> We already have ion drives Yes, indeed. > Your burn rate will increase dramatically depending on Mars/Earth alignment. Why? I am not talking about keeping

> We already have ion drives Yes, indeed. > Your burn rate will increase dramatically depending on Mars/Earth alignment. Why? I am not talking about keeping trip time constant. I am just talking about significantly reducing trip time compared to free fall. That is easily done with a low thrust ion drive; even a hundredth of a gee makes a significant reduction in trip time, and the longer the distance, the more the reduction in time even holding the thrust constant (since the trip time now increases only logarithmically with the distance instead of linearly). And that makes trips more feasible even when the planets aren't in alignment. > even with the best technology we can see ourselves building within the next decade or so we're talking 3 months on a good alignment Now it's you who aren't conferring much confidence in your knowledge. Below are some numbers (travel times are one-way). I've even included numbers for 1/1000 gee to show that even those numbers are much better than the free-fall numbers. Closest approach: about 60 million km (6 x 10^10 m) Free-fall travel time: 180 days 1/100 gee travel time: 18 days 1/1000 gee travel time: 57 days Furthest away: about 400 million km (4 x 10^11 m) Free-fall travel time: 1200 days 1/100 gee travel time: 47 days 1/1000 gee travel time: 148 days The free-fall transit time for closest approach is taken from the Wikipedia article on free-return Earth-Mars trajectories; I have used the shortest time that appears to have a reasonable return trip (though the return trip transit time is longer). [1] Free-fall transit times for furthest away are taken by multiplying the closest approach time by 6.67, which is probably somewhat of an underestimate (it assumes straight linear dependence of trip time on distance, without confirming that there actually is a viable transfer orbit with that transit time). 1/100 gee and 1/1000 gee transit times are obtained by using the formula t = 2 sqrt(d / a), where d is the distance and a is the acceleration. This formula assumes constant acceleration to the halfway point, then turnaround and constant deceleration the rest of the way. (I also adjusted the units in the answers above to days instead of seconds, which is what the formula I just gave gives; I rounded up all fractions.) I have ignored orbital maneuvers and takeoff/landing at both ends for all travel times. https://en.wikipedia.org/wiki/Free-return_trajectory#Earth%E2%80%93Mars https://en.wikipedia.org/wiki/Free-return_trajectory#Earth%E... > Columbus was on the sea for 36 days Which was by no means the longest, or even a particularly long, voyage for the time period. Vasco da Gama was at sea for more than three months at a time during his voyage around Africa to the Far East and back. So was Magellan during his round the world expedition.