Talk:Radial trajectory

In the last section "Orbit inside a radial shaft", it says that gravity is proportional to the distance from the center, and therefore supposedly results in simple harmonic motion like a spring. I am not an expert, but this seems obviously untrue. The spring force is indeed F=kx, but g ~ 1/x^2 . This means that the spring force is increasing with distance and the gravity force is decreasing. Moreover, the later effect is actually complicated by the fact that much of the mass of the planet will be "behind" the point mass as it approaches the center, making the force approach zero. The motion is going to be periodic, though, due to conservation of energy. -Evan — Preceding unsigned comment added by 69.123.96.13 (talk) 04:43, 28 September 2011 (UTC)[reply]

Some scaling seems wrong: $x(t)=\left({\frac {9}{2}}\mu t^{2}\right)^{\frac {1}{3}}$ , but the separation at time t the bodies would have if they were on a parabolic trajectory is also given as

p=\left({\frac {3}{2}}t\right)^{2/3}

Patrick (talk) 22:21, 19 January 2010 (UTC)[reply]

Thanks for pointing that out, it's all fixed now.

Apparently we both had been up all night editing the same file. I had to merge your last two edits with my mine, I hope I didn't screw up anything you were working on.

Be advised, the definitions given in escape orbit, and capture orbit are profoundly incorrect. The distinction made between an "escape" and a "capture" orbit is misleading at best.

For all open orbits, if the objects are moving closer when the time is negative, they will be moving apart when the time becomes positive, and vice versa. Flipping the sign of the velocity is a coordinate transform equivalent to flipping the sign of the time, it does not change the essence of the orbit type or or the mathematical analysis.

-- Norbeck (talk) 18:08, 20 January 2010 (UTC)[reply]

I restored some of my changes, it is not clear which of your reversals in [1] were accidental and which were deliberate.

Turning around the other body is not a radial trajectory, and bouncing back would be by a repellent force, not gravity.

For every trajectory there is also a reverse trajectory due to time-symmetry, but since the periapsis cannot be passed these are different cases (except for the elliptic trajectory), and you cannot say "time since" for the case where it should be "time until".--Patrick (talk) 06:36, 21 January 2010 (UTC)[reply]

so... the above formula doesn't seem compatible with the general formula here: http://en.wikipedia.org/wiki/Free_fall#Inverse-square_law_gravitational_field what's going on? 207.68.247.230 (talk) 21:53, 2 May 2010 (UTC)[reply]

The one above is for going to or from infinity with zero total energy, the other is for falling from a point.--Patrick (talk) 22:36, 2 May 2010 (UTC)[reply]

Trajectory vs Orbit

The topic is "radial trajectory", but we are really only talking about idealized kepler orbits. The first sentence of the article defines a "radial trajectory" as a kepler orbit, but this is not possible since "orbit" and "trajectory" are not synonymous. A trajectory describes a single body, an orbit describes two.

Since Hyperbolic trajectory, Parabolic trajectory, Radial trajectory exclusively describe orbits, I suggest they be renamed to "- orbit", and that "- trajectory" be redirected to "- orbit".

--Norbeck (talk) 18:08, 21 January 2010 (UTC)[reply]

It seems that trajectory is preferred if it is not periodic, and orbit if it is periodic. I am not sure about "A trajectory describes a single body, an orbit describes two.": A trajectory is relative to something else. Also, with two bodies there are two orbits/trajectories, plus that of one body with respect to the other, so a pair of bodies is not associated with one particular orbit.--Patrick (talk) 23:36, 21 January 2010 (UTC)[reply]

orbit

1 : a path described by one body in its revolution about another (as by the earth about the sun or by an electron about an atomic nucleus)

tra·jec·to·ry

1 : the curve that a body (as a planet or comet in its orbit or a rocket) describes in space

Just as there is a difference between a tensor and it's representation in an arbitrary reference frame, a trajectory is not the same as it's representation. Trajectories exist independent of reference frames. For example, the trajectory of a falling object as seen from a horizontally moving frame is a cycloid. In the frame of the center of mass it is a line. In both cases "trajectory" refers to the representation of the trajectory, not the trajectory itself. Similarly for orbits, two bodies have only one orbit, but there are an infinite number of representations of that orbit. There are three reference frames where these orbits are particularly easy to describe, the center of mass of the primary, the center of mass of the secondary, and the center of mass of the system. --Norbeck (talk) 23:11, 22 January 2010 (UTC)[reply]

Considering the meaning of the term revolution (see also Turn (geometry)) your definition of orbit would support the use of orbit if it is periodic.--Patrick (talk) 07:20, 23 January 2010 (UTC)[reply]

The statement: "trajectory is preferred if it is not periodic, and orbit if it is periodic", is unfounded, but rather than continue debating, let's mix it up a bit and try some research. Google "orbit trajectory", then click on the VERY FIRST LINK, then read the 2nd sentence to find out what NASA has to say: "Orbit is commonly used in connection with natural bodies (planets, moons, etc.) and is often associated with paths that are more or less indefinitely extended or of a repetitive character". (Lol, you do realize that you are debating the meaning of the word "orbit" with the guy who solved the 400 year old "Kepler problem for radial orbits"?)

