<p>Maybe, but the probability is <em>very</em> low.</p>
<p>There are excellent answers at the recent related question, <a href="https://astronomy.stackexchange.com/questions/58245/">Can it be predicted if an Interstellar Object will get bound to the solar system by knowing its speed and direction?</a> But I'll take this opportunity to give some actual numbers relevant to the trajectory of 'Oumuamua.</p>
<p><a href="https://ssd.jpl.nasa.gov/horizons/" rel="nofollow noreferrer">Horizons</a> has a file for 'Oumuamua. From the <a href="https://ssd.jpl.nasa.gov/api/horizons.api?format=text&amp;MAKE_EPHEM=NO&amp;COMMAND=%27Oumuamua*%27" rel="nofollow noreferrer">body data page</a>,</p>
<blockquote>
<p>1I/'Oumuamua (A/2017 U1)</p>
<p>NOTE: This orbit solution includes non-gravitational perturbations [...] The behavior of these accelerations outside the observed data arc from 2017 October 14 to 2018 January 2 can only be assumed. Predictions outside this time interval, especially prior to October 2017, could be much more uncertain than reported here.</p>
</blockquote>
<p>So please take the data in this answer with a grain of salt. ;)</p>
<hr />
<p>We can determine whether a body is bound to the Sun by calculating its <a href="https://en.wikipedia.org/wiki/Specific_orbital_energy" rel="nofollow noreferrer">specific orbital energy</a>, which is the sum of the body's kinetic and potential energies (relative to the Sun), per unit mass. The equation is
<span class="math-container">$$\varepsilon = \frac{v^2}{2} - \frac{\mu}{r}$$</span>
where <span class="math-container">$\varepsilon$</span> is the specific orbital energy, <span class="math-container">$v$</span> is the body's speed (relative to the Sun), and <span class="math-container">$r$</span> is its distance from the centre of the Sun. <span class="math-container">$\mu$</span> is the <a href="https://en.wikipedia.org/wiki/Standard_gravitational_parameter" rel="nofollow noreferrer">gravitational parameter</a> of the Sun.</p>
<p>Also,
<span class="math-container">$$\varepsilon = -\frac{\mu}{2a}$$</span></p>
<p>where <span class="math-container">$a$</span> is the semi-major axis  of the orbit; <span class="math-container">$a$</span> is negative for hyperbolic orbits.</p>
<p>In an isolated two body system, <span class="math-container">$\varepsilon$</span> is constant. If the body's orbit gets altered by interacting with a planet, then <span class="math-container">$\varepsilon$</span> can change.</p>
<p>If <span class="math-container">$\varepsilon$</span> is negative, the body is in a bound orbit. Otherwise, it's on an escape trajectory. If <span class="math-container">$\varepsilon$</span> is zero, the body's speed tends to the limit of zero as its distance approaches infinity. If <span class="math-container">$\varepsilon$</span> is positive, the body's speed tends to <span class="math-container">$v_{\infty} = \sqrt{2\varepsilon}$</span>, the hyperbolic excess velocity.</p>
<p>Using the data from the Horizons page, we get
<span class="math-container">$\varepsilon \approx 348.432 \, \rm{(km/s)^2}$</span> and <span class="math-container">$v_{\infty}  \approx 26.405 \, \rm{km/s}$</span>. So to capture 'Oumuamua we need to supply a fairly hefty amount of <span class="math-container">$\Delta v$</span>. By comparison, the mean orbital speed of Earth is <span class="math-container">$\approx 29.78 \, \rm{km/s}$</span>.</p>
<p>The specific orbital energy equation is closely related to the vis-viva equation,
<span class="math-container">$$v^2 = \mu\left(\frac2r - \frac1a\right)$$</span></p>
<p>(I have a derivation of the vis-viva equation at the end of <a href="https://physics.stackexchange.com/a/676872/123208">this answer</a>).</p>
<p>For a circular orbit, <span class="math-container">$r=a$</span>, so <span class="math-container">$v=\sqrt{\mu/r}$</span>. If a body at distance <span class="math-container">$r$</span> from the Sun has <span class="math-container">$v \ge \sqrt{2\mu/r}$</span>, then its <span class="math-container">$\varepsilon \ge 0$</span>, so it's unbound.</p>
<p>Wikipedia <a href="https://en.m.wikipedia.org/wiki/Gravity_assist" rel="nofollow noreferrer">Gravity assist</a> discusses the principles of the gravitational slingshot. Briefly, a slingshot can change a body's speed by upto twice the planet's speed, if the trajectories are in line. But that's <em>extremely</em> unlikely to occur for a body falling towards the Sun from the outer Solar System. It's much more likely for the body to cross the planet's orbit at roughly a right angle, and in that case, the maximum <span class="math-container">$\Delta v$</span> is roughly equal to the planet's speed. And in order for the body to be captured, its speed has to be reduced to less than <span class="math-container">$\sqrt{2}$</span> of the planet's speed.</p>
<p>The orbital radius of Jupiter is ~<span class="math-container">$5.2$</span> au, and its speed is ~<span class="math-container">$13.07 \, \rm{km/s}$</span>. At that radius the speed of 'Oumuamua is ~<span class="math-container">$32.2 \, \rm{km/s}$</span>, so an inline close Jupiter encounter <em>could</em> reduce its speed sufficiently, but a right-angle encounter wouldn't provide enough <span class="math-container">$\Delta v$</span>.</p>
<p>At <span class="math-container">$1$</span> au, 'Oumuamua speed is up to ~<span class="math-container">$49.7 \, \rm{km/s}$</span>, but Earth's speed is <span class="math-container">$29.78 \, \rm{km/s}$</span>, as mentioned earlier. So a right-angle slingshot off Earth <em>could</em> work! We get even better results with Venus or Mercury.</p>
<p>Those estimates are based on treating the bodies as points.
