1 NECAT, April 29, 2010 at 10:15 p.m.:

Hi,

Kepler himself has changed his reasoning on elliptical orbit by giving his third law about periods:(P1/P2)^2=(r1/r2)^3.He wanted to say that the elliptical orbits will be transformed to circular orbits with time.How a solid ellipse may be transformed to a circle? By being not an ellipse.Then what is the real shape of the orbits? They are spiraled.Newton's F*dt=m*dv,when solved, gives the equation of the motion: r=-a*t*(t-Tmax)+K.The polar graph of this equation shows billions of cardioidal looking spirals: like in the photos of the galaxies.So,orbits are not elliptical,the Sun is not at one of the focus of these ellipses,no aphelion,no perihelion,nor equality of swept out areas in equal interval of time.All these are old estimation of Kepler,proved by the mathematicians(!).Mathematicians should consider "spiraled orbits theory", and they will be surprised by the consequences they may discover: ligth year having no sense;years length different in the very beginning of the Earth (15 days a year for example) and ..etc .necattasdelen@ttmail.com

2 Eric Dong, July 2, 2010 at 6:03 a.m.:

Don't say stupid things like that. Kepler's 3rd law is: (P1/P2)^2=(a1/a2)^3. There is no such thing as R in the equation; it is a, the semimajor axis. And your "derivation" of the spiral orbit is sheer nonsense. Please show me the t, Tmax, and r in Fdt=mdv! And you are just throwing technobabble after that. Kepler's laws were not proved by the mathematicians, they were proved by Kepler himself by observation. Newton derived Kepler's 3rd law HIMSELF. And your reference to light year making no sense makes no sense. Light year is as concrete as meter: it's the distance light goes in a year. Please do not play sham astrophysicist here. If you want my derivation (using PHYSICS!!!) of Kepler's 3rd law or a demonstration of the stableness of elliptical orbits, please e-mail me: ericcesium133@aol.com.

3 Eric Dong, July 2, 2010 at 6:04 a.m.:

Don't say stupid things like that. Kepler's 3rd law is: (P1/P2)^2=(a1/a2)^3. There is no such thing as R in the equation; it is a, the semimajor axis. And your "derivation" of the spiral orbit is sheer nonsense. Please show me the t, Tmax, and r in Fdt=mdv! And you are just throwing technobabble after that. Kepler's laws were not proved by the mathematicians, they were proved by Kepler himself by observation. Newton derived Kepler's 3rd law HIMSELF. And your reference to light year making no sense makes no sense. Light year is as concrete as meter: it's the distance light goes in a year. Please do not play sham astrophysicist here. If you want my derivation (using PHYSICS!!!) of Kepler's 3rd law or a demonstration of the stableness of elliptical orbits, please e-mail me: ericcesium133@aol.com.

4 Peter L. Griffiths, May 13, 2011 at 5:37 p.m.:

Galileo's law of falling bodies v^2=d is effectively the same as Kepler's distance law v^2=1/r. Let L indicate a small change then we have
v^2+Lv^2=d+Ld=1/(r-Ld). This is the usual way of measuring elocity. For the reciprocal method of measuring velocity we have
v^2+Lv^2=r+Lr=1/(d-Lr). There are only three variables involved distance, time and velocity. There are two methods of measuring the same velocity, distance per unit time and time per unit distance , one method is the reciprocal of the other. d+r equals the major axis of the elliptical orbit.

5 Peter L. Griffiths, October 31, 2012 at 4:26 p.m.:

One of the simplest properties of all elliptical orbits is that in every elliptical orbit there is constant acceleration for half the orbit and constant deceleration for the other half. Velocity is a different matter but can be compared between planets in a different way.

6 Peter L. Griffiths, December 18, 2012 at 8:08 p.m.:

Further to my previous comments, the connection between Galileo's v^2=d at the empty focus end of the elliptical orbit and Kepler's v^2=(1/r) at the Sun focus end is mathematically very interesting and not at all straight forward. Kepler's version can be adapted for further research purposes by including a constant V being the maximum velocity, then the variable velocities can be expressed as as V/#r where # is my notation for square root. In this way the same velocity arises on both the accelerating side as well as the decelerating side, but in opposite directions. As one of the properties of all perfect ellipses d is the distance from the curve to the empty focus, and r is the distance from the curve to the Sun focus, d+r equals the major axis of the elliptical orbit.

7 Peter L. Griffiths, January 24, 2013 at 3:55 p.m.:

The time version of Kepler's distance law is t=r^(1.5). The time version of Galileo's law of falling bodies is t=d^(0.5). The time taken by Halley's comet to cover the decelerating half of the elliptical orbit is 38 years. Over half this decelerating distance, the comet will have taken 38/2#2 years which is 13.435 years decelerating. The time taken by Halley's comet to cover the accelerating half of the elliptical orbit is 38/#2 years which is 26.87 years. As previously # is my notation for square root.

8 Peter L. Griffiths, May 8, 2013 at 4:25 p.m.:

In the second edition 1713 page 540 Cajori edition of his Principia, Isaac Newton incorrectly calculates the major axis of the orbit of Halley's Comet as being 35 AU. Newton does this by taking the ratio of the time of the Comet's orbit to the time of the Earth's orbit round the Sun as being 75:1, and then applying Kepler's distance law to this 75 namely 75^(2/3) which gives about 17 AU which Newton correctly thought was far too low so he doubled it to approximately 35 AU. In fact the correct figure for the major axis of the orbit of Halley's Comet is about 60 AU involving far more detailed calculations.