Orbits are not trajectories. Every orbit is comprised of two trajectories and an interaction. This and the other two articles are about Kepler orbits and nothing else. They are misnamed and should be corrected to be consistent with elliptic orbit. --Norbeck (talk) 04:26, 24 January 2010 (UTC)[reply]

"Orbits are not trajectories" is not confirmed by the linked webpage. It says they are the same, but the choice of term varies more or less with the application.--Patrick (talk) 08:40, 24 January 2010 (UTC)[reply]

Kepler orbits have orbital energy and angular momentum, a path in space does not. A path in space does not have orbital energy so it cannot be classified as parabolic/hyperbolic/radial. The earth-sun barycenter is 466 km from the sun, the semi-major axis of the Earth's orbit is 466 km larger than the semi-major axis of it's trajectory. Another example: Connect two masses with a 1 meter string and let it spin freely. The diameter of the trajectories (there are two, one for each mass) is 1/2 meter. The radius of the orbit is one meter. Given the sum of the masses and the semi-major axis of the orbit, Newton's modification to Kepler's third law allows you to calulate the orbital period. If you use the radius of the trajectory you will the wrong answer. In a circular orbit the centripetal acceleration of a body is w^2*r, r = radius of the trajectory, the gravitational acceleration the same body is G*(M)/a^2, a = radius of the orbit. When the masses are equal a=2*r.

These are definitions I was taught in college. Trajectory: the path of body through space. Orbit: the relative motion of a body with respect to another body or system of bodies under mutual interaction. In most cases the distinction is not relevant and the two terms can be used interchangeably, but they are not equivalent. When refering to relative motion use "orbit", when refering to motion with respect to an implied reference frame use "trajectory". There are uses which violate this definition, the paths of particles in sourceless fields are called orbits because they have energy and (possibly) angular momentum. Perhaps energy is the defining characteristic of an orbit. —Preceding unsigned comment added by Norbeck (talk • contribs) 15:10, 26 January 2010 (UTC)[reply]

After all I do not mind changing the names of the three pages, and abandoning the distinction of terms based on periodicity, since we already use "Kepler orbit" as general term. I still wonder how "orbit" should be defined, e.g. as system of orbiting bodies (in that case an orbit has energy), or just as relative position as function of time. --Patrick (talk) 23:05, 27 January 2010 (UTC)[reply]

Periodicity of radial elliptic orbits

What happens at t=0? an think of four possibilities:

Inelastic collision, the particles slow down or stick together. The orbit is nonperiodic.
Perfectly elastic collision, the particles recoil. The orbit is periodic.
Particles "wrap around" each other, this indistinguishable from a perfectly elastic collision. The orbit is periodic.
Particles pass through each other. The orbit is periodic.

Kepler orbits can exist only in closed systems. Since the orbital constants must stay constant the orbit cannot decay or spontaneously change. This rules out inelastic collisions. The remaining possibilites are all periodic.

I can't find the reference but there is a source supporting #3 "wrap around". To rule out the other possibilities we must define radial orbits as having "insignificant but nonzero" angular momentum (L). This is equivalent to defining radial orbits as standard orbits in the limit of L->0, which is equivalent to claiming radial elliptic orbits are degenerate elliptic orbits.

Since no real orbit is perfectly straight the rigorous definition of a radial orbit is: "a radial trajectory is a Kepler orbit with insignificant but nonzero L". I felt this was too technical. On the other hand, the L=0 definition is strictly incorrect and potentially very misleading. -- Norbeck (talk) 23:11, 22 January 2010 (UTC)[reply]

The case L=0 should be clearly distinguished from the case of a small nonzero L. Considering the limit of some quantity as L tends to 0 can be interesting, but that should be done explicitly, not by mixing up the two cases.

For finite densities, hence masses with non-zero size, the trajectory/orbit is non-periodic in the case L=0 as well as in the case that L is small enough.--Patrick (talk) 07:48, 23 January 2010 (UTC)[reply]