# Finding the correct time to launch a space ship and have it successfully reach Mars

Recently for a project I have been trying to figure out the correct time to launch a space ship and have it successfully reach Mars. This is what I've managed so far
Because of Earth's orbit around the sun a ship would have a transfer orbit that looks like this Distance from starting point to ending point= 1+1.523 =2.523 AU (addition of orbits)
Semi major axis of transit orbit= 2.523/2=1.261
Kepler's third law [T]^{2} = [a]^{3}[/itex], therefore T= \sqrt{1.261^{3}} = 1.417454 years
This is a time for a fall orbit so time for a trip=1.417454/2= 0.70873 years
location of mars at time of launch= 360* (0.70873/1.88)= 135.555^{o}, so something like this Now the angle between Earth Sun and Mars = 180-135.6= 44.4^{o}
To find when this angle occurs I was thinking of using c^{2}=a^{2}+b^{2}-2ab CosC, where a is the distance from earth to the sun, b is the distance from Mars to the Sun, c is the distance between the two planets and C is the Earth Sun Mars angle. For a and b I used r=periapsis*{1+e}/{1+e cos(theta)} , I know this equation isn't perfect as it assumes Mars and Earth orbit at the same velocity and start at the same point but I'm going to adjust it later. Now my problem is I've been having trouble getting an equation for c, can anyone help me with that, or alternately does anyone know how I could find when this particular Earth Sun Mars angle is going to occur.

Thanks for any help in advance

Janus
Staff Emeritus
Gold Member

If you are looking for a rough date, you can do this: First you determine the date for an opposition of Mars (the next one will occur on Apr 8, 2014.)

Then you find the synodic period (opposition to opposition) of Mars by

$$P = \frac{1}{\frac{1}{P_E}- \frac{1}{P_M}}$$

this gives you a value of 780 days. From this you find that the Earth-Sun-Mars angle changes at a rate of 2.167 degrees per day. If you take this and divide it into the needed Earth-Sun-Mars angle, you get how many days before opposition you have to launch. For 44.4° this is 20.5 days or ~Mar. 19, 2014. (Though actually, I get an answer of ~60.5° and a launch date of Mar 11, 2014)

However, this is only approximate. Unfortunately, Mars has a fairly eccentric orbit, and its orbital velocity varies by a quite a bit. How many days it takes to cover a given number of degrees of its orbit depends on what part of its orbit it's in at the time. For example, because of this, the period between oppositions can vary by several weeks.

So to get an accurate answer, you need to know what point of its orbit Mars will be in at arrival and then compute how many degrees of its orbit it will travel during the transfer orbit.