<p>Your initial analysis is correct <strong>for a two body system</strong>.</p>
<p>In a 3 or more body system then yes, the total energy must be conserved, but how that is partitioned between the bodies can be varied by their mutual interactions.</p>
<p>e.g. Take a triple system where two objects orbit close together and a third one swings in on a highly eccentric (but still bound) orbit. The result of its interaction with the central pair could be that the central pair get more tightly bound - thus their orbital energy, <span class="math-container">$-GM_1M_2/a$</span> becomes more negative. To conserve energy, the outer object must become <em>less</em> bound and may even pick up enough kinetic energy to escape altogether.</p>

CC BY-SA 4.0
2024-11-18T18:52:37.373Z
So the extra energy came from the 3rd object that perturbed the protoplanet?
Miss Understands
CC BY-SA 4.0
2024-11-18T18:53:58.960Z
BTW, If it matters, your answers have been more helpful than any others at stack exchange.
Miss Understands
CC BY-SA 4.0
2024-11-18T23:04:13.700Z
Yeah, basically.  You can't really have a planetoid escape without a third body to provide the extra energy, unless the primary explodes or some such.  Most theories of how planets would get stripped off a star involve a second star swinging by to provide the enormous quantities of energy necessary to catapult a whole planet beyond escape velocity. (Which could be a binary that acts to lengthen and destabilize the orbit over millions of years, or could be a one time pass that just sends it blasting off into the void.)
Darth Pseudonym
CC BY-SA 4.0
2024-11-19T16:14:35.150Z
This is like the gravitational assist that some space probes use to reach the outer solar system, right?
Barmar
CC BY-SA 4.0
2024-11-19T16:24:38.730Z
@Barmar yes it is. Total energy is conserved but is shared out among the participating bodies in a different way.
ProfRob
CC BY-SA 4.0
2024-11-19T23:49:26.127Z
@MissUnderstands Any interaction that your main object (protoplanet) has with a third body (i.e. not the Sun) is effectively an orbital slingshot, also known as a [gravity assist](https://en.wikipedia.org/wiki/Gravity_assist). Based on the specific interaction between your main and third body, energy can be transferred between the two bodies (increase or decrease, depending on their relative positions as they come near each other). This _can_ add more kinetic energy to your protoplanet, enough so that it can overcome the Sun's gravity well, which it couldn't without the slingshot.
Flater
CC BY-SA 4.0
2024-11-18T18:41:47.723Z
2024-11-18T18:41:47.723Z
ProfRob
<p>You are correct in observing that, in the absence of third bodies or non-gravitational interactions, an object initially at standstill relative to the Sun and falling towards it under gravity can never escape the Sun's gravity well.</p>
<p>This statement has two major caveats, however:</p>
<ol>
<li>it only holds if the object is initially at standstill, and</li>
<li>it only holds in the Newtonian two-body model of gravity, where any third bodies or non-gravitational interactions are neglected.</li>
</ol>
<p>If either of these assumptions does not hold (and in practice they never do), it is possible for the object to escape the Sun's gravity.</p>
<p>In particular, an object on a <a href="https://en.wikipedia.org/wiki/Hyperbolic_trajectory" rel="noreferrer">hyperbolic trajectory</a> relative to the Sun can escape the Sun's gravity well because its <a href="https://en.wikipedia.org/wiki/Characteristic_energy" rel="noreferrer">characteristic energy</a> is positive, meaning that it was never gravitationally bound to the Sun in the first place.  Such an object is moving fast enough that, even though the Sun's gravity slows it down as it flies away, it never comes to a complete stop and starts falling in again.</p>
<p>Since the dynamics of the system are time-reversible, this also implies that the object was never (at least during the time period where this two-body approximation is valid) at a standstill relative to the Sun, but must've been moving towards the Sun even before the encounter.  But there's nothing unusual about that; very few if any objects in the universe are ever at a complete standstill relative to each other.  Indeed, any object that approaches the Sun from far away (where the Sun's gravity is negligible) must, by definition, have a non-zero initial velocity vector pointing at least approximately towards the Sun.  If it did not, the object would never get close to the Sun in the first place.  Thus, the fly-by trajectory of such an object is typically hyperbolic.</p>
<p>The other big assumption, of course, is that we can neglect the gravitational influence of any bodies other that the object and the Sun, as well as any non-gravitational interactions.  This is also never actually true.  In particular, the galaxy has plenty of other stars besides the Sun, and anywhere more than a couple of lightyears from the Sun's vicinity their gravitational pull is stronger than the Sun's, and thus cannot be neglected.</p>
<p>Even inside the solar system there are planets whose gravity may perturb the trajectory of an object flying past them, potentially converting a bound elliptic orbit into an unbound hyperbolic trajectory or vice versa.  And of course non-gravitational interactions cannot be entirely neglected either.  In particular, an object that started out <em>exactly</em> at standstill relative to the Sun would eventually interact non-gravitationally with the Sun by falling into it.  To actually stay in a bound orbit around the Sun, an object needs to have <em>some</em> initial (transverse) velocity even at aphelion — but not too much, or it will fly away from the Sun on a hyperbolic trajectory.</p>
<p>One important aspect to realize is that the further you get from the Sun, the weaker the Sun's gravitational pull is, and thus the smaller the <a href="https://en.wikipedia.org/wiki/Escape_velocity" rel="noreferrer">velocity needed to escape</a> the Sun's gravity well entirely.  Out in the <a href="https://en.wikipedia.org/wiki/Oort_cloud" rel="noreferrer">Oort cloud</a> it takes less than 1 km/s to escape the Sun's gravity, which also means that any objects that <em>are</em> in bound orbits around the Sun that far out must be moving slower than that.  Anything moving faster than that relative to the Sun that far away from it will escape the Sun's gravity well, either directly or after a hyperbolic fly-by.</p>

