Podcast #128: We chat with Kent C Dodds about why he loves React and discuss what life was like in the dark days before Git. Listen now.

# Tag Info

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Gravitational acceleration is not just one number for a given body Gravitational acceleration for any body is a function of the body's mass and the distance from the body's center of mass at which you are measuring it. It is proportional to mass and inversely proportional to the square of the distance; double the distance and acceleration divides by 4. The ...

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Once you're already in orbit you can increase your velocity and semi major axis even if your acceleration is a small fraction of the local gravity field. But long, gradual burns give you a spiral trajectory rather than elliptical transfer orbits. See Adler's answer to my question General guidelines for modeling a low thrust ion spiral? As for the sun's ...

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Raising an orbit with a weak form of propulsion Yes the Sun's gravity is stronger than the Earth's on each body's surface, but it drops like $1/r^2$. See the math below. A spacecraft in a heliocentric orbit around the Sun will just continue to orbit the Sun without any propulsion for millions or possibly billions of years because the spacecraft is ...

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With an elliptical orbit we can't have any fixed Lagrange points, not even the unstable ones aligned with the massive bodies, because the massive bodues are not fixed rekative to each other unless we make a very contrived reference frame. We can, however, conceive of trojan-like objects with the following properties: *The mean orbit period about a primary ...

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They do of course in the case of the circular orbit, but that is rather trivial. What about elliptical orbits with considerable eccentricity? Lagrange points aren't really defined for elliptical orbits. They are defined only in the Circular Restricted Three Body Problem (CRTBP or CR3BP). Two bodies have significant masses and the third does not (that's the ...

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According to Wikipedia, the general equation for inclination change is: $$\Delta{v_i}= {2\sin(\frac{\Delta{i}}{2})\sqrt{1-e^2}\cos(\omega+f)na \over {(1+e\cos(f))}}$$ Where: $e\,$ is the orbital eccentricity $\omega\,$ is the argument of periapsis $f\,$ is the true anomaly $n\,$ is the mean motion $a\,$ is the semi-major axis For circular orbits, this ...

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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-...

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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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In addition to Russell Borogove's good answer there is another factor here to keep in mind: When you are dealing with planets you can't just add together separate burns. If you add up the energy needed to escape Earth, the energy needed to go from Earth's orbit to Mars' orbit and the energy needed to enter Mars' orbit you will get more than 16 km/sec. In ...

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16 km/s is about right for Earth surface to low Mars orbit, summing up a few entries in this table. I am not sure as how he did it though, and was unsure if he used the escape velocity of Earth, the Hohmann Transfer, and also if that was only that velocity computed needed to be obtained once At a minimum, this would be split into ascent to LEO (~9.4 km/...

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After reading all your answers I'd like to summarise the situation. The black circles are the circular orbits and the red ellipse is the transfer orbit. Consider a spacecraft in the elliptical orbit. At the point P the velocity is greater than the circular orbital velocity, and that's why the distance from the centre increases. And at the point A the ...

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Imagine your spacecraft is on elliptical orbit around space body. The body is inside ellipse (in the focus, if more specifically). When spacecraft is on the vertex of ellipse and speed up in direction of movement, another vertex of trajectory is moving far from body. Vice versa, if the spacecraft is braking (by turning engines forward and burning) in the one ...

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To simplify the explanation and terminology, let's consider the case of a spacecraft orbiting Earth. All orbits are elliptical, with the center of mass of the system (Earth, in our example) at one focus. Circular orbits are a special case where the ellipse has no eccentricity and the focii are coincident. The orbit of our spacecraft has a perigee (closest ...

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To answer your title question: By using its engines. However you seems to be quite puzzled by the fact that velocity of an object can decrease and increase over the course of an orbit. If the orbit is perfectly circular, the speed will always remain the same (until thrusters are used). However, as is the case with Chandrayaan-2, most orbits are ...

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