3 ms·

I assume we use distance as proxy for “how hard is to get there” (time and energy needed). Actual momentary distance is a bit tangential, because you never move

I assume we use distance as proxy for “how hard is to get there” (time and energy needed). Actual momentary distance is a bit tangential, because you never move in a straight line.

Dylan16807 2y ago
From low Earth orbit, transfer orbits to Venus and Mars take 3.5 and 3.6 km/s of thrust respectively. According to https://en.wikipedia.org/wiki/Hohmann_transfer_orbit https://en.wikipedia.org/wiki/Hohmann_transfer_orbit
- pwatsonwailes 2y ago
  The problem with a Hohmann transfer orbit is that you rely on launch windows. You can only do it when the two planets are in a straight line from the sun (or more accurately, and the same equivalent points in each of their orbits at the same time). This is a pain as Earth-Mars launch windows only come along every two years. If you want to launch without that issue, you're then into the world of using more delta-v, which is catastrophically expensive, unless you've got some form of propulsion for your craft beyond what we've got now. So yes, but with caveats. Obviously, there's a giant pile of detail beyond this in terms of the calculations for trading off time of flight vs distance between bodies, vs delta-v to accelerate towards the body, and then accelerate away from it to slow your descent into an orbital trajectory, but that's more depth than I'm going into for a quick HN comment.
  - Dylan16807 2y ago
    If you're jamming the throttle then the difference between Venus and Mars is also not very much.
pwatsonwailes 2y ago
Distance and time are two of the fundamental constraints for how much delta-v you're going to need to move from point A to point B, with the third being how you control your velocity when you arrive. The other child comment and my answer to them goes into this at about the highest level possible.

3 ms·

I assume we use distance as proxy for “how hard is to get there” (time and energy needed). Actual momentary distance is a bit tangential, because you never move

by stoperaticless 2y ago

I assume we use distance as proxy for “how hard is to get there” (time and energy needed). Actual momentary distance is a bit tangential, because you never move in a straight line.

Dylan16807 2y ago
From low Earth orbit, transfer orbits to Venus and Mars take 3.5 and 3.6 km/s of thrust respectively. According to https://en.wikipedia.org/wiki/Hohmann_transfer_orbit https://en.wikipedia.org/wiki/Hohmann_transfer_orbit
- pwatsonwailes 2y ago
  The problem with a Hohmann transfer orbit is that you rely on launch windows. You can only do it when the two planets are in a straight line from the sun (or more accurately, and the same equivalent points in each of their orbits at the same time). This is a pain as Earth-Mars launch windows only come along every two years. If you want to launch without that issue, you're then into the world of using more delta-v, which is catastrophically expensive, unless you've got some form of propulsion for your craft beyond what we've got now. So yes, but with caveats. Obviously, there's a giant pile of detail beyond this in terms of the calculations for trading off time of flight vs distance between bodies, vs delta-v to accelerate towards the body, and then accelerate away from it to slow your descent into an orbital trajectory, but that's more depth than I'm going into for a quick HN comment.
  - Dylan16807 2y ago
    If you're jamming the throttle then the difference between Venus and Mars is also not very much.
pwatsonwailes 2y ago
Distance and time are two of the fundamental constraints for how much delta-v you're going to need to move from point A to point B, with the third being how you control your velocity when you arrive. The other child comment and my answer to them goes into this at about the highest level possible.