How long would it take to go to Mars going at the speed of light

Let's make a big assumption:

The orbits of Mars and Earth are so close to being circular that we don't have to fuss around with a whole bunch of equations on ellipses (it makes it nicer for us, even if it isn't completely accurate).

Now let's say that the 2 planets start out in a perfect line with the sun and they begin to orbit. Mars takes 1.8808 earth years to complete 1 orbit, so let's figure out the orbital resonance:

1.8808:1 =>
18808:10000 =>
4702:2500 =>
2351:1250

So, for every 1250 martian years, earth passes through 2351 years. The average distance from the sun to mars is 1.524 AU

Now, let's plot the parametric functions for the positions of earth and mars

x[e] = 1 * cos(t)
y[e] = 1 * sin(t)
x[m[ = 1.524 * cos(t * (1250 / 2351))
y[m] = 1.524 * cos(t * (1250 / 2351))

The formula for the distance between them at some value of t

d^2 = (y[m] - y[e])^2 + (x[m] - x[e])^2
d^2 = 1.524^2 * cos(t * (1250/2351))^2 - 2 * 1.524 * cos(t * 1250 / 2351) * cos(t) + cos(t)^2 + 1.524^2 * sin(t * 1250/2351)^2 - 2 * 1.524 * sin(t * 1250/2351) * sin(t) + sin(t)^2
d^2 = 1.524^2 + 1 - 2 * 1.524 * (cos(1250t/2351)cos(t) + sin(1250t/2351)sin(t))
d^2 = 3.322576 - 3.048 * (cos(t * (1250/2351) - t))
d^2 = 3.322576 - 3.048 * cos(t * (1250 - 2351) / 2351)
d^2 = 3.322576 - 3.048 * cos(-1101 * t / 2351)

Since cos(-t) = cos(t)

d^2 = 3.322576 - 3.048 * cos(1101 * t / 2351)
d = sqrt(3.322576 - 3.048 * cos(1101 * t / 2351))

Now for the fun part. We're going to use the MVT to determine the average distance from t = 0 to t = 2351

f(t) = (3.322576 - 3.048 * cos(1101 * t / 2351))^(1/2)

(F(2351) - F(0)) / (2351 - 0)

I'm going to use Wolfram Alpha to do this, because quite frankly, I don't feel like it (I'm not even sure I could do it)

http://www.wolframalpha.com/input/?i=int…

3978.32 / 2351 =>