I decided to do a bunch of numbers about getting off each respective rock:
Using a = G * M/r^2, we get the following:
For Mercury:
AMercury = G * 3.29e23/2.44e6 = 3.68 m/s^2
ASun@Merc = G * 1.99e30/5.79e10 = 0.039 m/s^2
Atotal (if you launch off the dark side) = 3.68 + 0.0039 = 3.719 m/s^2
For Earth:
AEarth = G * 5.97e24/6.37e6 = 9.81 m/s^2 (duhh)
ASun@Earth = G * 1.99e30/1.49e11 = 0.0059 m/s^2
ATotal (off the dark side) = 9.81 + 0.0059 = 9.816 m/s^2
So even if you launch the 'hard way' from both planets (shooting away from the Sun), the gravity well you're in to get off Mercury is less (37%) than that of Earth, thanks to Mercury's far smaller mass. This isn't surprising, given that the local body completely dwarfs the influence of the Sun in both cases.
I don't have my lecture notes at work so I can't do it fully, but off the top of my head this is a rough first pass at the relative difficulty of transfers between each planet:
Earth orbital velocity: 29.78 km/s
Mars orbital velocity: 24.08 km/s
Mercury orbital velocity: 47.87 km/s
Venus orbital velocity: 35.02 km/s
d(Earth-Mars) = 5.70 km/s
d(Earth-Venus) = 5.24 km/s
d(Earth-Mercury) = 18.09 km/s
So in short, you need to change your velocity by 3x - 3.5x as much to get between Earth-Mercury, as you would between Earth-Mars or Earth-Venus (which are quite similar). Given that kinetic energy = 0.5mv^2, that's a 9x - 12.25x factor of energy to get to Mercury vs. the other two.
In summary, to get off Mercury is easy compared to Earth (duhh) and the Sun doesn't make any difference there. To get between the planets however is a huge difference and will be the limiting factor on regular Earth-Mercury transfers of matter.
Regardless of which way you go (towards or away from the Sun), you need to either shed or add the respective velocities to change from a circular orbit at Earth's orbit to a circular orbit at the other bodies' orbits.
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.
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.
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.
Using a = G * M/r^2, we get the following: For Mercury:
AMercury = G * 3.29e23/2.44e6 = 3.68 m/s^2
ASun@Merc = G * 1.99e30/5.79e10 = 0.039 m/s^2
Atotal (if you launch off the dark side) = 3.68 + 0.0039 = 3.719 m/s^2
For Earth:
AEarth = G * 5.97e24/6.37e6 = 9.81 m/s^2 (duhh)
ASun@Earth = G * 1.99e30/1.49e11 = 0.0059 m/s^2
ATotal (off the dark side) = 9.81 + 0.0059 = 9.816 m/s^2
So even if you launch the 'hard way' from both planets (shooting away from the Sun), the gravity well you're in to get off Mercury is less (37%) than that of Earth, thanks to Mercury's far smaller mass. This isn't surprising, given that the local body completely dwarfs the influence of the Sun in both cases.
I don't have my lecture notes at work so I can't do it fully, but off the top of my head this is a rough first pass at the relative difficulty of transfers between each planet:
Earth orbital velocity: 29.78 km/s
Mars orbital velocity: 24.08 km/s
Mercury orbital velocity: 47.87 km/s
Venus orbital velocity: 35.02 km/s
d(Earth-Mars) = 5.70 km/s
d(Earth-Venus) = 5.24 km/s
d(Earth-Mercury) = 18.09 km/s
So in short, you need to change your velocity by 3x - 3.5x as much to get between Earth-Mercury, as you would between Earth-Mars or Earth-Venus (which are quite similar). Given that kinetic energy = 0.5mv^2, that's a 9x - 12.25x factor of energy to get to Mercury vs. the other two.
In summary, to get off Mercury is easy compared to Earth (duhh) and the Sun doesn't make any difference there. To get between the planets however is a huge difference and will be the limiting factor on regular Earth-Mercury transfers of matter.
Regardless of which way you go (towards or away from the Sun), you need to either shed or add the respective velocities to change from a circular orbit at Earth's orbit to a circular orbit at the other bodies' orbits.