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7

This is not a recurrence for a Legendre polynomial $P_{n}(\mu)$ but an Associated Legendre Function (ALF). This1 states much more than is needed here. ALFs are often denoted $P_{n,m}(\mu)$ in geophysics. The degree $n$ and order $m$ are non-negative integers $0, 1, 2, \dots$ with $0 \le m \le n.$ Argument $\mu$ is the sine of the latitude (measured north or ...


6

To calculate the new orbit, assuming an instantaneous burn you would: calculate the state vectors for the burn point add the velocity burn (assuming a prograde burn it'll be in the direction of the velocity state vector) calculate the orbital parameters from the state vectors If the burn isn't instantaneous, you'll have to do some calculus for the duration ...


5

The time in transfer will be a good approximation - replacing the long burns with impulsive. LEO altitude 160 km + 6,371km Earth average radius, so periapsis of the transfer orbit is 6531km from Earth center. Geostationary orbit radius: 42,164 km Semi-major axis is the average between the two. a=24347km $T=2 \pi \sqrt{a^3 \over GM}$ The period of that ...


5

The telescope was placed into a heliocentric orbit when its helium supply was depleted On 29 April 2013, ESA announced that Herschel's supply of liquid helium, used to cool the instruments and detectors on board, had been depleted, thus ending its mission. At the time of the announcement, Herschel was approximately 1.5 million km from Earth. Because ...


5

Good question, I’m also interested if someone has a more specific answer to share! On the following table, you can find the required delta-v per year for different orbits. And about formula, I'm not sure but maybe this page can give you an idea: https://en.wikipedia.org/wiki/Delta-v


5

Since you are specifically asking for a reference, I'll refer you to the Appendices in History of Space Shuttle Rendezvous, where these topics are covered in some detail. Appendix G describes the relative frame used in the diagram. Appendix H describes the burn nomenclature (the squares in the picture are burns, and this appendix decodes their names). For ...


4

The optimal trajectory for reaching the cheapest ("altitude zero" circular) orbit around a perfectly spherical, airless body involves turning immediately after lift-off so that the vertical component of thrust only offsets gravity drag (weight minus centrifugal force) providing no vertical acceleration, and all the remainder of thrust goes towards horizontal ...


4

There are multiple ways to approach this question. You could take the trajectory parameters from this postflight document. It doesn't have a full set of orbital parameters for all the cases you are looking for, but it is some actual tracking data which would have been uplinked to the AGC. Then, as you are familiar with Orbiter, you probably know about the ...


3

The Antares will have a long coast phase due to engineering related trade offs rather than orbital-mechanical advantages. However from an an orbital-mechanical view alone, there would always be a long coast phase. Ignoring the atmosphere and engineering limitations, you would apply all your "first stage" impulse as close to instantaneously as possible to ...


3

The escapement (the part that actually measures time) uses a spring and balance wheel so it will operate the same way no matter the presence or absence of linear acceleration in any direction. There is no pendulum or other form of eccentric mass involved in this part of the watch. However, the self-winder is driven by an eccentric mass. It depends on the ...


2

[ This is going to be a draft answer as I learn more and get more time to fill it out and edit with symbolic math -- right now its going to be super rough and this answer will evolve a lot over time ] Basic Simplified Problem The generic problem is to take a spacecraft from a set of initial conditions $\vec{r}(0), \vec{v}(0), m(0)$ to a set of terminal ...


2

I just want to add that the relativistic correction term mentioned in the answer by uhoh, which is the "post-Newtonian expansion" at the "1PN" level, approximate relativistic effects by introducing a repulsive $1/r^3$ term. The expression is used by the JPL so you have to use it if you want to get as close to there ephemeris as possible. Even though you get ...


2

Looks like they've used some figures that are very close to the sample calculations shown in the original paper with Katherine Johnson, which is available from NASA archives. The figure is for the distance of the vehicle from the center of the earth at the time of the retro rocket firing (or burnout) which initiates reentry. They've changed the original ...


2

(This answer continues the one above.) A table of $P_{n,m}(\mu)$ values can be envisioned as a square array. Rows down the page are indexed by degree $n$, and columns to the right by order $m$. Due to the underlying math, the only sensible $P_{n,m}(\mu)$ values occur for $0 \le m \le n$. The effect is that the table is lower-triangular instead of square; $...


2

I'm afraid that your whole approach is wrong due to starting with contradictory assumptions. If the two bodies are connected by a rigid tether, then in general, they will not move in circular orbits, since in addition to Earth's gravity, forces from the tether will act on them. (As uhoh pointed out, there are some configuration where the bodies do move in ...


2

A gravity turn is the most optimal path to orbit, assuming no atmosphere or staging. On Earth, launch vehicles use a modified flight path accounting for those differences, generally resulting in a more vertical trajectory than would be the case otherwise


2

The most common method, by far, is brute force, by propagating the relevant satellite and then calculating the visibility. There are however many ways to optimize this brute force: Large delta-t until "close" and then smaller delta-t after that Using Multi-threading to reduce processing time Few github examples: https://github.com/shupp/Predict https://...


1

--Formulas-- radius = sqrt(x^2 + y^2 + z^2) latitude = cotan(z/(sqrt(x^2+y^2))) Actually, the link you have shows $$ \Phi'(t) = \tan^{-1} \frac{z(t)}{\sqrt{x^2(t) + y^2(t)}}$$ That's the inverse tangent function, not the cotangent. For the figures you show, the inverse tangent would be $-0.25598\, \text{rad}$ or $-14.666\,\text{deg}$ That said, there's ...


1

Despite watching a few times it looks like I linked the use of the term "capture" at this point in the video with the use of the term "capture" at 08:20 in the video. But that's about the mechanical capture of Snoopy during a docking maneuver and this is about the gravitational capture of the combined pair in the Earth's gravitational field. What Manley is ...


1

'Fundamentals of astrodynamics' discourages using patch conics... While I would also discourage you from relying on patched conics I would certainly encourage you to try using patched conics as a very useful, educational exercise! Patched conics is a methodology based on Keplerian orbits. Remember that in this case the Earth and Moon move in elliptical ...


1

The video portion captioned "Propulsive Passivation" appears to answer the question by 2 noteworthy observations. First, the exhaust nozzle is venting propellant to space - the "Passivation" part. This is confirmed by the absence of any flames at the rocket's exhaust. Second, the propellant is expelled under pressure (as opposed to "free" venting) - which ...


1

This I believe is in reference to John Glenn's flight. It seems that this is some real numbers, but simplified in a way that doesn't let everything be seen. The speed is almost orbital. I believe what is actually being calculated is the location where a very late abort would put John Glenn landing, although it is somewhat difficult to tell from the limited ...


1

Yes it's possible, and known ... at Pluto. What are officially called the (four known) smaller moons of Pluto are actually orbiting the center of mass formed by Pluto and its large moon Charon, which is outside both Pluto and Charon. Wikipedia includes an animation of all the known moons orbiting Pluto. The small moons are all much farther away from the ...


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