# Tag Info

## New answers tagged orbital-motion

6

The orbit of a satellite is determined by 6 parameters, the orbital elements: The eccentricity $e$ The semimajor axis ($a$) The inclination ($i$) The longitude of the ascending node ($\Omega$) The argument of periapsis ($\omega$) The mean anomaly at a specific time ($M_0$) The plane of reference is the equator, and the reference direction is the Vernal ...

5

Actually, both are quite possible. In the general case, for an arbitrary elliptical orbit, what you'll tend to find is that B is true (granted there will be some precession, but not usually in line with the Sun). However, it is possible to set up an orbit (such as a Sun-synchronous one) in which A is true. But orbits such as this require planning and precise ...

4

The velocity of the orbiting space junk is a vector, with both a radial and a tangential component. $$\vec{v}_f = \dot{r}_f\hat{r} + r_f\dot{\theta}_f\hat{\theta}$$ (my $r_f$ is your $r$) The equation for conservation of angular momentum involves only the tangential component of velocity, because it comes from the cross product of the radius vector and the ...

7

$\hat r, \hat \theta, \hat z$ form an orthogonal, right-handed triad of basis vectors in 3d. Taking the cross-product with them is no more exotic or unusual than doing so for Cartesian basis vectors. $\hat r \times \hat \theta = \hat z$, and so on.

0

For each body, you keep its velocity, position, and any other changing properties in variables. Whether you organize them as having all properties of each object in a struct (or object) associated with that body, as in Brandon Enright's comment, or keep an array of all x coordinates, an array of all y coordinates, etc. each indexed by body, doesn't matter. ...

2

No, the right ascension of the mean Sun is NOT zero at the vernal equinox. It is in fact nearly identical to the ecliptic longitude of the mean Sun (the difference is due to UT vs ephemeris time), and this is defined such that it coincides with the ecliptic longitude of the apparent Sun when the Earth is at perihelion. So that should be the starting time to ...

1

The trajectories of the two bodies each trace out separate ellipses. The barycenter is located at the left focus of one trajectory, and the right focus of the other. The orbit of the two bodies is the ellipse generated by the polar graph of the re distance between the bodies as a function of the azimuthal angle. This ellipse is larger than either of the ...

0

There are many different types of orbits according to General Relativity - more that that according to Newtonian mechanics. For example: http://arxiv.org/abs/1207.7041 Characterization of all possible orbits in the Schwarzschild metric revisited (skip right to the figures in the end of the article). Elliptical path is one particular case of these.

0

The Oberth effect takes advantage of when you add something to a quantity that is squared then the effect will be larger, the larger the quantity is initially. It works when a body is under the gravitational influence of another, usually larger e.g. a spacecraft around the Earth. In the central body approximation, the energy equation of a (keplerian) orbit ...

1

This link answers the question well: http://en.wikipedia.org/wiki/Bertrand%27s_theorem#Radial_harmonic_oscillator It sidesteps some of the difficulties you would otherwise have with this class of problem by looking at the potential, which is a metric of the square of the radius, which eliminates the square root, which can then be isolated with each ...

3

What you're looking for is Bertrand's theorem which states that for central force field in $1/r^p$, only $p=2$ (newtonian field) and $p=-1$ (harmonic field) give closed orbit whatever the initial conditions are (and assuming you are bounded to the field). For sure you can find some initial conditions that give closed orbits (for example, all $p\in[-1;2]$ ...

0

Look at it this way. An electron is a charged particle. A moving, electrically charged particle creates a magnetic field, and the particle itself already has an electric field. If the particle is accelerating, then you're going to have a ripple effect from the electric and magnetic field, or an electromagnetic ripple, ie an electromagnetic wave. So an ...

2

On a uniform circular orbit, even if the speed does not change in norm, it does change in direction so that the speed vector change over time and $\frac{d\vec{v}}{dt}\neq\vec{0}$. In fact, in polar coordinates, you have $$\vec{a} = \frac{d\vec{v}}{dt} = -\frac{v^2}{R}\,\vec{e_r}$$ Imagine a car taking a turn at constant speed: if the turn is left, you feel ...

5

In uniform circular motion: $a=\frac{v^2}{r}$

2

Let's generalize your idea and see if it can be more propellant-efficient, at least in principle. Call your two orbits $\mho_1$ and $\mho_2$. Both have semi-major axis $a_2$ and $a_2$ and inclinations $i_1$ and $i_2$ respectively. As per the problem, $$a_1<a_2$$ $$i_1=i_2$$ and any phasing issues may be ignored. Also, $$r_{p1} = r_{a1}$$ r_{p2} = ...

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