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OK, I cannot give you the answer for the ISS, MEO and GEO off the top of my head. However, if you do insist for any specific objects, please post their catalogue numbers (five numbers at the beginning of every TLE line, i.e. columns 03-07) in comments. But here we go. First of all, I suggest having a look at how we represent multidimensional uncertainty as ...


11

It depends on a lot of factors, including: Time since epoch Altitude Atmospheric density variations Spacecraft maneuvers Accuracy, number and distribution of observations used to fit the TLE Fit span used for differential corrections Typical errors for a TLE for a non-maneuvering spacecraft at an altitude higher than 400km and with good observation data ...


10

Hill and Peterson "Mechanics and Thermodynamics of Propulsion", third printing, November 1970, page 385, has a diagram that agrees with your intuition. (sorry for poor scan quality) You are correct - as the surface area increases, so does the mass flow rate. There must be other factors in play in the graph from Wikipedia.


9

Underneath the diagrams in the Wikipedia article you link, there's a mention of the BATES grain geometry: Circular bore: if in BATES configuration, produces progressive-regressive thrust curve https://en.wikipedia.org/wiki/BATES In which there is combustion occurring on both ends of a cylindrical segment as well as the central bore. In such a case, ...


9

This question was asked on Physics.SE with the question Amateur moon laser ranging. The answer is that in order to do so, you'd have to have a really precise laser beam (1 mRad divergence), along with a very accurate time measurement system (Nanosecond scale), and need a reasonably large telescope (On the order of a meter or two seems to be large enough). It'...


8

The first formula gives you the altitude at a particular point in the orbit, assuming that the position vector is the satellite's current position relative to the center of the Earth. The second formula is the altitude of the periapsis (lowest point) of an elliptical orbit.


7

The current accuracy of the JPL Developmental Ephemerides released in September of 2013 is given by its authors in their article The Planetary and Lunar Ephemerides DE430 and DE431, quoted below: The present-day lunar orbit is known to submeter accuracy through fitting lunar laser ranging data with an updated lunar gravity field from the Gravity ...


7

If you have no other infomration about the orbit of your satellite (e.g. the orbit is circular), I believe you have to solve this problem with the Lambert's theorem assuming an elliptic transfer orbit (see Wikipedia). However, as far as I know there is no analytical solution and either numerical methods or series expansions need to be used. In this answer I ...


7

From my experience, the fundamental theorem of space curves isn't all that fundamental with regard to spaceflight. This is just a movie, and even the most historically accurate of movies get fundamental things wrong. That said, there are a couple of places where the Frenet-Serret frame (or the Serret-Frenet frame) might well have been useful in the early ...


7

An exercise that was left unsolved from last year's class gives me this equation : $$ t-t_{p} = \sqrt{\frac{a^3}{\mu}}*(\arcsin(X) - e*X) $$ where : $$ X = \frac{\sqrt{1-e^2}*\sin(v)}{1+e*\cos(v)}. $$ This is just Kepler's equation $M = E-e\sin E$, but written in terms of $X = \sin E$, where $E$ is the eccentric anomaly. We don't have a derivation of ...


7

Sorry it took so long to post an answer. I've spent the past day researching this question, reading over two dozen PDFs (at least 1000 pages of material) and reading the Apollo 13 flight journal (which sadly, is incomplete). It can't be emphasized enough just how important the guidance computers were to guidance and navigation. Every "normal" procedure ...


6

First we have to go back to the chemical equations, and this time, include the standard enthalpy of combustion. Hydrogen: 2 H$_2$ + O$_2$ → 2 H$_2$O + 572 kJ/mol Methane: CH$_4$ + 2 O$_2$ → CO$_2$ + 2 H$_2$O + 889 kJ/mol Dodecane: 2 C$_{12}$H$_{26}$ + 37 O$_2$ → 24 CO$_2$ + 26 H$_2$O + 15,026 kJ/mol Ethanol: C$_2$H$_5$OH + 3 O$_2$ → 2 CO$_2$ + 3 H$_2$O + ...


6

The example initial conditions do not result in an orbit. The energy is positive and the eccentricity is greater than one. It is a hyperbola. Also you over-specified the problem by claiming it is in perifocal coordinates, but then providing initial conditions that result in a periapsis that is not on the p-axis. The computed eccentricity is close, but ...


6

One thing that may be tripping you up is that the d term in the gravitational formula is the distance between the centers of mass of the objects, not the altitude above Earth's surface. The other thing to keep an eye on is your units. The big G gravitational constant is ~6.67 x 10-11 m3 kg-1 s-2; if you're using that value, make sure you're consistently ...


