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In 1845 W. R. Hamilton demonstrated [1] by the use of the hodograph representation that the velocity of any Keplerian orbiter is the simple addition of two uniform velocities, one of rotation plus one of translation. Consequently, if we believe the laws of geometry, the derived acceleration from Hamilton's velocity must be centripetal, but not attractive like Newton pretends. Many authors[2-9] have confirmed since this geometric property of the Keplerian motion revealed by Hamilton.

Because a centripetal acceleration causes a rotation, while an attractive acceleration causes a translation, there is a conflict between Hamilton's Keplerian velocity and the (non demonstrable) postulate of Newton.

Therefore, who should I believe, Hamilton's kinematics which is fully demonstrated, or the non-demonstrable postulate of Newton?

Refs:

[1] W. R. Hamilton, The hodograph, or a new method of expressing in symbolic language the Newtonian law of attraction, Proc. R. Ir. Acad. III , 344353 (1845).

[2] David Derbes, Reinventing the wheel: Hodographic solutions to the Kepler problems. American Journal of Physics 69, 481 (2001).

[3] Orbit information derived from its hodograph, J. B. Eades, Tech. Rep. TM X-63301, NASA (1968)

[4] W. R. Hamilton, The hodograph, or a new method of expressing in symbolic language the Newtonian law of attraction, Proc. R. Ir. Acad. III , 344353 (1845). orbits, Am. J. Phys. 43 , 579-589 (1975).

[5] H. Abelson, A. diSessa and L. Rudolph, Velocity space and the geometry of planetary orbits, Am. J. Phys. 43 , 579-589 (1975).

[6] A. Gonzalez-Villanueva, H. N. Nunez-Yepez, and A. L. Salas-Brito, In veolcity space the Kepler orbits are circular, Eur. J. Phys. 17 , 168-171 (1996).

[7] T. A. Apostolatos, Hodograph: A useful geometrical tool for solving some difficult problems in dynamics, Am. J. Phys. 71 , 261-266 (2003).

[8] E. I. Butikov, The velocity hodograph for an arbitrary keplerian motion, Eur. J. Phys. 21 (2000) 1-10

[9] R. H. Battin, An Introduction to the Mathematics and Methods of Astrodynamics, Revised Edition, American Institute of Aeronautics and Astronautics, Inc., Reston, 1999.

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  • $\begingroup$ Centripetal force is a force directed from a body in circular motion to the centre of rotation. If there is "something" that causes that force to exist and that "something" is in the centre of rotation, the force can be said to be attractive. But this regards forces, an acceleration itself is not "attractive". Also, since you are referring to "Newton' postulate" in celestial mechanics, I assume you're talking about gravitational attraction. Please, correct me If I misunderstood. If $F_G=G\frac{m_1m_2}{r^2}$ is "Newton's postulate" then I don't understand why you call it "non-demonstrable". $\endgroup$
    – Alessandro Bertoli
    Commented Feb 11 at 10:55
  • $\begingroup$ @Alessandro Bertoli Newton pretends to the "universal attraction", the Earth attracts the apple, and the apple attracts the Earth. If we see the apple falling to the Earth, no one can prove by the measure that the apple is attracting the Earth. Furthermore, no one can prove by the measure that the apple falls on a straight line, rather than on an ellipse of major axis equal to the Earth radius and minor one of the size of an atom. This is however crucial as a straight line is not a special case nor a limit of the conic equation $p = (1+e \; cos \theta) r$, imposed by the first law of Kepler. $\endgroup$
    – Hervé Le Cornec
    Commented Feb 11 at 14:21
  • $\begingroup$ The mutuality of attraction is given by the so-called Third Newton Law (action-reaction) this law is strongly tied to the concept of conservation of momentum and it is mathematically rooted in Noether's Theorem. Furthermore, Kepler laws are a description of a phenomena based on some hypotheses one of them is having point like particles. The earth is clearly not point-like when compared to an apple... $\endgroup$
    – Alessandro Bertoli
    Commented Feb 11 at 17:32
  • $\begingroup$ I don't know if it's your English but "pretend" has the specific meaning of "to behave in a particular way, in order to make other people believe something that is not true" [Oxford Dictionary] I would suggest you stop repeating that word since it's, at very best, not appropriate. A proper alternative may be "assumes", otherwise you will look as one of those person that believe gravity does not exist and they can "debunk" physics... I would not suggest you to appear as one of those. $\endgroup$
    – Alessandro Bertoli
    Commented Feb 11 at 17:36
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    $\begingroup$ The whole point of the conversation is that you make confusion between "centripetal" and "attractive". I would suggest you to try to understand this point before tackling Hamiltonian mechanics. Newton's principle and the articles you posted are not mutually exclusive but Hamiltonian formulation assumes a Newtonian gravitational potential. I would suggest you to clarify your doubt about forces using a physics textbook like Principle of Physics by Halliday and Resnick (circular motion is at chapter 4.5). $\endgroup$
    – Alessandro Bertoli
    Commented Feb 11 at 20:22

2 Answers 2

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I am not sure to understand the question. Everything depends on the meaning of “attractive”.

