Solving for Mars Encounter Orbit w/Limited Info

[skiboka33]

Ok, say there is a space probe orbitting the sun at 1 AU and it needs to be put in an orbit that will encounter Mars using the least amount of energy. The orbit hsa perihelion at Earth's orbit and apelion at mar's orbit. In the problem we are asked to solve for stuff like the eccentricity of the orbit and change in orbital velocity to achieve the orbit. No need to do the question for me, but I'm just a little confused about what information we can get from the original question. Seems like it's not enough information to solve the problems. Any ideas on what I'm missing? thanks.

 

[Janus]

Well, you know the aphelion and perhelion of the tranfer orbit, you can solve for the eccentricity with that information.

You can solve for the total energy of your probe in this new orbit, and you can solve for the total energy in the original orbit. From this you can solve for the energy difference between the orbits, which at perhelion will be due to kinetic energy alone. This will allow you to get the change in orbital velocity needed.

 

[SpaceTiger]

Seems like it's not enough information to solve the problems.

In real life, you would need phase information -- that is, you'd need to know the positions of Mars and the satellite at a particular time. The fact that the orbit has aphelion at Mars' orbit doesn't mean you'll encounter it in any reasonable amount of time. In fact, if your satellite's orbit was resonant with that of Mars, you'd likely never encounter it at all!

You almost certainly don't need to worry about these things, however. The problem should be solvable using the information given, along with the procedure Janus described.

 

[ek]

This was a question on my assignment due today. Odds are this guy is in my class (Uvic).

Last week of school, already started studying for finals, an astronomy assignment isn't exactly on my radar. Went to school this morning, oh **** I forgot to do this assignment. And this one was actually tough! Flipped it off in an hour and a half. 60% at most. Oh well. I pwn that class anyway.

Sorry this isn't a more helpful post, although the thing was due five hours ago so I guess help now wouldn't help you too much.

 

[Janus]

In real life, you would need phase information -- that is, you'd need to know the positions of Mars and the satellite at a particular time. The fact that the orbit has aphelion at Mars' orbit doesn't mean you'll encounter it in any reasonable amount of time. In fact, if your satellite's orbit was resonant with that of Mars, you'd likely never encounter it at all!

True, and in addition, in real life you would need to know at what position Mars will be in it's orbit in relation to its perhelion when intercept is met. Mars' orbit is eccentric enough that if you based your \Delta V on its average distance from the Sun you could find yourself missing Mars by some 21 million km.

As a result, when you launch a probe from Earth, it takes the least \Delta V to intercept Mars when you launch in the last half of February and the most if you launch in August. (this is just for intercept and doesn't take account the \Delta V needed to match orbit with Mars once you get there.)

This leaves a few best launch "windows" where both the energy needed is the least and Mars and Earth are in the proper relative positons of their orbits for intercept.

But again, for this question, you are allowed to simplify and assume that Mars is in a circular orbit.

 

[Romanus]

This is pretty straightforward, in the simplest case. Just calculate the parameters of an orbit with its perihelion at Earth's orbit, and aphelion where it intersects Mars' orbit. Take half the orbital time (say 190 days), and calculate how far Mars would move in that interval. Launch on the trajectory when Mars is the requisite number of degrees ahead of the rendevous point, and boom--simple Hohmann transfer. With the set of orbit equations in hand, you can calculate a trajectory to just about any object in the Solar System. Of course, in real-life it would be complicated by orbital inclination, but anyways...