[jessriedel]

The gravity of the Earth absolutely changes the speed of the probe.

In terms of the basic momentum transfers, non-propulsive gravity assists are essentially the same as elastic collisions with balls of non-equal mass. In particular, energy can be transferred, and that is mediated by the interaction forces: if a very heavy ball is rolling along at speed v and I place a tiny ball at rest in front of it, the tiny ball will bounce off at about 2v. We could certainly say “the atomic forces between the heavy ball and the tiny ball during the collision propel the tiny ball to its new destination”. This is true even though the tiny ball’s speed is constant in the center-of-mass frame.

[JumpCrisscross]

> non-propulsive gravity assists are essentially the same as elastic collisions with balls of non-equal mass

From the Earth’s frame of reference there is no change in delta-v other than a change in direction. It’s only from the Sun’s frame of reference that there is velocity added in the speed component (v_infinity, commonly). If you can find a single measurement to the contrary, that’s novel enough to be worth publishing.

That’s why you can’t gravity assist around the Sun to get around the Solar System faster.

[nick238]

I don't know what you mean by "gravity assist around the Sun":

1. Using the perihelion in an orbit "around the sun" as a gravity assist?: spacecraft usually care about their speed relative to the sun (characteristic energy, C3), and a (free) gravity assist around the sun won't do much. Dropping close to the sun to perform a powered bi-elliptic transfer could be a thing if you wanted to travel extreme distances (e.g. put a telescope at 500 AU to use the solar gravitational lens)

2. Using other bodies that are "around the sun" to get a gravity assist?: spacecraft do this all the time.

Also "get around the solar system faster":

1. Decrease the orbital period (lower orbits orbit faster): This is exactly what Messenger and Parker Solar Probe is doing flying by Venus/Mercury. They're 'bouncing' off of the planets, trading orbital energy and raising the planets' orbit around the sun while dropping their own.

2. Get to places faster: This is what outer planets probes (Voyagers 1/2, Cassini, New Horizons) do. If Jupiter wasn't there, these missions might not be possible.

[JumpCrisscross]

> If Jupiter wasn't there, these missions might not be possible

If Jupiter weren’t there and moving relative to the destinations. The gravity isn’t the critical piece, it’s the relative motion.

Just having a massive object does nothing because gravity isn’t doing any work, it’s just coupling you to a moving object.

[nick238]

Oh, sorry, I didn't specify which Jupiter I was referring to, the real, moving Jupiter that orbits the sun vs. the stationary, straw-man Jupiter that jumps out at you in bad faith retorts.

[JumpCrisscross]

> I didn't specify which Jupiter I was referring to, the real, moving Jupiter that orbits the sun

This is why I said you can't gravity assist around the Sun to travel around the Solar System. The Sun is moving around the galaxy at a terrific speed. But so is the Solar System. Dropping into and out of the Sun's gravity well does nothing other than change your trajectory.

[ahazred8ta]

If you fire rockets at closest approach to get more delta-V, that's an Oberth Maneuver - https://en.wikipedia.org/wiki/Oberth_effect

And as you say, an unpowered planetary flyby is a gravity assist.

[bgirard]

> From the Earth’s frame of reference there is no change in delta-v other than a change in direction.

But no one is talking about the Earth's frame of reference.

> But “the gravity of Earth” imparts no net delta-v and wouldn’t on its own allow the craft to reach Venus.

That's statement is untrue. Gravity assists with planets can provide net delta-v allowing spacecrafts to reach other planets. See the Voyager 2 gravity assists for one example.

[JumpCrisscross]

> Gravity assists with planets can provide net delta-v allowing spacecrafts to reach other planets. See the Voyager 2 gravity assists for one example

If you were on Jupiter measuring Voyager’s speed before and after it interacted with Jupiter, you wouldn’t measure a net effect. It’s only if you’re standing on the Sun (or somewhere else where you can see Jupiter revolving) that you see Jupiter “pull” the spacecraft along, thereby imparting velocity.

Gravity doesn’t do any work. The gravitational potential energy of the Voyager-Jovian system is entirely conserved in a flyby. Jupiter’s orbital energy about the Sun is what’s stolen.

This is a common misconception when it comes to gravity assists. It’s why I think that language could be tighter.

[bgirard]

> If you were on Jupiter measuring Voyager’s speed before and after it interacted with Jupiter, you wouldn’t measure a net effect.

Again that's not something I'm claiming.

[saagarjha]

You’ve retreated to a different argument nobody was making.

[JumpCrisscross]

> You’ve retreated to a different argument nobody was making

Respectfully, I have a background in aerospace engineering. This is the number one popular misconception about gravity assists/slingshot maneuvres.

[nick238]

If jessriedel hadn't said

> The gravity of the Earth absolutely changes the speed of the probe.

but instead

> The gravity of the Earth absolutely changes the characteristic energy with respect to the sun of the probe.

Would you be happy?

[JumpCrisscross]

No. The gravity of the Earth doesn't do any work on the probe. It's a coupling mechanism. The work that is done is by Earth's orbital energy around the Sun. That is the energy that is less after the spacecraft encounters the Earth than before. The Earth is "dragging" the spacecraft along with it around the Sun. Not trebucheting it.

[jessriedel]

The gravity of the Earth absolutely does net work done on the probe. That’s an unambiguous mathematical statement. It’s just that this work is frame dependent. It sounds like you want it to frame-independent and to be given by the value in the center-of-mass frame. But it’s not.