An attractive force does not necessarily cause translation.

The motion of a material point subjected to a force of given functional form is the solution of a differential equation.

The solution depends both on the type of force, it’s functional form, and the initial conditions.

A central force is said to be attractive when it is always directed toward the center.

That is the case of Newton’s gravitational force and of the Coulomb one for charges with opposite sign, when one mass or charge has a fixed given position: the center of the force exerted on the other material point.

The attractive gravitational force in Newton’s form, together with an initial velocity which is not parallel to the said force may give rise to a closed orbit: an ellipse with the center as one of its centres.

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  • $\begingroup$ The trajectory due to an attractive acceleration is collinear to the acceleration, while the trajectory due to a centripetal acceleration is perpendicular to the acceleration. If my car is towed by a tow truck, it experiences an attractive force and moves in the direction of this force. It does not orbit around the tow truck. The kinematics makes a clear difference between attractive and centripetal acceleration, definitely. $\endgroup$
    – Hervé Le Cornec
    Commented Feb 11 at 15:12
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    $\begingroup$ I disagree. Sorry, you have misconceptions about basic facts of elementary mechanics. $\endgroup$
    – Valter Moretti
    Commented Feb 11 at 16:44
  • $\begingroup$ Thanks for the compliment, but I would prefer a scientific argument rather than an authority argument. $\endgroup$
    – Hervé Le Cornec
    Commented Feb 11 at 18:47
  • $\begingroup$ I gave you a scientific argument, but you do not understand. Please stop here. $\endgroup$
    – Valter Moretti
    Commented Feb 11 at 18:48
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Rotation and translation describe how the trajectory of an object looks: The trajectory of a rotation is a circle, the trajectory of a translation is a straight line. Attractive and centripetal are adjectives describing the behavior of forces: Attractive forces always point to a specific object (the "attractor"), while centripetal forces always point to the center of a circular trajectory. If the attractor is at the center of a circular trajectory, then the attractive force is a centripetal force.

Note that the behavior of a force does not prescribe the exact trajectory of an object. An attractive force can lead to translations (if the attracted object starts at rest relative to the attractor, or with a velocity pointing to or away from the attractor), and it can lead to rotations (if the attracted object starts with a specific velocity perpendicular to a line bewtween attractor and attractee). In the latter case, the attractive force is automatically a centripetal force.

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  • $\begingroup$ The trajectory due to an attractive acceleration is collinear to the acceleration, while the trajectory due to a centripetal acceleration is perpendicular to the acceleration. If my car is towed by a tow truck, it experiences an attractive force and moves in the direction of this force. It does not orbit around the tow truck. The kinematics makes a clear difference between attractive and centripetal acceleration, definitely. $\endgroup$
    – Hervé Le Cornec
    Commented Feb 11 at 15:05
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    $\begingroup$ Well, simply put, no, kinematics does not make a "clear difference". If you tie the side of your car to a pole with a rope and accelerate, you will cover a round trajectory around the pole. This is a prime example of "centripetal acceleration" but the force is also attracting the car towards the pole since the rope will be stretched and it's elasticity will attract the car back on the radius with length of the rope. $\endgroup$
    – Alessandro Bertoli
    Commented Feb 11 at 17:24
  • $\begingroup$ Yes, a car in a roundabout feels a centripetal acceleration, and then has a circular trajectory perpendicular to the acceleration. A car towed by a tow truck feels an attractive force, its trajectory is collinear to the acceleration. Of course the centripetal acceleration can be incorrectly called "attractive", because we could consider that it "retains" the car which would go straight ahead otherwise. But if there are two words, centripetal and attractive, this is for the good reason that they describe two different kinematics. Words have a meaning. $\endgroup$
    – Hervé Le Cornec
    Commented Feb 11 at 19:11
  • $\begingroup$ Still, you are making confusion, the centripetal acceleration can be correctly called "attractive", because we could consider that it "retains" the car which would go straight ahead otherwise $\endgroup$
    – Alessandro Bertoli
    Commented Feb 11 at 20:00
  • $\begingroup$ What Newton calls attraction is the apple falling to the ground on a straight line. Newton's attractive acceleration causes an accelerated translation. He never assumed that the apple was submitted to a centripetal acceleration. $\endgroup$
    – Hervé Le Cornec
    Commented Feb 11 at 21:56

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