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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 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 ...

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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 $a$ is not defined. In the case of a vertical bounded free-...

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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 period, and $i$ is the inclination of the ...

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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 $$\mathbf{e} = {v^2 \mathbf{r} \over {\mu}} - {(\mathbf{r} \cdot \... 6 There are two major issues that I can see. Whatever you're using to calculate the arcsin is indeed giving you a value in degrees. The parameters you've provided are basically that of a satellite racing by an object ad a moderate distance, moving far above escape velocity all along its trajectory. It neither gets close enough, nor hangs around long enough ... 5 The equations for the position in a hyperbolic trajectory contain the hyperbolic sine, cosine and tangent. A hyperbola is defined by the equation:$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$It can be described by several parametric equations: Using the hyperbolic sine and cosine functions, (1), plot cyan:$$ \boxed{x = \pm a \cosh(t) \\y = b \sinh(t) \\ ...

5

In the case where the spacecraft can achieve relativistic speeds and no gravity is involved, $\Delta v$ is actually $c$ times the maximal change of rapidity which can be achieved. So the maximal speed the spacecraft can accelerate to, starting from zero, is $c \tanh \frac{\Delta v}{c}$. For example, if the maximal rapidity is $22$, then the maximal speed is $... 5 One method of calculating the angle involves using the ellipse reflection law. Light from one focus reflects off the ellipse into the other focus. Thus in the picture below (by the author), the radial vector from the focus$F_1$is reflected at$P$onto the second focus$F_2$, forming a triangle whose third side is the line between the foci. Your flight ... 4 The potential of fly-by ping pong is pretty much unlimited, provided you have enough time to your disposal. Given an initial transfer with a perihelion slightly lower than the orbit of Venus, a Venus flyby can increase the aphelion to a bit further out than the Earth's orbit. On the following Earth flyby, the perihelion can be lowered, and you can just ... 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 Let's go through this one thing at a time. First off, your value for$\delta$is clearly in degrees. With an eccentricity of$200$, the reciprocal$1/e$is so small it's nearly equal to it's own inverse sine -- in radians.$\delta$is basically$2/200$radians$=0.573°$. Next, realize that geometrically, the eccentricity of a hyperbola equals the center ... 4 There are really two questions here: Do there exist$n$-body systems with long-term stability? Can a third body (massive or not) be shown, a-priori, to be bounded or to escape—without resorting to numerical simulation? 1. Stability of$n$-body systems It is widely known that$n$-body systems are "chaotic" when$n>2$. However, this must be unpacked ... 4 If your ellipse is a circle, the Flight Path Angle is 0. You’re done. Otherwise, for an elliptical orbit, start with the polar equation that relates radial distance$r$, true anomaly$\theta$, semimajor axis$a$, and orbital eccentricity$e$: $$r=\frac{a(1-e^2)}{1+e\cos\theta}$$ Solving for$\theta$gives us the following:$$\theta = \arccos\left({\frac{-... 3 Does J22 represent the rest of Earth's quadrupole moment? With a more usual notation ldegree, morder, there are the zonal l2m0, tesseral l2m1 and sectoral l2m2 spherical harmonics, but you probably already know. What does J22 look like? What is its shape and symmetry? Here's what I get for l2m2: For comparison, here's what I get for your "acceleration ... 3 1) and 2) are easy to show, the bonus is very hard and I will not attempt it. A$L$iberation point can be seen as a balance between three accelerations in a rotating frame of reference. Gravity from$M_1$Gravity from$M_2$Centrifugal acceleration. For$L_2$, the first two are$-\frac{(1 - \delta)M_1}{(R + r_2)^2}$and$-\frac{M_2}{r_"^2}$respectively. ... 3 A numerical integration is the best option for me. Using the JPL’s DE/LE438 ephemerides and the NAIF’s SPICE library, I found the dates when the angle between the Earth-Sun and the Earth-Moon vectors is about 180, 0 and 90 degrees and the Moon/Earth distance is about the same: Date Angle Distance 2000-11-12 175.54 371281 2008-03-08 4.25 ... 