Miss Dada  🏳️‍⚧️

You are currently moving at nearly 30 km/s around the sun. That is so fast that you need to go barely 25% faster to leave the solar system and never come back. Cancelling that velocity is *really* hard.

Now that's not to say that reaching Mercury isn't hard. It is. Mercury is going even faster than we are (nearly 50 km/s relative to the sun!) Only one satellite in history has orbited Mercury, which was MESSENGER in 2004. It needed 4 gravity assists to do so (1 earth 3 venus)

I phrased the question as "requires the least energy" specifically to avoid clever gravity assists. In theory, you can reach all three destinations just by getting to the moon. Enough gravity assists there gets you to solar orbit, earth to venus, venus to jupiter, etc.

But that phrasing was nonsense because so much of this is kinetic vs potential energy, which doesn't really work with that phrasing. A lot of folks asked "orbiting mercury or reaching a similar orbit of Mercury?" As I phrased it it actually doesn't matter.

Ultimately the only difference between orbiting a planet and being in the same solar orbit as that planet is whether you're next to it or not. While orbiting Mercury might take less Δv than getting to the same solar orbit, the laws of physics say that can't take less energy.

The difference here is ultimately that when performing orbital insertion, you're stealing energy from Mercury relative to the sun, slowing it down slightly. That's where the difference in Δv costs comes from. (Physics degrees don't @ me about potential energy plz)

This is also why all of this is cheaper if you start from Earth instead of heliocentric orbit. When considering only 2 bodies, you're just manipulating your exhaust's kinetic energy to go faster. When considering 3 bodies, you're slowing down Earth relative to the Sun.

You can remove this from the equation by just starting from a heliocentric orbit with the same SMA as Earth, and looking at the Δv required to get to 142Mmx29 vs 58x58. You'll see that getting to the sun still takes far more Δv. (it's close though, the plane change might tip it)