I ...
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A direct insertion into a polar orbit gives you only one option for an orbital plane: one that's initially pointed straight at the Sun. This is sub-optimal for an observation mission, since you're more or less directly over the sunrise/sunset line, and can only see deeply shadowed ground. It will be several months before the planet moves far enough in its ...
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If you have all the orbital parameters of the new orbit, and you also have the radial distance from the central body of the spacecraft after the maneuver, and the orbit isn't circular you have enough information to determine the new true anomaly.
Given:
Radial distance: $r$
Orbital Eccentricity : $e$
Semimajor axis: $a$
The polar equation of a Keplerian ...
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I think you're overthinking this. Won't it behave almost as if it were a single mass at it's center of mass? Set your periapsis to the distance between the center of mass of the spacecraft and set the rotational velocity to what your velocity will be at periapsis, you can cut the tether when it gently deposits you on the surface.
However, there are two ...
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Turns out the answer is just a plain old ellipse, the numerics were wrong.
Construction:
With a point mass $A$ and location $B$, one calculates the greatest apoapsis obtainable in the degenerate case, the point $D$.
For all orbits, the second focus, $A_2$, must lie on the circle $c$ with centre $B$ and radius $BD$.
For all locations of $A_2$, the extreme ...
answered Jul 12 at 21:29
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2
Partial result:
I've come to realise that cutting the tether at periapsis is not the optimal way of going about this.
Because, as long as the angle between the tether and the velocity vector of the payload is larger than 90 degrees, the payload would still be bleeding energy, and thus end up in a tighter capture orbit.
$$\angle\vec{A B}\vec{v_{A}} \geq \frac{...
answered Jul 12 at 19:41
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2
Not the reference you are looking for, but a response to this:
Any references and insights would be deeply appreciated!
I would like to note that low-thrust phasing in elliptical orbits is "boring", in the sense that the optimal strategy is conceptually simple.
For sufficiently low thrust, the phasing orbit does not have time to noticeably ...
answered Jul 12 at 17:55
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1
For a Moon mission, the difference is very small.
Even if one makes no attempt at picking launch sites other than the equator, or don't use any inclined orbit, the resulting lunar transfer orbit isn't all that bad:
A apogee velocity of 190 m/s tangential to lunar orbit, instead of parallel to lunar orbit increases lunar orbital injection (and escape) cost by ...
answered Jul 10 at 9:18
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4
I'm leaning towards the "just coincident" explanation here.
The semi-major must be >54.5 million km (and even that is too hot), as the aphelion must touch the orbit of Venus to enjoy the repeated gravity assists to reach this orbit, and the perihelion must be outside the sun. So ~57 million km is right were the trajectory constraints would ...
answered Jul 9 at 21:53
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3
When just throwing something together, I use GeoGebra.
2D and 3D diagrams can be constructed from mathematical primitives, both declaratively or through drawing.
There's a bit of everything in it. A computer algebra system, LaTeX, spreadsheets, scripting and so on. None of it works particularly well, but when you don't feel like making a custom graphic, you ...
answered Jul 9 at 21:13
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5
This is an Oberth maneuver, getting the most out of your delta-v budget by adding velocity on top of an already high velocity.
Nowhere in the solar system does one achieve greater orbital velocities than during perihelion of a Sun dive, as close as thermal management allows. At that ~100km/s velocity, every km/s of velocity added corresponds to ~14km/s at ...
answered Jul 9 at 8:04
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3
Let's say you want to do something "reasonable", like collecting second-by-second conjunctions for a hundred year period, for all the objects you can get your hands on state vectors for (a hundred thousand or so?)
You have an $O(S\cdot N^2)$ approach, so... about $\approx10^{20}$
Yes, I can see there's a problem.
For the solar system, things move ...
answered Jul 8 at 20:26
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1
Goal: lower Phobos's orbit.
Status: ACHIEVED!
By the time you read this line, Phobos' orbital altitude is already lower than when you started reading this question.
Tidal deceleration is dropping Phobos by about 2cm per year, and in less than 50 million years Phobos will impact Mars.
Well, actually it won't. In only about 20-25 million years, Phobos will ...
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First, one assumptions:
The acceleration is so low that instantaneous impulse solutions are out of the question, and the trajectory can be modelled as a very gentle spiral.
This is quite reasonable, as an absolutely enormous amount of thrust would be necessary to provide high acceleration to a $1.0659×10^{16} kg$ rock.
So let's get started then. First, we ...
answered Jul 7 at 11:56
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I was looking at something like this (asteroid mining) but I was more concerned with the size of the asteroid. I use the small body ferret database to get a smaller list of asteroid names. Then I looked at the orbits and more data on the JPL Small-Body Database Browser.
After looking at several asteroids orbital diagram I starting to wonder about their speed ...
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I don't have the time right now to post a comprehensive answer, so I'll do a provisional one with some resources that might help you out. At some point I will might edit this into a full answer.
Here's a couple of papers that you'll be interested to read:
https://dspace.mit.edu/handle/1721.1/73467
https://planet4589.org/jcm/pubs/sci/papers/2018/Taylor18.pdf
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