# Return to Answer

 5 deleted 215 characters in body edited Feb 25 at 12:50 Rikki-Tikki-Tavi 7,4692525 silver badges5050 bronze badges It's not necessary to take an entirely different route every time. There is the minimum energy transfer, that you mentioned, but you can use more fuel(or better yet, use a more efficient engine) to take a similar, but different route that takes you there faster. The raise in energy requirement is moderate at first, but becomes uncomfortable if you want to cut your travel time by more than a month or so. I think there is a table in a book I have at home ("Astronautics" by Ulrich Walter). I will let you know. Also, there are entirely different courses you can take, but I don't think any of them are faster. For example, in some constellations a Venus flyby gives you the chance to return after only a short stay at Mars, rather than waiting out a full cycle. Edit: I misremembered. In Astronautics I found the following two figures (which are even better; read below): The image above makes three examples of orbits we could fly. You could fly even more extreme courses than the 70 day one, but you can see how the required earth escape speed $$v_\infty$$ becomes uncomfortable. This graph gives us the the same in exact numerical terms. You can see that shortening the time a bit is quite cheap, but then it quickly becomes expensive. $$t_x$$ is the transfer time. $$t_H$$ is the time for the Hohmann-transfer. I won't explain all variables involved, I don't think anyone has the patience to read that. It is true that you also need an burn to turn into Mars' orbit, but I disagree with HopDavid about this being the major factor: The $$\Delta v$$ for the for the injection is $$v_\infty\sqrt{\frac{1-e}{2}}$$ where $$e$$ is the eccentricity of the orbit around mars and $$v_\infty$$ is the excess velocity with regard to mars now. This can be very high, because you can slowly circularize the orbit for free by aerobraking. For $$e=0.95$$ the Mars injection takes only 16% of $$v_\infty$$. Of course, there is a limit to how eccentric your orbit can be, because you may escape Mars' sphere of influence if you come in too fast. It's not necessary to take an entirely different route every time. There is the minimum energy transfer, that you mentioned, but you can use more fuel(or better yet, use a more efficient engine) to take a similar, but different route that takes you there faster. The raise in energy requirement is moderate at first, but becomes uncomfortable if you want to cut your travel time by more than a month or so. I think there is a table in a book I have at home ("Astronautics" by Ulrich Walter). I will let you know. Also, there are entirely different courses you can take, but I don't think any of them are faster. For example, in some constellations a Venus flyby gives you the chance to return after only a short stay at Mars, rather than waiting out a full cycle. Edit: I misremembered. In Astronautics I found the following two figures (which are even better; read below): The image above makes three examples of orbits we could fly. You could fly even more extreme courses than the 70 day one, but you can see how the required earth escape speed $$v_\infty$$ becomes uncomfortable. This graph gives us the the same in exact numerical terms. You can see that shortening the time a bit is quite cheap, but then it quickly becomes expensive. $$t_x$$ is the transfer time. $$t_H$$ is the time for the Hohmann-transfer. I won't explain all variables involved, I don't think anyone has the patience to read that. It is true that you also need an burn to turn into Mars' orbit, but I disagree with HopDavid about this being the major factor: The $$\Delta v$$ for the for the injection is $$v_\infty\sqrt{\frac{1-e}{2}}$$ where $$e$$ is the eccentricity of the orbit around mars and $$v_\infty$$ is the excess velocity with regard to mars now. This can be very high, because you can slowly circularize the orbit for free by aerobraking. For $$e=0.95$$ the Mars injection takes only 16% of $$v_\infty$$. Of course, there is a limit to how eccentric your orbit can be, because you may escape Mars' sphere of influence if you come in too fast. It's not necessary to take an entirely different route every time. There is the minimum energy transfer, that you mentioned, but you can use more fuel(or better yet, use a more efficient engine) to take a similar, but different route that takes you there faster. The raise in energy requirement is moderate at first, but becomes uncomfortable if you want to cut your travel time by more than a month or so. I think there is a table in a book I have at home ("Astronautics" by Ulrich Walter). I will let you know. Also, there are entirely different courses you can take, but I don't think any of them are faster. For example, in some constellations a Venus flyby gives you the chance