# Tag Info

18

Given the mass costs in terms of consumables and the risk and support costs of keeping humans in space for longer, it seems unlikely that the multiple Earth-Venus flybys used by a lot of robot probes to get out to Jupiter or in to Mercury will ever be a sensible choice for humans. A Jupiter flyby on the way to Saturn is probably a no-brainer apart from maybe ...

8

If you define perfection as absolutely zero eccentricity then perfection is impossible. There will always be eccentricity in an orbit, even if it is very small. Orbits vary due to: Spacecraft system inaccuracy: no spacecraft is perfect, no matter how accurate Changes in the density of the orbited planet Gravitational influence of other celestial bodies: my ...

7

You don't have enough information. You need a minimum of one more value to determine the true anomaly such as the time of perihelion passage. In general six values and a timestamp are needed to fully specify a Keplerian orbit. The reason six values and a timestamp are needed in the general case is that the two body problem is a second order differential ...

7

As a new user I cannot comment or tweak the original answer, so I'll try my own. Happy update per any recommendations that come up. @uhoh's answer is close, but a few things to note. The TLE tells us the number of orbits per day is 15.50995519 (line 2 columns 53–63) $$\frac{24 \frac{hours}{day} * 60 \frac{mins}{hour} }{15.50995519 \frac{orbits}{day}} = ... 4 You would need a perfect two body system, a perfect central body and a perfect spacecraft. No other planets or stars influencing the orbit. The central body should be a sphere with constant density or at least spherical symmetry. You need a perfect measurement of orbital parameters. The thrusters of the spacecraft needed for circularation of the orbit ... 4 A few things come to mind: 1) where you put in the radius 8.76×10^{10} I think you meant the eccentricity, which for Earth's orbit is about 0.0167. Putting in a distance parameter there makes the equation dimensionally inconsistent. 2) Do not forget to convert your angles to radians. 3) To deal with the sign: Measure the angle from the initial ... 4 The Registration Convention provides a list of objects in earth orbit. You'd need to use this list to make sure that you aren't going to collide with anything currently in orbit. A Collision On Launch Assessment (COLA) must also be performed prior to launch to make sure you don't hit anything on the way up. From there, NASA's guidelines on cubesats, ... 4 This answer doesn't talk about how to do it Python at all: rather how to deal with the rotation. I think once you can do that then turning the maths into Python is simple. Initially I'll assume that you are computing positions in terms of three orthogonal axes, and the positions look like (x, y, z), and you're just projecting these down onto the (x, y) ... 4 Start with answers to How can I plot a satellite's orbit in 3D from a TLE using Python and Skyfield? Plotting in 3D makes my head hurt too, but for some reason I like it when my head hurts. If you like you can paste python into your question; blocks of text that are indented by 4 spaces appear as a "code block". You can have a look at the 3d plotting in ... 3 Your user need to supply you with information about mean anomaly at fixed time (epoch) M(t_0), and then mean anomaly at the moment t_1. Or about known mean motion n instead of current mean anomaly M(t_1). You can also get the mean motion as square root of gravitational parameter \mu divided by cube of semi-major axis, and also as 2\pi divided by ... 3 Mission design is very hard because it has so many different constraints. Finding the path to solar escape with minimum delta v within a given calendar window is well defined and will probably have the solution you suggest. Then someone will say that is too long for the astronauts to be on the whole mission and suggest the last pass rendezvous that you ... 3 Is it possible to represent an orbit with just a periapsis position and 2 angular velocities? This works, except for singularities at the poles. In general, an epoch time and six scalar values that completely represent the state at the epoch time are needed. Only five scalar values are needed if the epoch time is known to be the periapsis time. For example, ... 3 I will use a series of images of the Earth taken by the Apollo 11 crew during the flight to the Moon. I use the formulas developed by tfb to determine the percentage of the visible Earth. Image AS11-36-5325 from a distance of 18,700 km showing 37.3 % using a 80 mm lens Image AS11-36-5330 from a distance of 87,000 km showing 46.6 % using a 80 mm lens Image ... 3 This is not an answer as such, but an attempt to provide the mathematical apparatus you need to think about this (and it was way too long for a comment). There are two sensible things that might be meant by 'seeing the entire Earth': one is that the Earth would be visible in the field of view of a human eye; the other is that you can see a given amount of ... 3 Seeing that you already have the orbital period, you can use the mean anomaly directly to calculate the time parameter. This should work if you set time_of_periapsis to 0, as you have all the other required parameters. (GM is 1 since your python program assumes a unit mass system) Combine everything to a wrapper function, containing nothing but parameter ... 3 The answer depends: how accurate is the physics you consider? Considering the symmetric system postulated in other answers/comments, one with two spherically-symmetric bodies of equal masses, orbiting their barycenter: if you use only Newtonian physics, empty the universe of all masses other than the ones you're focusing on, and make the bodies perfectly ... 3 Given the radial distance r, velocity v, flight path angle \gamma in radians, gravitational parameter \mu, and specific orbital energy \epsilon, the specific relative angular momentum is$$h= \|{\overrightarrow{r} \times \overrightarrow{v}}\| = rv\sin\left(\frac{\pi}{2}+ \gamma\right) =rv\cos\left( \gamma\right) Then the orbital eccentricity ...

