# Tag Info

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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 ...

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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 ...

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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 ...

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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 ...

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Your formulas are using a base-10 logarithm, not the natural logarithm. This causes all your delta-v values to be off by a factor of 2.3 In google sheets, you can supply the base of the logarithm as a second argument to the LOG function. For other systems, the identity $log_a(x) = \frac{log_b(x)}{log_b(a)}$ may be useful if you need to obtain a logarithm in ...

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Discrete Fourier techniques introduce errors in their terms that track with the sinc function. Any target in the image will produce side lobes like those in the following graph. Strong reflections will create side lobes that have a higher amplitude than the background of the image. This is why the target appears to have a larger spatial extent than it ...

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Which is nowhere near the speed the rocket has here near the stage separation. The video is reporting velocity in km/hour, not in m/s. 4383 m/s is 15780 km/hour -- over twice as fast as the Falcon 9 is moving at stage separation. In general, the rocket equation isn't going to give you directly usable results for the initial ascent phase of a rocket. Drag ...

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There are several ways to do this. The easiest and most straightforward is to break it into two sets by including velocity as a variable, and solve together. Instead of a single second order differential equation $$\ddot{\mathbf{r}} = -\frac{\mu}{r^3}\mathbf{r}$$ We can solve the following pair of first order differential equations in parallel $$\dot{\mathbf{... 7 As someone who computes simple orbits for a living, I would say your question has meaning only in the original Keplerian model of projectile motion: gravitational fields of central bodies are point symmetric, barycenters are at the geometric centers, and other bodies are not present. Then the symmetry arises from the straightforward, simple math used to ... 6 First, you appear to have the following misunderstanding of the solar sail force vectors: Tilt it at 45 degrees to make the thrust tangential Thrust is not tangential at 45 degrees. In fact, a solar sail always has thrust perpendicular to the sail, and can thus not achieve thrust perfectly tangential to the Sun, since the cross section would then be zero. ... 6 No, in general, they don't close. (Though, as you say, some geodesics do.) Consider, for example, an oblate ellipsoid of revolution, and take two points on its equator such that the angle \alpha between them is not a rational multiple of \pi. On a sphere, the only geodesic passing through tho points on the equator is the equator; but in our case, since ... 5 That parameter is known as the Effective Exhaust Velocity because... Sutton, 4th edition: When [the exit plane pressure is equal to the ambient pressure], the effective exhaust velocity is equal to the average actual exhaust velocity of the propellant gases...the effective exhaust velocity is usually close in value to the actual exhaust velocity. 5 As \Delta v is just change in velocity, we can just integrate the norm of the acceleration function over time:$$\Delta v = \int|\mathbf{a}(t)| dt$$You're out of luck getting a closed form of that integral though. As far as analytical solutions goes, we can note that at t = \frac{\pi}{2}, all of a_x, a_y and a_z are maxed out, and hence \Delta v &... 5 Are bicomplex numbers including tessarines ever used in spaceflight (as an alternative to quaternions)? Without the parenthetical remark, the answer is "I don't know". But with that parenthetical expression, the answer is all-caps "NO". Unit quaternions are used in space exploration precisely because they map nicely to rotations in three ... 5 I'm sorry, I should have included more explanatory links in what I wrote earlier. When reading what I put together below, please keep in mind that each of these paragraphs is normally an entire grad school course in mathematics. I have attempted to clarify my meaning, but if I haven't succeeded, please note that it took me several years to understand all ... 5 The answer has to be 'not necessarily', because, in general, as you go along, you're free to adjust the solar sail angle and thus the trajectory. In addition, the trajectory need not lie in a single plane since the sail can produce out of plane forces. I posted an analysis in the comments yesterday for a weak solar sail and a shallow spiral orbit and ... 4 Matrix manipulation is mainly for systems of linear equations, and this is a system of nonlinear equations. You need to either change variables to form a linear system, or use a tool designed for nonlinear systems. If the system is overdetermined (has more equations than unknowns), or there is error in the measured values you plug into the equations, there ... 4 Yes, they're all just forms of the same relation that are specific to different values of eccentricity. This is one of the few topics treated in greater detail in Richard Battin's An Introduction to the Mathematics and Methods of Astrodynamics (chapter 4, pages 141–173) than in Vallado, including more biographical anecdotes about the mathematicians who ... 4 There are several approaches, only differentiated by how big a budget you have and how accurate you want the answer - quite plausibly according to where you are in the programme / development proposal etc. First iteration: Just assume the thruster nozzle governs the mass flow rate and its completely expanded to ambient and that its discharge coefficient is 1.... 4 For coplanar orbits, a bi-elliptical transfer is more efficient than an Hohmann transfer when the ratio of the initial and final radii is greater than 15.58. When the ratio is less than 11.94, an Hohmann transfer is more efficient. (Thanks to notovny for correcting me.) A bielliptic transfer is effectively two subsequent Hohmann transfers. Section 6.3.2 of &... 4 Just to add to the existing established relativistic doppler shift is a continuous process. The equation is on the order of:$$1 + z=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$Which contains no quantizing terms. The notion of redshift quantization (very neat! Thanks for telling me about this!) is only based on a few observations, whereas continuous redshift has ... 4 This is largely dependent on the radial velocity of the orbit of Mars, since its eccentricity is pretty large. Radial velocity is maximised at a distance of:$$r = \frac{r_P(2a - r_P)}{a}Where r_P is perihelion distance and a is the semi-major axis. This part of the orbit is 156 days before perihelion. At this distance, the radial velocity of Mars ... 4 I think the premise is wrong: in physics "everything is similar" unless proven otherwise. It's one of the principles of physics. So you say that the orbit (shape? Or did you mean velocity, or something else) should't be "similar": but in what way should't it be similar? You should specify that, what did you expect? If your premise is: &... 3 The CSAR (coherent synthetic aperture radar) did not use very high frequencies and short wavelengths like 3 GHz (0.1 m) or 30 GHz (0.01 m) allowing small narrow beam directional antennas. Very low frequencies of 5 , 15 and 150 MHz and 60 , 20 and 2 m wavelength were used. These low frequencies were selected to image not only the lunar surface but also the ... 3 ...ensure the center of gravity is behind the point generating thrust Should read center of drag is behind the center of gravity relative to the direction of flight. In the air, rockets follow the same directional stability rules as their aircraft cousins, and indeed, arrows. Putting a crude stick on a rocket will increase its drag but enhance its ... 3 I believe I have worked out my own question using http://www.dept.aoe.vt.edu/~cdhall/courses/aoe4140/attde.pdf This uses the TRIAD algorithim in order to determine a rotation matrix between the body and inertial frames by knowing two vectors in both frames. Usefully, it also weights one as being more accurate than the other which in this case is likely to be ... 3 Provided that "enveloped" means the radially-varying density spheres have their mass distribution additively combined, and the rate of rotation is equal to that of a two body system, the setup is identical to the CR3BP, for regions outside the bodies. ...which doesn't help much since L_1, L_3, L_4 and L_5 are now inside m_1. But even L_2... 3 A diagram showing the cross section (longitudinal section) of the earth at the same longitude as the geostationary satellite is shown below. A location from the the geo stationary satellite aligns with the horizon is marked. Its latitude is also marked in the picture. At higher latitudes, the satellite is below the horizon. cos(\mathrm{lat}) = \frac{6400}...

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Partial answer only: I hope a more specific answer will also be posted! I see you've nicely updated your question. It's certainly still on-topic here but I think you can also abstract it just a bit and ask a related question in Math SE or SciComp SE as well. A math question might look like: Given two period functions $g_1, g_2$ of the independent variable \$...

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