to return after only a short stay at Mars, rather than waiting out a full cycle. Edit: I misremembered. In Astronautics I found the following two figures (which are even better; read below): The image above makes three examples of orbits we could fly. You could fly even more extreme courses than the 70 day one, but you can see how the required earth escape speed $$v_\infty$$ becomes uncomfortable. This graph gives us the the same in exact numerical terms. You can see that shortening the time a bit is quite cheap, but then it quickly becomes expensive. $$t_x$$ is the transfer time. $$t_H$$ is the time for the Hohmann-transfer. I won't explain all variables involved, I don't think anyone has the patience to read that. It is true that you also need an burn to turn into Mars' orbit, but I disagree with HopDavid about this being the major factor: The $$\Delta v$$ for the for the injection is $$v_\infty\sqrt{\frac{1-e}{2}}$$ where $$e$$ is the eccentricity of the orbit around mars and $$v_\infty$$ is the excess velocity with regard to mars now. This can be very high, because you can slowly circularize the orbit for free by aerobraking. For $$e=0.95$$ the Mars injection takes only 16% of $$v_\infty$$. Of course, there is a limit to how eccentric your orbit can be, because you may escape Mars' sphere of influence if you come in too fast. 4 broken images fixed (click 'rendered output' to see the difference); for more info, see https://gist.github.com/Glorfindel83/9d954d34385d2ac2597bbe864466259f edit approved Feb 24 at 13:20 Glorfindel 25911 gold badge22 silver badges1010 bronze badges It's not necessary to take an entirely different route every time. There is the minimum energy transfer, that you mentioned, but you can use more fuel(or better yet, use a more efficient engine) to take a similar, but different route that takes you there faster. The raise in energy requirement is moderate at first, but becomes uncomfortable if you want to cut your travel time by more than a month or so. I think there is a table in a book I have at home ("Astronautics" by Ulrich Walter). I will let you know. Also, there are entirely different courses you can take, but I don't think any of them are faster. For example, in some constellations a Venus flyby gives you the chance to return after only a short stay at Mars, rather than waiting out a full cycle. Edit: I misremembered. In Astronautics I found the following two figures (which are even better; read below): The image above makes three examples of orbits we could fly. You could fly even more extreme courses than the 70 day one, but you can see how the required earth escape speed $$v_\infty$$ becomes uncomfortable. This graph gives us the the same in exact numerical terms. You can see that shortening the time a bit is quite cheap, but then it quickly becomes expensive. $$t_x$$ is the transfer time. $$t_H$$ is the time for the Hohmann-transfer. I won't explain all variables involved, I don't think anyone has the patience to read that. It is true that you also need an burn to turn into Mars' orbit, but I disagree with HopDavid about this being the major factor: The $$\Delta v$$ for the for the injection is $$v_\infty\sqrt{\frac{1-e}{2}}$$ where $$e$$ is the eccentricity of the orbit around mars and $$v_\infty$$ is the excess velocity with regard to mars now. This can be very high, because you can slowly circularize the orbit for free by aerobraking. For $$e=0.95$$ the Mars injection takes only 16% of $$v_\infty$$. Of course, there is a limit to how eccentric your orbit can be, because you may escape Mars' sphere of influence if you come in too fast. It's not necessary to take an entirely different route every time. There is the minimum energy transfer, that you mentioned, but you can use more fuel(or better yet, use a more efficient engine) to take a similar, but different route that takes you there faster. The raise in energy requirement is moderate at first, but becomes uncomfortable if you want to cut your travel time by more than a month or so. I think there is a table in a book I have at home ("Astronautics" by Ulrich Walter). I will let you know. Also, there are entirely different courses you can take, but I don't think any of them are faster. For example, in some constellations a Venus flyby gives you the chance to return after only a short stay at Mars, rather than waiting out a full cycle. Edit: I misremembered. In Astronautics I found the following two figures (which are even better; read below): http://www.jealousjellyfish.de/stackeximages/hohman_orbit.png The image above makes three examples of orbits we could fly. You could fly even more extreme courses than the 70 day one, but you can see how the required earth escape speed $$v_\infty$$ becomes uncomfortable. http://www.jealousjellyfish.de/stackeximages/transfer_speed.png This graph gives us the the same in