5

In his text Solar sailing: Technology, Dynamics and Mission Applications (see here for some of his open access titles), Colin R. McInnes derives the pressure on a perfect sail face on to Sol as $$P(r) = (L/(3 \pi c R^2)) * (1 -[1 - (R/r)^2]^{1.5})$$ $P$ is photon pressure $c$ is speed of light $r$ is distance from center of Sol $L$ is luminosity of Sol $R$ ...


5

Aside from numerical issues, "With Sun as centre" may be part of your problem. Get all the data from Horizons relative to the Solar System Barycenter, not the Sun, which moves relative to the barycenter. That barycenter is an inertial frame of reference, whereas the center of the Sun is not. Also make sure that you are putting in the initial position and ...


4

You can find a C++ method contained in the source code of Andrew Holme homemade GPS receiver project. The method is called GetXYZ and is the EPHEM (ephemeris) namespace and looks as the following: void EPHEM::GetXYZ(double *x, double *y, double *z, double t) { // Get satellite position at time t // Time from ephemeris reference epoch double t_k = ...


3

edit: @DavidHammen has just posted a much more thorough and insightful answer, which also points out some problems applying Newton's method to the current form. I'm pretty sure there are has never been an analytical expression discovered to solve $\nu(t-t_p)$, but solving using Newton's method applied to $$ \sqrt{\frac{a^3}{\mu}}*(\arcsin(X) - e*X) - (t-...


3

I agree with Hobbes about TM being a subset of data, but from my experience I would define it differently: Spacecraft data: just all data that is inside the spacecraft at some time, no matter whether it is stored, transmitted or a transitional state information. Telemetry: data that is transmitted on a monitoring & control link in the "monitoring" ...


3

Telemetry is a subset of data. Telemetry: data related to the status of the spacecraft itself. Fuel level, temperature, engine speed, position information etc. This data is needed for the operation of the satellite. In addition to telemetry, the spacecraft's payload generates data. This is data that is useful for people other than the operator of the ...


3

That would be the Jacobi integral ($C_\text{J}$, or $C_\text{H}$ in Hill's problem): In celestial mechanics, Jacobi's integral (also The Jacobi Integral or The Jacobi Constant; named after Carl Gustav Jacob Jacobi) is the only known conserved quantity for the circular restricted three-body problem. Unlike in the two-body problem, the energy and ...


3

The easiest way to get this information is to use a tool such as the Satellite Tool Kit (STK). Essentially, this information varies over time, especially for high inclination orbits. A good orbital simulator will allow you to enter mean altitude, inclination, and eccentricity (Or alternatively apogee/ perigee/inclination) If you already have the satellite ...


3

This might help to get you started. It's just a 1D radial solver, but you can play with the math. I included a 3D version of the derivative function to show one way to make the acceleration a vector in NumPy. Don't forget to add the rotation of the earth. The idea about starting on top of a mountain like Mt. Kilimenjaro is worth exploring for sure. I just ...


3

I just ran it, and mine look pretty much like those in the paper. See some coordinates at the bottom. Here are some {x,y} coordinates at the times in the left column: 0. {1.,3.} {-2.,-1.} {1.,-1.} 5. {2.46917,-1.22782} {-2.2782,-0.20545} {0.34106,0.901049} 10. {0.77848,0.141392} {-2.02509,0....


3

No, your assumptions are not correct. That's Vi, inertial velocity. NASA doesn't use the V1/"first orbital velocity" terminology, or at least didn't in Shuttle. If you zoom in on the image you can see the second character is an "i" not a "1". I don't know if Orion uses English or metric units, but the units are either kilofeet/sec (like Shuttle) or km/...


2

The usual proportionality law for solar sails is $$\frac{\cos^2\left(\alpha\right)}{r^2}$$ where $\alpha$ is the sails angle from zenith, and $r$ is the distance from the Sun. This assumes that the Sun a point source of light, and that the size of the sail is negligible compared to the Sun. This can simply be decomposed into $$\frac{\cos^3\left(\alpha\...


2

There is another parameter known as Tisserand's parameter which stays constant for a given body in 3 body probelm. This is more useful in identifying any given asteroid or comet. Given by $T = \dfrac{a_e}{a}+2\sqrt{\dfrac{a}{a_e}(1-e^2)}cos(i)$ where e subscript represents parameters of perturbing body and no subscript for small body. It stays constant ...


2

If the spacecraft is in orbit around the star, its velocity is below escape velocity. There is no radius above which this changes.


2

First of all the eccentricity is a constant, so it should not change with the true anomaly. You could use it as an temporary variable, bit it is not required to calculate $e$. If you use different definitions for the angular momentum you can find the eccentricity using only the position and velocity at one given time, $$ h=r v_{perp} = \sqrt{\mu a (1-e^2)}....


2

I can access it without problems. There's no requirement to register and log in, that means it's a public website.


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