CC BY-SA 4.0
2024-11-19T17:28:42.193Z
2024-11-19T17:28:42.193Z
Ilmari Karonen
<p>I should know this, and it scares me that I don't. How is it ever possible for anything to gain enough speed from falling toward the sun to be thrown out of the solar system?</p>
<p>When an object falls from a great distance and swings around the sun,it can only have the energy had gained from falling toward the sun.</p>
<p>Right?</p>
<p>But isn't that exactly the same as the potential energy it had when it was in orbit way out in the cold?</p>
<p>How could it gain enough extra energy to be thrown farther than it already was when it started out?  Isn't the maximum distance it could gain from falling towards the sun and swinging around exactly equal to its original distance?</p>
<p>I know that a planetary flyby craft gains more energy than it had because when Voyager swung around Jupiter, Jupiter slowed down and fell slightly closer toward the sun. But in the present case, the sun isn't moving.</p>
<p>And I also know that the maximum acceleration is proportional to how close to the surface you can get. If the sun had the same mass but was a neutron star, you could swing VERY close to it and gain a huge amount of energy. But can you ever gain more energy than you had stored in the potential energy of your distance?</p>
<p>This certainly isn't a mass into energy issue.  And it's not a special relativity issue, otherwise I'd understand it. SR explains where the extra energy comes from when the ball jumps from my hand and slams against the ground. When that happened one time when I was a little kid, it made me giggle, and I vowed that someday I'd understand it.</p>
<p>Now I do. The energy is subtracted from the object's momentum through time. That's what causes time dilation. The falling object is experiencing imaginary rotation in 4 dimensions.</p>
<p>Okay, cool. But there's only a finite amount of energy that can be converted that way, because it's the energy that was stored by virtue of the  distance between the two objects.</p>
<p>I'm overlooking something really obvious. Something is just not occurring to me. But what is it?</p>
<p>It IS the case that falling objects can be swung out of the solar system, right? I'm starting under wonder if maybe that's what I'm mistaken about. And the only thing worse than not understanding is being mistaken.</p>

CC BY-SA 4.0
2024-11-18T19:24:45.137Z
You may also be interested in the time-reversed problem, of an initially unbound body becoming bound to the Solar System. See https://astronomy.stackexchange.com/q/58245/16685 & https://astronomy.stackexchange.com/q/58419/16685
PM 2Ring
CC BY-SA 4.0
2024-11-18T19:37:40.257Z
@PM2Ring Yeah, that's another problem I was thinking about. I assume that objects falling from  far enough away can never be captured, like the Umagooma space turd that passed through our system.  and since almost everything is "far enough away," not to mention moving fast, it seems to me that everything in the solar system must have been there from the start. I'm tempted to ask that in another question, but instead I'll go to the link you gave.
Miss Understands
CC BY-SA 4.0
2024-11-18T19:41:50.687Z
@PM2Ring Yes, the answer that you provided to the other guy also answers my new question. Thank you!
Miss Understands
CC BY-SA 4.0
2024-11-19T13:11:47.440Z
Note that the reverse occurs by the same mechanism. Bodies starting outside the solar system would normally pick up too much speed to stay in orbit as they fall toward the sun, but in particular circumstances they can get captured.
Nuclear Hoagie
CC BY-SA 4.0
2024-11-19T15:04:03.790Z
BTW, a billion miles isn't *that* far, just a bit farther than Saturn.
PM 2Ring
CC BY-SA 4.0
2024-11-19T22:35:27.353Z
I really don't think special relativity could have any observable effect on a ball that you throw at the ground; something else is happening there. That might be worth a separate question?
JounceCracklePop
<orbit><gravity><orbital-mechanics>
When a protoplanet falls toward the Sun, even from a billion miles, how could it ever be flung out of the Solar System?
Miss Understands
# When a protoplanet falls toward the Sun, even from a billion miles, how could it ever be flung out of the Solar System?