[nick238]

How about:

> The gravitational coupling of the probe to the Earth absolutely changes the characteristic energy with respect to the sun of the probe.

[jessriedel]

Your first paragraph is exactly what I already said in my comment.

[adolph]

> The gravity of the Earth absolutely changes the speed of the probe.

Wait! Wouldn't Earth's gravity take away when departing just as much as given when arriving? However, the probe's direction could change based on how close it passes Earth.

As the probe passes Earth, a mass proportionate amount of Earth's velocity would be shared to the probe. I have a distant grade-school memory of an analogy of two people on a roller-skate rink. The passing and passed persons link hands and some of the passed person's velocity is emparted to the passing person's velocity.

[mgsouth]

The magnitude of the probe's "average" velocity relative to the Earth-object barycenter remains the same. If the universe was just the Earth and it, then you'd see the object making nice ellipical orbits around the Earth, and the Earth wobbling a bit.

However, the barycenter is moving relative to Venus. Imagine just the three things--the Earth, Venus, and this little object. Now imagine the object is coming almost directly from Venus, loops in a tight ellipse around the Earth, and goes shooting back almost directly towards Venus. The velocity relative to Venus changes enormously. Even if you're just concerned with the magnitude, some of the Earth - Venus relative motion gets added to the probe. Think bouncing a rubber ball against a wall that's moving towards you. The wall slows down a tiny amount, and almost all of the wall's velocity is added to the ball when it shoots back towards you.

[adolph]

That is a nice explanation. However, it fails to answer the question posed to the assertion regarding gravity:

  >> The gravity of the Earth absolutely changes the speed of the probe.

  > Wait! Wouldn't Earth's gravity take away when departing just as much as given when arriving?

As I understand your contribution, it is congruent with Earth's velocity altering the speed of the probe, not Earth's gravity.

The flyby of Earth reduced Juice’s speed by 4.8 km/s relative to the Sun, guiding Juice onto a new trajectory towards Venus. Overall, the lunar-Earth flyby deflected Juice by an angle of 100° compared to its pre-flyby path.

I can see how your characterization of "goes shooting back" doesn't mean 180° change, but a change relative to where Venus will be as the probe interacts with Earth and arrives back to Venus.

I have a hard time understanding how the flyby fits in the overall plan however. This is the first Earth flyby, right? "Flybys en route: August 2024 Lunar-Earth, August 2025 Venus, September 2026 Earth, January 2029 Earth" [0] If so, the probe is not going "back" to Venus, because it hasn't been there yet. It has been on an orbit of a different ellipse than Earth and so this flyby is where it takes a left turn to head towards Venus for the first time.

As JUICE starts its first elliptical solar orbit, its distance to the Sun decreases. This results in an increase in speed—according to Kepler's second law of planetary motion—and the spacecraft overtakes Earth. [1]

0. https://www.esa.int/Science_Exploration/Space_Science/Juice

1. https://sci.esa.int/web/juice/-/58815-juices-journey-to-jupi...

[mgsouth]

Well, it's the gravity that alters the spacecraft's trajectory. If the Earth was massless then the craft would just sail by.

Here's a nice short-ish web page from NASA with nice details [1].

Yeah, "back" seems a little wonky. Maybe somebody got confused by earlier plans? Your second link, to the ESA video, is from 2017 and evidently isn't what they wound up with. According to that video the Vensus flyby was supposed to have been last October, and next encounter would be Mars. Or maybe they're thinking "back down to Venus's orbit."

[1] https://science.nasa.gov/learn/basics-of-space-flight/primer...

[jessriedel]

> Wouldn't Earth's gravity take away when departing just as much as given when arriving?

Only in the Earth’s center-of-mass frame. In the solar system rest frame, the probe leaves with a different speed than it entered with.

> However, the probe's direction could change based on how close it passes Earth.

Both the magnitude and the direction of the velocity vector change.

[wongarsu]

Everything you gain from gravity you have to give back on the way out. You can change the shape of your orbit that way, for example rotating it or making it more elliptical. But if we are talking about gaining energy from a gravity assist it would be more accurate to describe that as stealing some of the rotational energy of the planet.

[jessriedel]

> Everything you gain from gravity you have to give back on the way out.

This is true only in the center-of-mass frame, and that's true for all conservative forces.

If I have a bowling ball covered in springs and I throw it at a marble at speed v, the marble will traveling at a speed 2v after the collision. It would be completely correct and not misleading to say "the springs on the bowling ball accelerated the marble to 2v" even though the marble's speed in the center-of-mass frame is the same before and after the collision.

It's also true that the energy gained by the marble comes from the bowling ball in the rest frame. That doesn't make the first statement wrong or misleading. It just means that you like thinking about things in the center-of-mass frame.

[JumpCrisscross]

> would be completely correct and not misleading to say "the springs on the bowling ball accelerated the marble to 2v"

I suppose this is where I disagree. The springs transferred momentum. Your throw (and the resulting motion of the bowling ball) did the work. (Tyres don't accelerate a car, its engine does.)

You could slingshot around a moving body magnetically, the fundamental principles remain the same. The effect of a gravity assist comes from the motion of the body, not gravity per se.

[jessriedel]

“Work done” is the integral of a force.

https://en.wikipedia.org/wiki/Work_(physics)