3

I looked more carefully at my source, which is plots using STK, and here is a recent closely zoomed in plot. Note that for some reason it uses a different perigee and apogee on a regular basis, which seems a bit odd, to say the least... I'm using the classical orbital parameters, mean of epoch. I switched this from the "Apogee Altitude" to the "Apogee ...

3

They are some weird processing artifacts. Here is the correct plot for KMS-4 (ID 41332) obtained from the TLEs downloaded from www.space-track.org and processed with the CSpOC library (downloadable from the same link): the shape is incredibly smooth and there are no evident secondary perturbations (surely there are several small components of perturbations,...

2

Is it possible to update an existing TLE using new data? Of course it is. How do you think TLEs are updated? That said, it's not easy. You'll need to Compute the Jacobian of your new data with respect to the elements of a TLE you might want to modify. Come up with a weighting matrix that indicates which of those new data are better than others. For ...

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I'm not an expert and this is not an expert answer, but these points may be helpful. Don't even think about using SGP4, per my comments below this question and Wikipedia it's a circa 1980's clever approximation to get approximate state vectors within a few weeks of any given TLE's epoch. It's an approximator based on evolution of orbital elements, not a ...

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Crew Dragon with Falcon Heavy can theoretically be sent to a lunar free return trajectory, so pretty much any apogee could be done. As for just a Falcon 9, it's hard to know for certain. Putting in a desired orbit of 2000 km with the NASA Performance Vehicle Calculator, I got the following plots. Note that 9000 km is the launch mass of Crew Dragon. The ...

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Partial answer (to What is the maximum apogee a Space Shuttle (and which one of the five space-flying Shuttles?) ever reached ) The maximum altitude reached by a shuttle orbiter on an unclassified1 mission was on STS-82 (Discovery) after the 3rd reboost of Hubble. This resulted in a 335.1 X 321.0 NM orbit. (620.6 x 594 km) 1 Published orbits for the ...

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This paper by Rene Schwarz could be useful in this regard - it covers the direct conversion from Keplerian Orbital Elements (a, e, i, etc.) to Cartesian State Vectors (R and Rdot, which is actually X/Y/Z and X./Y./Z.) It can be found here. The steps are enumerated there as follows, but the equations are substantial and transcribing each here would run a high ...

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I've started a Community Wiki answer so that we can consolidate all the best answers as links here, accompanied by short explanations. I've chosen this question to do so because the question is simple and short and so it requires the most general answer. From @OrganicMarble's comments: Does this answer your question? Converting Orbital Elements to ...

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I can answer the question of why Semimajor Axis and Eccentricity aren't included in the calculation, at least. What you have here are the main Keplerian Orbital Elements that define the plane of your orbit around Earth. Every orbit with the same inclination and Longitude of the ascending node lies in exactly the same plane, which intersects the globe of ...

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