# Changes

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Then I guess the Heliocentric model is wrong, huh Pete? LOL. Seriously, don't introduce Heliocentric theories as facts which do not rebut the inconsistency I have shown with the model.

The moon's speed is constant, period. 2 cars are traveling on a highway the same direction. One at 60 MPH one at 65 MPH. They have a different relative velocity to each other and to the pedestrian watching them go by, but they still have a constant velocity. If you know of some source claiming the speed of the moon varies to maintain a relative velocity to the Earth of 2,288 MPH throughout its orbit, you need to post that source. The moon would have to come to a stop and travel backwards (the opposite direction it was traveling during full moon phase) at 64,712 MPH during the new moon phase to maintain a relative velocity with the Earth of 2,288 MPH. Remember Earth (our frame of reference) is flying in a linear direction at 67,000 MPH. It would then have to stop and change direction and velocity again to maintain its relative velocity to Earth.(Unsigned by JasonZ) :The moon's speed is constant in the ''earth's'' frame of reference. I.e. in the earth's frame of reference its speed is measured to be constant. In the sun's frame of reference it's apparent speed varies. In the sun's reference frame: At full moon: <math>v = 2,288 \text{MPH}</math> At first quarter <math>v = \sqrt{2288^2 + 67000^2} = 67039 \text{MPH}</math> At new moon <math>v = \sqrt{(67000 - 2288)^2} = 64712 \text{MPH}</math> At last quarter <math>v = \sqrt{2288^2 + 67000^2} = 67039 \text{MPH}</math> In the earth's frame of reference, the speed is always 2288 MPH. It's velocity does change as its direction changes. Yes in the sun's frame of reference, at new moon the moon does change direction and travel in the opposite direction to that at full moon. This can be seen in the video your reference. For the earth, the moons position <math>\vec{r}</math> at time <math>t</math> may be parameterised (assuming it is circular as seen from the earth) as: <math>\vec{r}_m(t) = a_m \cos{\omega_m t} \, \vec{i} + a_m \sin{\omega_m t} \, \vec{j}</math> where <math>a_m</math> is the moon's orbital distance and <math>\omega_m</math> is the moon's angular frequency as seen from earth. This is just the parameterisation of a circle. From this the velocity of the moon is: <math>\dot{\vec{r}}_m(t) = - a_m \omega \sin{\omega_m t} \, \vec{i} + a_m \omega \cos{\omega_m t} \, \vec{j}</math> Hence the speed which is <math>a_m \omega_m</math>, which is constant. From the sun, the earth appear to follow a circular path, so its position is: <math>\vec{r}_e = a_e \cos{\omega_e t} \, \vec{i} + a_e \sin{\omega_e t} \, \vec{j}</math> For <math>a_e</math> being the sun earth distance and <math>\omega_e</math> being the angular velocity of the earth around the sun. Adding these give the path the moon follows as seen in the sun's reference frame: <math>\vec{r}(t) = (a_e \cos{\omega_e t} + a_m \cos{\omega_m t}) \, \vec{i} + (a_e \sin{\omega_e t} + a_m \sin{\omega_m t}) \, \vec{j}</math> From this, it is clear that the velocity and speed of the moon in the sun's reference frame do change. Remember the moon is accelerating as it is undergoing circular motion so it's velocity does change. Though there is no need for the moon to stop in either reference frame. Just as a side question what is your background in physics? I just don't want introduce concepts too advanced or to simplify my replies if you already have that sort of understanding. [[User:PeterIceHockey|PeterIceHockey]] ([[User talk:PeterIceHockey|talk]]) 16:03, 16 December 2016 (EST)

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