(Asked by: username_0 on 2024-11-18)

I should know this, and it scares me that I don't. How is it ever possible for anything to gain enough speed from falling toward the sun to be thrown out of the solar system?

When an object falls from a great distance and swings around the sun,it can only have the energy had gained from falling toward the sun.

Right?

But isn't that exactly the same as the potential energy it had when it was in orbit way out in the cold?

How could it gain enough extra energy to be thrown farther than it already was when it started out? Isn't the maximum distance it could gain from falling towards the sun and swinging around exactly equal to its original distance?

I know that a planetary flyby craft gains more energy than it had because when Voyager swung around Jupiter, Jupiter slowed down and fell slightly closer toward the sun. But in the present case, the sun isn't moving.

And I also know that the maximum acceleration is proportional to how close to the surface you can get. If the sun had the same mass but was a neutron star, you could swing VERY close to it and gain a huge amount of energy. But can you ever gain more energy than you had stored in the potential energy of your distance?

This certainly isn't a mass into energy issue. And it's not a special relativity issue, otherwise I'd understand it. SR explains where the extra energy comes from when the ball jumps from my hand and slams against the ground. When that happened one time when I was a little kid, it made me giggle, and I vowed that someday I'd understand it.

Now I do. The energy is subtracted from the object's momentum through time. That's what causes time dilation. The falling object is experiencing imaginary rotation in 4 dimensions.

Okay, cool. But there's only a finite amount of energy that can be converted that way, because it's the energy that was stored by virtue of the distance between the two objects.

I'm overlooking something really obvious. Something is just not occurring to me. But what is it?

It IS the case that falling objects can be swung out of the solar system, right? I'm starting under wonder if maybe that's what I'm mistaken about. And the only thing worse than not understanding is being mistaken.

**Tags:** orbit, gravity, orbital-mechanics

> **Comments:**
> **username_1**: You may also be interested in the time-reversed problem, of an initially unbound body becoming bound to the Solar System. See https://astronomy.stackexchange.com/q/58245/16685 & https://astronomy.stackexchange.com/q/58419/16685  
> **username_0**: @username_1 Yeah, that's another problem I was thinking about. I assume that objects falling from  far enough away can never be captured, like the Umagooma space turd that passed through our system.  and since almost everything is "far enough away," not to mention moving fast, it seems to me that everything in the solar system must have been there from the start. I'm tempted to ask that in another question, but instead I'll go to the link you gave.  
> **username_0**: @username_1 Yes, the answer that you provided to the other guy also answers my new question. Thank you!  
> **username_2**: Note that the reverse occurs by the same mechanism. Bodies starting outside the solar system would normally pick up too much speed to stay in orbit as they fall toward the sun, but in particular circumstances they can get captured.  
> **username_1**: BTW, a billion miles isn't *that* far, just a bit farther than Saturn.  
> **username_3**: I really don't think special relativity could have any observable effect on a ball that you throw at the ground; something else is happening there. That might be worth a separate question?  

---

## 2 answers

### Answer by username_8 on 2024-11-19. Score: 15 | Accepted answer

You are correct in observing that, in the absence of third bodies or non-gravitational interactions, an object initially at standstill relative to the Sun and falling towards it under gravity can never escape the Sun's gravity well.

This statement has two major caveats, however:

1. it only holds if the object is initially at standstill, and
2. it only holds in the Newtonian two-body model of gravity, where any third bodies or non-gravitational interactions are neglected.

If either of these assumptions does not hold (and in practice they never do), it is possible for the object to escape the Sun's gravity.

