How to go to mars

Since I'm currently preparing a report for School about Mars Exploration, I thought it would be cool to share with you some of the knowledge that I acquired while collecting information for this report. Today, I will be diving into how spacecraft travel to another celestial body.

Computing a transfer window

Because both Mars and Earth are moving at different speeds, not just any moment in time is suited for traveling to Mars efficiently. Approximately every two years, the two planets line up perfectly and allow the most efficient transfer to mars. This time is called a transfer window.

To simplify the calculations in this post, we will assume that both Earth and Mars have perfectly circular orbits at a distance from the Sun of 1 AU and 1.5 AU, respectively. Our transfer orbit on the way to mars would probably look about like this: (Marked in gray)

To compute our transfer window, we must know how long it takes to go from Earth to Mars, or, in other words, how long it takes to complete 1/2 of that orbit.

Computing the orbital period

As described here, we know that the constant

C = 1{year^2 \over AU^3}

and that the orbital period

T^2 = CR^3

So, to calculate T, we must first solve for R:

R = { perihelion + aphelion \over 2 }
  = { 1 AU + 1.5 AU \over 2 }
  = { 2.5 AU \over 2 }
  = 1.25 AU

Now, we can solve for T:

T^2 = CR^3

T^2 = (1 {year^2 \over AU^3})(1.25 AU)^3

T^2 = (1 {year^2 \over AU^3})(1.953125 {AU^3})

T^2 = 1.953125 { years^2 }

T = 1.39754248594 years

The orbital period of our spacecraft appears to be 1.39754248594 years. However, this is the time it takes to go to mars and back, but we only want to go one way, so our orbital period is only half as long:

{1 \over 2}T = 0.69877124296 years

Computing the ideal position of mars at launch time

For our craft to reach Mars's orbit at the correct moment to enter it's SOI (Sphere Of Influence), we must make sure that mars is exactly the distance it travels in 0.69877124296 years away from the point of intersection.

Because we know that Mars travels at a speed of about 24km/s over a time of 22051090.5731 seconds, we can say that

d = v * t = 24 km/s * 22051090.5731 s = 529226173.754 km = 3.537658465843391 AU

Which is about 1/3 of Mars's orbit. That means we need to leave Earth 254 Days before Mars meets us on the other side of the sun.