trig, as I said, I will not attack you on the centripetal force matter, even though you left out the mass from the equation.

You are ever ready to send everybody here back to high school, but where should we send you, given the fact you used the term CENTRIPETAL FORCE, and left out the m (mass) term in the equation?

Your explanation cannot hide the fact that the centripetal force of the Sun is A HUGE QUANTITY, given the fact we have to multiply the w^2r term with its (the Sun's) mass.

You wrote this:

*It is totally useless to consider the whole Sun as one single mass upon which all the centripetal force is applied because the direction of the force is not the same and the amount of the force is not the same in every place of the Sun. With this faulty reasoning, I could say that a typical Merry-go-round, which weighs about 10000 kilograms and turns once every 15 seconds and has a radius of 4 meters will produce a devastating 7018 Newtons (equivalent to 716 kilograms of weight) of centripetal force, thereby killing any brave child who rides it.*trig, you should go back to kindergarten, not only to high school...

Let us suppose now, the sake of our discussion, a child (mass 25 kg) on a merry-go-round is moving with a speed of 1.35 m/s when 1.20 m from the center of the merry-go-round.

Let us calculate the acceleration and THE NET HORIZONTAL FORCE EXERTED ON THE CHILD:

a. a = v^2/r = (1.35 m/s)^2/(1.2 m) = 1.52 m/s^2

b. F = ma = (25 kg)(1.52 m/s/s) = 38 N = centripetal force on the child (F = ma = mv^2/r)

You take into account the MASS OF THE CHILD, NOT THAT OF THE MERRY GO ROUND...where did YOU learn physics trig?

Now, let us calculate the total

**CENTRIPETAL FORCE EXERTED ON THE ROUND EARTH**:

Some preliminary calculations: (365.26 days)(24 hr/day)(3600 s/hr) = 31558464 s

For convenience, R (radius of earth) = 1.5 x 10^11 m

Mass of Earth is 5.98 x 10^24 kg

a = v^2/r = (2πr/T)^2/r = 4π^2r/T^2 , π = pi

So now,

F = (5.98 x 10^24 kg)(0.005946 m/s^2) = 3.55566x10^22 N =

**3.6 x 10^22 N toward the Sun**trig, this is a HUGE QUANTITY, a very large magnitude of the centripetal force indeed.

So, THE CENTRIPETAL FORCE OF THE SUN, trig, MUST TAKE INTO ACCOUNT the mass of the Sun =

**1,98892 x 10^30 kg**As I said, I will not attack you as I should be doing it, for your mistakes...which are typical of an amateurish approach to physics...please go back to a course in basic mechanics, and read up the facts...

Dr. Neville Jones is no crackpot; not by any stretch of the imagination, he is one of the most talented physicists I have seen in a very long time...you call other people crackpots, YET YOU LEFT OUT THE MASS IN THE CENTRIPETAL FORCE EQUATION which is Fcp = - mw^2r, no serious physicist would do that...you cannot even describe correctly the centripetal force exerted on a child on a merry go round...

Don't you know that in my theory satellites exist very well? You should know that by now; I do not accept some the facts stated in the official FAQ.

http://www.geocentricuniverse.com/Restoring%20forces.htm (I invite all of you to read this carefully, a superb analysis of atmospheric physics)

The official view of current astrophysics, for your information, is that in the solar system the combination of the Earth's spinning and revolution and the Earth's motion toward Vega forms a right-handed super-helical motion.

Therefore, trig, this fact immediately contradicts Kepler's first law...