# Tag Info

293

Since the Apollo 11 code is on GitHub, I was able to find the code that looks like an implementation of sine and cosine functions: see here for the command module and here for the lunar lander (it looks like it is the same code). For convenience, here is a copy of the code: # Page 1102 BLOCK 02 # SINGLE PRECISION SINE AND COSINE ...

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Nonrelativistic solution The variables used will be $x$ for the distance travelled $v$ for velocity $a$ for acceleration ($1~\mathrm{g}$) $t$ for the time $c$ for the speed of light. Non braking Assuming the velocity you arrive at does not matter we take the equation $$x = \frac12 a t^2\ .$$ Solve for $t$: $$t = \sqrt{\frac{2x}{a}}\ .$$ (Let’s ...

35

You also asked for the logarithm, so let's do this as well. As opposed to the sine and cosine functions, this one is not implemented with a Taylor series-like approach only. The algorithm is based on shifting the input and counting the number of shifts needed to arrive at the required scale. I don't know the name of this algorithm, this answer on SO ...

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The authors of the paper are Harold A. Hamer, Katherine G. Johnson, and W. Thomas Blackshear. Of these, the name Katherine Johnson might ring a bell with people, as she was one of the protagonists in the 2016 movie Hidden Figures, honouring the women that were instrumental in the early days of the US space program. Katherine Johnson and her colleagues were ...

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The pendulum fallacy is the belief that rockets would be passively stable with engines at the top, with the rocket "hanging" from them. The error lies in expecting gravity to pull the body of the rocket down while the engines pull it up. In reality, gravity acts on the body of the rocket and the engines equally, exerting no torque (except for ...

27

Was standard Newtonian mechanics sufficient or were relativistic effects included? Relativistic effects didn't have to be modeled; other sources of error would have swamped the effects of relativity, and midcourse corrections were made. Were the Earth, Moon, and spacecraft modelled as point masses or more complicated bodies? The moon's gravity was ...

23

This is a large question, but we can certainly boil it down. You need several levels of requisite knowledge. I'll break it down as so: Relevance of Delta v for propellent budget Conversion between gravitational potential and its corresponding velocity The basic physics of Hohmann transfers Non-ideal factors going from surface to orbit Not all of these ...

23

In the inverted pendulum problem: gravity exerts a vertical force on the pendulum, at the center of gravity the support of the pendulum (like the finger under the pencil) exerts a vertical force on the pendulum, at the bottom of it In a rocket: gravity is the same engines exert a force along the long axis of the rocket, where the engine is (which doesn't ...

16

If you want to understand how the 'seconds' value fits the greater image, there's this rather contrived definition (which nobody uses because it's contrived and mostly useless but evocative enough.) 0) $I_{sp}$ in seconds is equal to the amount of time a rocket must be fired to use a quantity of propellant with weight (measured at one standard gravity) ...

14

This is not a complete answer. It is instead an extended comment to the following: I understand I will have to retard the gravity from each source by its particular light-time, as well as correct for the light time for the HST images. While you do want to correct for light time travel with regard to seeing a moving remote object, you definitely do not ...

14

The eccentricity is 1.0. The eccentricity $e$ of an orbit can be found from the radius of apoapse and periapse as: $$e=\frac{r_a-r_p}{r_a+r_p}$$ and the semimajor axis $a$ can as well, from: $$a=\frac{r_a+r_p}{2}$$ If you throw an object horizontally (velocity perpendicular to position vector) you will end up in a closed orbit if you throw at slower ...

13

First, we need an equation for the acceleration: $$F = \frac{F_0}{R^2}$$ One thing to note is that this equation is missing a factor (the units don't work out). Technically, $F_0$ needs to be multiplied by $1\ AU^2$, that way the units cancel out properly. Assuming an optimal angle, so: $$A = \frac{F_0 \cdot 1AU^2}{M\cdot R^2}$$ Where $M$ is the mass of our ...

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Full disclaimer: I'm the author and main developer of poliastro. The most important step before doing anything is somehow retrieving the positions and velocities of the planets of the Solar System. Astropy, one of the core dependencies of poliastro, ships medium-precision approximate models described in Simon et al "Numerical expressions for precession ...

13

If you're in a circular orbit, your velocity is $\sqrt{\mu\over r}$. Escape velocity at that distance is $\sqrt{2\mu\over r}$. So the impulsive $\Delta V$ to reach escape velocity starting from that orbit is the difference of those two: $$\Delta V_{ei}=\left(\sqrt 2-1\right)\sqrt{\mu\over r}$$ Now we escape by thrusting infinitesimal amounts in the orbit ...