exact numerical terms. You can see that shortening the time a bit is quite cheap, but then it quickly becomes expensive. $$t_x$$ is the transfer time. $$t_H$$ is the time for the Hohmann-transfer. I won't explain all variables involved, I don't think anyone has the patience to read that. It is true that you also need an burn to turn into Mars' orbit, but I disagree with HopDavid about this being the major factor: The $$\Delta v$$ for the for the injection is $$v_\infty\sqrt{\frac{1-e}{2}}$$ where $$e$$ is the eccentricity of the orbit around mars and $$v_\infty$$ is the excess velocity with regard to mars now. This can be very high, because you can slowly circularize the orbit for free by aerobraking. For $$e=0.95$$ the Mars injection takes only 16% of $$v_\infty$$. Of course, there is a limit to how eccentric your orbit can be, because you may escape Mars' sphere of influence if you come in too fast. It's not necessary to take an entirely different route every time. There is the minimum energy transfer, that you mentioned, but you can use more fuel(or better yet, use a more efficient engine) to take a similar, but different route that takes you there faster. The raise in energy requirement is moderate at first, but becomes uncomfortable if you want to cut your travel time by more than a month or so. I think there is a table in a book I have at home ("Astronautics" by Ulrich Walter). I will let you know. Also, there are entirely different courses you can take, but I don't think any of them are faster. For example, in some constellations a Venus flyby gives you the chance to return after only a short stay at Mars, rather than waiting out a full cycle. Edit: I misremembered. In Astronautics I found the following two figures (which are even better; read below): The image above makes three examples of orbits we could fly. You could fly even more extreme courses than the 70 day one, but you can see how the required earth escape speed $$v_\infty$$ becomes uncomfortable. This graph gives us the the same in exact numerical terms. You can see that shortening the time a bit is quite cheap, but then it quickly becomes expensive. $$t_x$$ is the transfer time. $$t_H$$ is the time for the Hohmann-transfer. I won't explain all variables involved, I don't think anyone has the patience to read that. It is true that you also need an burn to turn into Mars' orbit, but I disagree with HopDavid about this being the major factor: The $$\Delta v$$ for the for the injection is $$v_\infty\sqrt{\frac{1-e}{2}}$$ where $$e$$ is the eccentricity of the orbit around mars and $$v_\infty$$ is the excess velocity with regard to mars now. This can be very high, because you can slowly circularize the orbit for free by aerobraking. For $$e=0.95$$ the Mars injection takes only 16% of $$v_\infty$$. Of course, there is a limit to how eccentric your orbit can be, because you may escape Mars' sphere of influence if you come in too fast. 3 added 24 characters in body edited Aug 28 '14 at 23:42 Rikki-Tikki-Tavi 7,4692525 silver badges5050 bronze badges It's not necessary to take an entirely different route every time. There is the minimum energy transfer, that you mentioned, but you can use more fuel(or better yet, use a more efficient engine) to take a similar, but different route that takes you there faster. The raise in energy requirement is moderate at first, but becomes uncomfortable if you want to cut your travel time by more than a month or so. I think there is a table in a book I have at home ("Astronautics" by Ulrich Walter). I will let you know. Also, there are entirely different courses you can take, but I don't think any of them are faster. For example, in some constellations a Venus flyby gives you the chance to return after only a short stay at Mars, rather than waiting out a full cycle. Edit: I misremembered. In Astronautics I found the following two figures (which are even better; read below): http://www.jealousjellyfish.de/stackeximages/hohman_orbit.png The image above makes three examples of orbits we could fly. You could fly even more extreme courses than the 70 day one, but you can see how the required earth escape speed $$v_\infty$$ becomes uncomfortable. http://www.jealousjellyfish.de/stackeximages/transfer_speed.png This graph gives us the the same in exact numerical terms. You can see that shortening the time a bit is quite cheap, but then it quickly becomes expensive. $$t_x$$ is the transfer time. $$t_H$$ is the time for the Hohmann-transfer. I won't explain all variables involved, I don't think anyone has the patience to read that. It is true that you also need an burn to turn into Mars' orbit, but I disagree with HopDavid about this being the major factor: The $$\Delta v$$ for the for the injection is $$v_\infty\sqrt{\frac{1-e}{2}}$$ where $$e$$ is the eccentricity of the orbit around mars and $$v_\infty$$ is the