I just noticed this. It's pretty good, but you have a few errors:

Q: Why hasn't the Earth reached the speed of light if it's accelerating?

A: Contrary to popular belief, acceleration is not linear. In Einstein's Theory of General Relativity, there is an equation for adding vectors in the form of a recursive formula:

` `*v* + *w*

*u* = ~~ ~~

(1 + *vw*) ÷ *c*^{2}

In which *u* is your velocity, *v* is you rate of acceleration, *w* is your previous velocity, and *c* is the speed of light.

And if you were to work it out, you would find that it would take forever to reach the speed of light.

This isn't right. First of all, it's the wrong formula. This should be obvious since the units are wrong: you are adding a dimensionless number (1) to a number with dimensions of speed

^{2} (

*vw*). The correct equation is:

` `*v* + *w*

*u* = ~~ ~~

1 + (*vw*/*c*^{2})Secondly, if you add a velocity and an acceleration, you get a meaningless physical quantity. That equation only makes sense when v and w are both velocities. Really what it says is that if two objects are moving directly away from each other at speeds of v and w in the reference frame of some fixed observer, then the speed of one object relative to the other object is given by u in the formula.

To adapt it to accelerations, you have to do a little work. If we started with a speed of

*v* and increase the speed by

*at*, say by accelerating at a rate of

*a* for a duration

*t*, then we would get a new speed:

` `*v* + *at*

*u* = ~~ ~~

1 + (*vat*/*c*^{2})This is actually inaccurate for large

*t*, since it assumes the acceleration is constant in the inertial reference frame in which the Earth is instantaneously fixed, but is correct in the limit as

*t* approaches 0. We can therefore compute the rate of change of

*u* when

*t*=0 by the quotient rule:

*u'* =

*a*(1-

*v*^{2}/

*c*^{2})

This tells us that if the acceleration of the Earth is given by

*a* in a reference frame where the Earth is instantaneously stationary (so a is the strength of the acceleration felt by inhabitants of the earth, the acceleration is given by

*a*(1-

*v*^{2}/

*c*^{2}) in a reference frame in which the Earth is already moving at speed

*v*.

To find speed in one fixed reference frame, starting out with u=0 at t=0, we have to solve the differential equation:

*u'* =

*a*(1-

*u*^{2}/

*c*^{2})

Some simple analysis shows that as

*u* approaches

*c*,

*u'* approaches 0, and therefore at a constant acceleration from the point of view of the Earth, the Earth will approach but never reach light speed as we expected.

So your answer had some of the correct features, but made some incorrect claims and should probably be corrected.

Q: With a little bit of math, I can show that the sun should be at different angles than what is observed. What's up?

A: The sun's light refracts as it enters the atmosphere and continues downward, thus causing light to be bent. Also notice that in cold areas this effect will have a greater effect on the sunlight because the cold air is denser. This also causes light to be bent in the way it is on the Equinoxes and Solstices.

Q: How do sunrises/sunsets occur?

A: The sun gets to far away for the sunlight to reach us, also caused by the refraction as stated above.

If your diagram shows the correct refraction due to sunlight, it's in the wrong direction. According to your diagram, the sun appears higher than it really is, but in actuality the sun appears lower in the sky than the FE model says it is (and rises and sets in the wrong places - a good deal further South than it should under the FE model, but that's another issue entirely). So if your diagram is right, it doesn't fix the problem, and in fact makes it work.

If the refraction instead went the opposite direction, that could be a solution to the problem. Unfortunately if this is true, then sunlight behaves the opposite of the way we expect it too when going from less dense into denser atmosphere, or else the atmosphere is actually denser higher up than it is lower down (which is contrary to what we observe when we go to a high elevation, and what we would predict.)