In particular, an object on a [hyperbolic trajectory](https://en.wikipedia.org/wiki/Hyperbolic_trajectory) relative to the Sun can escape the Sun's gravity well because its [characteristic energy](https://en.wikipedia.org/wiki/Characteristic_energy) is positive, meaning that it was never gravitationally bound to the Sun in the first place. Such an object is moving fast enough that, even though the Sun's gravity slows it down as it flies away, it never comes to a complete stop and starts falling in again.

Since the dynamics of the system are time-reversible, this also implies that the object was never (at least during the time period where this two-body approximation is valid) at a standstill relative to the Sun, but must've been moving towards the Sun even before the encounter. But there's nothing unusual about that; very few if any objects in the universe are ever at a complete standstill relative to each other. Indeed, any object that approaches the Sun from far away (where the Sun's gravity is negligible) must, by definition, have a non-zero initial velocity vector pointing at least approximately towards the Sun. If it did not, the object would never get close to the Sun in the first place. Thus, the fly-by trajectory of such an object is typically hyperbolic.

The other big assumption, of course, is that we can neglect the gravitational influence of any bodies other that the object and the Sun, as well as any non-gravitational interactions. This is also never actually true. In particular, the galaxy has plenty of other stars besides the Sun, and anywhere more than a couple of lightyears from the Sun's vicinity their gravitational pull is stronger than the Sun's, and thus cannot be neglected.

Even inside the solar system there are planets whose gravity may perturb the trajectory of an object flying past them, potentially converting a bound elliptic orbit into an unbound hyperbolic trajectory or vice versa. And of course non-gravitational interactions cannot be entirely neglected either. In particular, an object that started out *exactly* at standstill relative to the Sun would eventually interact non-gravitationally with the Sun by falling into it. To actually stay in a bound orbit around the Sun, an object needs to have *some* initial (transverse) velocity even at aphelion — but not too much, or it will fly away from the Sun on a hyperbolic trajectory.

One important aspect to realize is that the further you get from the Sun, the weaker the Sun's gravitational pull is, and thus the smaller the [velocity needed to escape](https://en.wikipedia.org/wiki/Escape_velocity) the Sun's gravity well entirely. Out in the [Oort cloud](https://en.wikipedia.org/wiki/Oort_cloud) it takes less than 1 km/s to escape the Sun's gravity, which also means that any objects that *are* in bound orbits around the Sun that far out must be moving slower than that. Anything moving faster than that relative to the Sun that far away from it will escape the Sun's gravity well, either directly or after a hyperbolic fly-by.


---

### Answer by username_4 on 2024-11-18. Score: 32

Your initial analysis is correct **for a two body system**.

In a 3 or more body system then yes, the total energy must be conserved, but how that is partitioned between the bodies can be varied by their mutual interactions.

e.g. Take a triple system where two objects orbit close together and a third one swings in on a highly eccentric (but still bound) orbit. The result of its interaction with the central pair could be that the central pair get more tightly bound - thus their orbital energy, $-GM\_1M\_2/a$ becomes more negative. To conserve energy, the outer object must become *less* bound and may even pick up enough kinetic energy to escape altogether.

> **Comments:**
> **username_0**: So the extra energy came from the 3rd object that perturbed the protoplanet?  
> **username_0**: BTW, If it matters, your answers have been more helpful than any others at stack exchange.  
> **username_5**: Yeah, basically.  You can't really have a planetoid escape without a third body to provide the extra energy, unless the primary explodes or some such.  Most theories of how planets would get stripped off a star involve a second star swinging by to provide the enormous quantities of energy necessary to catapult a whole planet beyond escape velocity. (Which could be a binary that acts to lengthen and destabilize the orbit over millions of years, or could be a one time pass that just sends it blasting off into the void.)  
> **username_6**: This is like the gravitational assist that some space probes use to reach the outer solar system, right?  
> **username_4**: @username_6 yes it is. Total energy is conserved but is shared out among the participating bodies in a different way.  
> **username_7**: @username_0 Any interaction that your main object (protoplanet) has with a third body (i.e. not the Sun) is effectively an orbital slingshot, also known as a [gravity assist](https://en.wikipedia.org/wiki/Gravity_assist). Based on the specific interaction between your main and third body, energy can be transferred between the two bodies (increase or decrease, depending on their relative positions as they come near each other). This _can_ add more kinetic energy to your protoplanet, enough so that it can overcome the Sun's gravity well, which it couldn't without the slingshot.  

---

