Miss Dada  🏳️‍⚧️

I'm not going to go into detail about the Oberth effect, but this particular case is easy to think about. Your spaceship's engine makes you go faster or slower the same amount no matter what. If you're in a *really* high orbit, you're going very slowly

You can think about it like throwing a ball high up, almost straight up but not quite. That's basically what your orbit looks like right before you escape the solar system. So at the highest point, you're going only a few hundred meters per second.

Which means that completely stopping is trivial. And if you're not moving, you're not in orbit. So you will fall straight down (what goes up must come down after all. The only difference is now you're not moving fast enough to miss the ground).

So if you answered "orbit of mercury" thinking about this exact maneuver, you were right. But it's not how I meant the question. The point of all of this is that most folks don't realize just how fast we are moving around the sun right now.

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)

Anyway, for anyone who I lost 20 tweets ago... The answer is the sun. Parker solar probe is incredible, its mission is insanely ambitious, and probably harder than anything else we'll do in our lifetimes. You should go read more about it.

For the KSP players who knew the insane Δv requirements of Mercury and also remembered that bi-elliptic transfers exist (unlike me when making this poll), you are correct. If you don't care about perihelion, that is the hardest place in this list to reach.

For those with physics degrees who are noticing that my aside about Oberth would interact with the Sun's velocity, and I should have considered Earth/AC transfer from galactic orbit by my own rules... I couldn't find enough concrete numbers to do the math.

All I could really find is that we are travelling 20km/s relative to stars in our local neighborhood, which may or may not apply to AC, and by itself doesn't imply a large enough maneuver to beat orbit of Mercury. I would love to be proven wrong here though!

I'm done. Folks with physics degrees, you can @ me now. I'm sorry for subtweeting you so hard.