excess velocity with regard to mars now. This can be very high, because you can slowly circularize the orbit for free by aerobraking. For $$e=0.95$$ the Mars injection takes only 16% of the $$\Delta v$$ you needed to get away from earth$$v_\infty$$. Of course, there is a limit to how eccentric your orbit can be, because you may escape Mars' sphere of influence if you come in too fast. It's not necessary to take an entirely different route every time. There is the minimum energy transfer, that you mentioned, but you can use more fuel(or better yet, use a more efficient engine) to take a similar, but different route that takes you there faster. The raise in energy requirement is moderate at first, but becomes uncomfortable if you want to cut your travel time by more than a month or so. I think there is a table in a book I have at home ("Astronautics" by Ulrich Walter). I will let you know. Also, there are entirely different courses you can take, but I don't think any of them are faster. For example, in some constellations a Venus flyby gives you the chance to return after only a short stay at Mars, rather than waiting out a full cycle. Edit: I misremembered. In Astronautics I found the following two figures (which are even better; read below): http://www.jealousjellyfish.de/stackeximages/hohman_orbit.png The image above makes three examples of orbits we could fly. You could fly even more extreme courses than the 70 day one, but you can see how the required earth escape speed $$v_\infty$$ becomes uncomfortable. http://www.jealousjellyfish.de/stackeximages/transfer_speed.png This graph gives us the the same in exact numerical terms. You can see that shortening the time a bit is quite cheap, but then it quickly becomes expensive. $$t_x$$ is the transfer time. $$t_H$$ is the time for the Hohmann-transfer. I won't explain all variables involved, I don't think anyone has the patience to read that. It is true that you also need an burn to turn into Mars' orbit, but I disagree with HopDavid about this being the major factor: The $$\Delta v$$ for the for the injection is $$v_\infty\sqrt{\frac{1-e}{2}}$$ where $$e$$ is the eccentricity of the orbit around mars. This can be very high, because you can slowly circularize the orbit for free by aerobraking. For $$e=0.95$$ the Mars injection takes only 16% of the $$\Delta v$$ you needed to get away from earth. Of course, there is a limit to how eccentric your orbit can be, because you may escape Mars' sphere of influence if you come in too fast. It's not necessary to take an entirely different route every time. There is the minimum energy transfer, that you mentioned, but you can use more fuel(or better yet, use a more efficient engine) to take a similar, but different route that takes you there faster. The raise in energy requirement is moderate at first, but becomes uncomfortable if you want to cut your travel time by more than a month or so. I think there is a table in a book I have at home ("Astronautics" by Ulrich Walter). I will let you know. Also, there are entirely different courses you can take, but I don't think any of them are faster. For example, in some constellations a Venus flyby gives you the chance to return after only a short stay at Mars, rather than waiting out a full cycle. Edit: I misremembered. In Astronautics I found the following two figures (which are even better; read below): http://www.jealousjellyfish.de/stackeximages/hohman_orbit.png The image above makes three examples of orbits we could fly. You could fly even more extreme courses than the 70 day one, but you can see how the required earth escape speed $$v_\infty$$ becomes uncomfortable. http://www.jealousjellyfish.de/stackeximages/transfer_speed.png This graph gives us the the same in exact numerical terms. You can see that shortening the time a bit is quite cheap, but then it quickly becomes expensive. $$t_x$$ is the transfer time. $$t_H$$ is the time for the Hohmann-transfer. I won't explain all variables involved, I don't think anyone has the patience to read that. It is true that you also need an burn to turn into Mars' orbit, but I disagree with HopDavid about this being the major factor: The $$\Delta v$$ for the for the injection is $$v_\infty\sqrt{\frac{1-e}{2}}$$ where $$e$$ is the eccentricity of the orbit around mars and $$v_\infty$$ is the excess velocity with regard to mars now. This can be very high, because you can slowly circularize the orbit for free by aerobraking. For $$e=0.95$$ the Mars injection takes only 16% of $$v_\infty$$. Of course, there is a limit to how eccentric your orbit can be, because you may escape Mars' sphere of influence if you come in too fast. 2 added 1405 characters in body edited Aug 26 '14 at 23:47 Rikki-Tikki-Tavi 7,4692525 silver badges5050 bronze badges 1 answered Aug 26 '14 at 11:06 Rikki-Tikki-Tavi 7,4692525 silver badges5050 bronze badges