5 ms·

Most of the "velocity change" in your math is done by the Sun as heliocentric potential energy converts to kinetic energy or vice versa. The minimum energy traj

by DavidSJ 13y ago

Most of the "velocity change" in your math is done by the Sun as heliocentric potential energy converts to kinetic energy or vice versa. The minimum energy trajectory from Earth to Mercury is about the same delta V as Earth to Mars. The problem with Mercury is you can't aerobrake, so add either a Venus slingshot or significantly increased delta V requirements to decelerate and land.

NamTaf 13y ago
Sure, but you still have to shed that kinetic energy, which is what I was getting at the end. It's no use gaining a huge amount of kinetic energy if it just means you're on hyperbolic trajectory past the body. You need to still shed all of that energy with (traditionally) rocket thrust or equivalent which is completely equivalent to accelerating by that amount. I was also basing it off a standard Hohmann transfer because sling-shotting and all of that jazz was out of my reach without my notes :) I am sure you can get a lower delta-v transfer from more exotic paths than just a Hohmann transfer but it's been a few years since I've crunched those numbers! Nevertheless, aerobraking is one option of shedding energy when approaching smaller-orbit bodies but I neglected it when doing this first-pass analysis. However I agree with you that approaching smaller orbits gives you benefits that approaching larger orbits does not, by way of using the Sun's gravitational well.
- DavidSJ 13y ago
  My point is simply that you can't subtract the velocities of the two planets and call that your delta V. Delta V is the amount of actual powered acceleration your rocket does, which is only very indirectly related to the differences in velocities between the origin and destination planets. For example, to get from an Earth-like solar orbit to an asteroid orbiting the sun very far away, at nearly 0 velocity, you'd need nearly solar escape velocity at 1 AU, or about 42 km/sec, minus the Earth's 30 km/sec. Even without taking advantage of gravitational slingshots or the Oberth effect, the delta V requirement is 12 km/sec, not 30 km/sec. So subtracting velocities has nothing to do with delta V.

by DavidSJ 13y ago

NamTaf 13y ago

Sure, but you still have to shed that kinetic energy, which is what I was getting at the end. It's no use gaining a huge amount of kinetic energy if it just means you're on hyperbolic trajectory past the body. You need to still shed all of that energy with (traditionally) rocket thrust or equivalent which is completely equivalent to accelerating by that amount. I was also basing it off a standard Hohmann transfer because sling-shotting and all of that jazz was out of my reach without my notes :) I am sure you can get a lower delta-v transfer from more exotic paths than just a Hohmann transfer but it's been a few years since I've crunched those numbers! Nevertheless, aerobraking is one option of shedding energy when approaching smaller-orbit bodies but I neglected it when doing this first-pass analysis. However I agree with you that approaching smaller orbits gives you benefits that approaching larger orbits does not, by way of using the Sun's gravitational well.

DavidSJ 13y ago
My point is simply that you can't subtract the velocities of the two planets and call that your delta V. Delta V is the amount of actual powered acceleration your rocket does, which is only very indirectly related to the differences in velocities between the origin and destination planets. For example, to get from an Earth-like solar orbit to an asteroid orbiting the sun very far away, at nearly 0 velocity, you'd need nearly solar escape velocity at 1 AU, or about 42 km/sec, minus the Earth's 30 km/sec. Even without taking advantage of gravitational slingshots or the Oberth effect, the delta V requirement is 12 km/sec, not 30 km/sec. So subtracting velocities has nothing to do with delta V.