3 This is a supplementary answer with calculations (for 1) two bodies and 2) three bodies) problem which confirms that @MatthewWells answer is correct. Would the Moon hit the Earth or not? Yes, it would hit the Earth. Do we have to do detailed numerical integration to find the answer, or can we use some simple equations that involve energy and/or ... 3 There are no limits on the direction. The Vis-viva equation will give you a speed. Assuming point-masses and sticking to classical mechanics, the Vis-Viva equation does not care at all about what direction you point your velocity in; It's merely an equation based on how the total Orbital Energy (which is the same for all orbits with the same semimajor ... 3 The article assumes that the Earth's shape is an oblate spheroid, i.e., a "squashed" sphere. If we "unsquash" everything back, including satellites' positions (which means multiplying the$z$-coordinates of the satellites by$1/\sqrt{1-\epsilon_e^2}$), the line segment connecting the new positions would intersect the sphere if and only if the line segment ... 3 Short answer: you stop nutation the way you stop any oscillation, by giving it an appropriate opposite impulse as it crosses zero, the oscillations central, zero energy point. Longer answer: The terminology in this area can be messy. So let's start with a simple model. The classic motions of a spinning top or (Navy) gyroscope are "spin", "precession", and "... 3 I will give my current best shot at this problem, and others should feel free to strengthen the argument with additional mathematics. (Or poke holes!) You ask two questions, I will answer the first as the second has been partially answered by the update. Are there other inclination change strategies that are more efficient for some values of$\alpha$? ... 3 note: This is a very helpful extended comment that may be of use to the OP but it can't currently be posted as a comment until this user reaches 50 reputation points. Oh this is a fantastic question. It is common to fall into the following trap when making these types of calculations. Check carefully your reference frames. Celta V numbers are all relative,... 3 How can Earth-Centered Inertial (ECI) coordinates be inertial if Earth's orbital motion is always accelerating? It is true that "Earth-Centered Inertial" is a bit of a misnomer. What this means is that one has to account for the fictitious acceleration that results from the acceleration of the frame of reference. Unlike the fictitious accelerations that ... 3 Earth's sphere of influence has a radius of about 924000km. A highly eccentric orbit with a perigee at 100km altitude and an apogee at the SOI radius has a semimajor axis of 465239km. Throwing that into the vis-viva equation$v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right)$where$G$is the gravitational constant,$M$the mass of earth,$r$the orbital ... 3 Brief answer: yes, it's possible. Here's a somewhat scrappy answer: this is pretty much a transcription of what I wrote down when working it out, so it's a bit messy: sorry. First of all I'll use what I think is the mathematicians' version of spherical polar coordinates (apparently physicists use the two angle names swapped). So starting from a right-... 3 The flight path angle is simply the angle between the velocity vector and the vector perpendicular to the position vector. An easy way to visualise this: If the orbit was a circle, this angle would be zero. The angle is therefore due to the contribution of the inward/outward motion of the object away from the focal point. The semi major axis ($a$) and ... 2 In his excellent answer Notovy used the vis viva equation to demonstrate same speeds implies r = a. A basic property points on an ellipse: sum of distance from one focus plus distance to the other focus is 2a. So if the radius vector has the same length as semi major axis a, that implies the end of the radius vector lies on the end of the ellipse's semi-... 2 All right, I'd approach this in the following manner. Part 1: True Anomaly$\theta$at$r$in terms of Elliptical Orbit Eccentricity$e$The first place I'd start is the Vis-Viva Equation, which, for all orbits around a particular body with a particular gravitational parameter$\mu$links relative speed$v$with radial distance$r$and semimajor-axis$a\$. ...

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Dimensional analysis is a zeroth-order way to approximate (thank you Mr. Ross; 9th grade Physics). There will likely be better answers, but let's see what happens. meters/sec^2 x seconds = meters/sec 9.8 m/sec^2 x 150 sec (MECO, where you're going mostly sideways) gives ~1500 m/s delta-v People usually give something like 0.9 to 1.5 km/s when forced to ...

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