Optimizing Orbital Trajectories for Efficient Mars Encounter

[olyviab]

A space probe initially moving in a 1 AU circular orbit around the sun (i.e. moving with the earth). The aim is to put this space probe in an orbit that will encounter Mars, using the least possible expenditure of rocket fuel (energy). This orbit, it turns out, is one in which perihelion (closest approach to the sun) is at the Earth's orbit (r=1AU), and aphelion (furthest approach from the sun) is at Mars' orbit.Suppose that the rockets had fired so that the change in velocity was in the radial direction (outwards). Calculate a (semi major axis) Will this orbit ever reach Mars?

 

[gneill]

A space probe initially moving in a 1 AU circular orbit around the sun (i.e. moving with the earth). The aim is to put this space probe in an orbit that will encounter Mars, using the least possible expenditure of rocket fuel (energy). This orbit, it turns out, is one in which perihelion (closest approach to the sun) is at the Earth's orbit (r=1AU), and aphelion (furthest approach from the sun) is at Mars' orbit.


Suppose that the rockets had fired so that the change in velocity was in the radial direction (outwards). Calculate a (semi major axis) Will this orbit ever reach Mars?

Do you mean that the original velocity change intended for the Hohmann orbit injection was mistakenly made in a radial direction? If so, then no, the orbit would not reach Mars. The Hohmann transfer is the least energy direct transfer orbit. If it's not the Hohmann orbit, then it can't reach Mars with the same energy expenditure.

 

[olyviab]

Do you mean that the original velocity change intended for the Hohmann orbit injection was mistakenly made in a radial direction? If so, then no, the orbit would not reach Mars. The Hohmann transfer is the least energy direct transfer orbit. If it's not the Hohmann orbit, then it can't reach Mars with the same energy expenditure.

Yeah i assume it was the least energy in the Hohmann orbit directed outwards. I understand if its the 3 km/s (i calculated that number as being the least v) being directed outward that it won't reach the Mars orbit. How am i able to calculate the semi-major axis?

 

[gneill]

Yeah i assume it was the least energy in the Hohmann orbit directed outwards. I understand if its the 3 km/s (i calculated that number as being the least v) being directed outward that it won't reach the Mars orbit. How am i able to calculate the semi-major axis?

If you have the new velocity (after the ∆V) and the radius, then you can calculate the total mechanical energy, ξ. Then 2a = -µ/ξ.

 

[olyviab]

If you have the new velocity (after the ∆V) and the radius, then you can calculate the total mechanical energy, ξ. Then 2a = -µ/ξ.

Great! Thanks for all the help :)