11

Rockets produce thrust by ejecting reaction mass at some velocity. The fundamental quantities involved are mass flow rate and exhaust velocity, thrust is the consequence of these. It's no coincidence that specific impulse in units of velocity equals exhaust velocity, that's what specific impulse is. The exhaust velocity gives you the impulse per unit of ...

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Using Perifocal polar coordinates, where the x-axis points from the central body to the periapsis, and the polar equation for conic sections: $$r=\frac{a(1-e^2)}{1+e \cos(f)}$$ Provided parameters $\mu$ Standard Gravitational Parameter of the central body $r_p$ Periapsis Distance (Point P in the diagram) $v_p$ Speed at Periapsis $r$ Radial distance at ...

10

First off, you need to know where all the bodies in the solar system are as a function of time. I would start by getting the SPICE toolkit from JPL, supported in C, Fortran, IDL, and MATLAB. You will also need the planetary and lunar ephemerides and the latest leap-second kernel. (The planetary kernel linked is good from 1550 AD to 2650 AD, and the leap ...

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(This is adapted from my question/answer at Day-to-day tasks of human computers, ala Hidden Figures movie - History of Science and Mathematics Stack Exchange) I was also fascinated by the film Hidden Figures, and a related article from New Scientist magazine "Gifted and black: The brilliant woman who got the US into space". Compressible Flow In the ...

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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

There are some subtleties here. The fields where the concept of nadir are most important are nadir-pointing Earth observation satellites, satellites formation flying, and rendezvous and proximity operations. The latter two almost inevitably use some form of a local vertical / local horizontal frame (aka a Hill frame), in which vertical (i.e., nadir and ...

9

In simplest terms it is just the thrust produced divided by the propellant flow rate. "How much thrust am I getting for the propellant I am expending?" So bigger is better - you are getting more thrust for the same propellant flow rate. Bigger numbers = more efficient engine = extracting more thrust from the same amount of propellant. It is a key ...

9

Breaking that down: launch due East site in the Ecuadorean Andes sometime before local midnight on a July 4 when there's a new moon Launch due east. With the exception of launch sites that cannot launch due east lest a failed launch rain debris on some other country, this is the preferable direction. Launching to the east takes advantage ...

8

There are two ways. One is to numerically integrate from the current time to the future time. That is conceptually the easiest, as well as the simplest to implement (given a good integration routine). On the other hand, it is the most compute intensive, most prone to accumulated numerical error depending on how far you have to go, and provides no insight ...

8

If we assume a perfect two-body problem, absent perturbations from external bodies or non-spherical gravity sources (i.e., perfect conic orbits with no precession or variation), your constraints regarding inclination are actually unnecessary, as we may, without loss of generality, examine this problem in a perifocal reference frame. Perifocal coordinates ...

8

How to calculate terminal velocity in general: $$V_t = \sqrt\frac{2W}{\rho C_d A}$$ where $V_t$ = terminal velocity $W$ = weight (mass times local gravity) $C_d$ = the coefficient of drag of the object $\rho$ = atmospheric density $A$ = frontal area of the object Comparing Mars to Earth, weight is $\approx 0.38$ and atmospheric density is $\approx ... 8 The expression on the right is meant to give the eccentricity vector but the vector notation has been lost. Here it is in this answer: $$e = {v^2 r \over {\mu}} - {(r \cdot v ) v \over{\mu}} - {r\over{\left|r\right|}}$$ and the vector nature is not clear either. We should write it as$$\mathbf{e} = {v^2 \mathbf{r} \over {\mu}} - {(\mathbf{r} \cdot \... 8 The eccentricity of a radial orbit is$1$, regardless of its energy. This is a class of orbits where the type of orbit cannot be inferred from the eccentricity alone. With a "traditional" parabolic orbit of$e=1$, the angular momentum$L$has a well defined value, but the semi-major axis$ais not defined. In the case of a vertical bounded free-... 8 tl;dr: use a parametric equation. If the Earth were not rotating, then we would have something like \begin{align} x & = \cos \omega (t-t_0)\\ y & = \sin \omega (t-t_0) \ \cos i\\ z & = \sin \omega (t-t_0) \ \sin i\\ \end{align} where the radius of the orbit is 1,\omega$is$2 \pi/T$and$T$is the orbital period, and$i\$ is the inclination of ...

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