After further thoughts and rough calculations, it looks like the asteroid would need to get closer than 1000 km to the planet centre to get sufficient deflection. Obviously, that's not possible. On the other hand, perhaps it can lose enough speed by atmospheric drag.</p>
<hr />
<p>However, the probability of such an encounter is <em>very</em> small. Space is big, and a planet is very small compared to the perimeter of its orbit. For example, it takes Earth a year to traverse its orbit, but it covers a distance equal to its diameter in around 7 minutes. So even if an asteroid's trajectory crosses Earth's orbit, the probability that Earth will be in the vicinity at that time is ~(7 minutes) / (1 year).</p>
<p>But it's unlikely that an asteroid's trajectory <em>will</em> cross a planet's orbit. It could be falling towards the Sun from any angle. In the case of 'Oumuamua, its orbital plane is inclined to the ecliptic by ~<span class="math-container">$122.7417°$</span>. It dropped into the Solar System from the north, whipped around the Sun, and headed back towards the north. Here's a diagram, produced using Horizons. In this diagram,  'Oumuamua is pink, approaching from the left, Jupiter is purple, Earth is blue. The grey plane is the ecliptic, and the pale grey line is the equinox direction. (The timestep is 50 days).</p>
<p><a href="https://i.sstatic.net/oTuyikMA.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/oTuyikMA.png" alt="'Oumuamua, Jupiter &amp; Earth, in 3D" /></a></p>
<p>Here are some other plots produced using Horizons. Here's its distance and speed near the approach time, with a 1 day timestep.</p>
<p><a href="https://i.sstatic.net/9nEaGyEK.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/9nEaGyEK.png" alt="Asteroid distannce" /></a></p>
<p><a href="https://i.sstatic.net/MBI89Ngp.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MBI89Ngp.png" alt="Asteroid speed" /></a></p>
<p>Here's its specific angular momentum, relative to the Solar System barycentre (SSB). It actually picked up a little bit of orbital energy relative to the SSB by getting a slingshot off the Sun.</p>
<p><a href="https://i.sstatic.net/2fGefz8M.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/2fGefz8M.png" alt="Specific angular momentum, relative to the SSB" /></a></p>
<p>Here's its specific orbital energy, relative to the SSB.
<a href="https://i.sstatic.net/TrrTurJj.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/TrrTurJj.png" alt="Specific orbital energy, relative to the SSB" /></a></p>
<p>And here's its specific orbital energy, relative to the Sun. Both of these calculations use the <span class="math-container">$\mu=GM$</span> value for the Sun from <a href="https://ssd.jpl.nasa.gov/astro_par.html" rel="nofollow noreferrer">this JPL page</a>.</p>
<p><a href="https://i.sstatic.net/kEL9o3db.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/kEL9o3db.png" alt="Specific orbital energy, relative to the Sun" /></a></p>
<p>Here's a plot of its distance from Earth near the time of its closest approach
<a href="https://i.sstatic.net/mGaCcKDs.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/mGaCcKDs.png" alt="Distance from Earth" /></a></p>
<p>Here's a close-up of the Sun, Earth, and 'Oumuamua from the 3D plot, from 2017-Aug-1 to 2017-Dec-1, with a 5 day timestep. Earth is moving anticlockwise, 'Oumuamua comes in from the upper right. This plot uses the rainbow mode, where the dot colour of the moving bodies is determined by the time, so dots with the same colour show the position at the same time. Note that this plot is rotated relative to the previous plot, in an attempt to make the 3D relationship more clear. But it's easier to see what's going on in the interactive 3D view.</p>
<p><a href="https://i.sstatic.net/ErGtvTZP.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ErGtvTZP.png" alt="Earth &amp; asteroid, 3D closeup" /></a></p>
<hr />
<p>Here's a little Python / Sage <a href="https://sagecell.sagemath.org/?z=eJxVUMtunDAU3d%2DvOIJFIWLGDzzDUIlFpEpVFlUXUbpk5BKnuI1hagPq%5FH1lTyOlK%2Dv6PO%5FNIR7Yh6%2DrW7VbNYp7Jrlo8CTKCg%2Df8o84cFUrIQVRjvuw%2DHmanR30K54mu1RwpFd0EKo9tM2p4Q3nRMEM4Xwx%5Fvysr%2DhwOqr4mz%2DuEz5%5FqeD6moVekkvKWjZCKsW5EnvZtEq0po1hj8ZZp3%5FOHvqPDaTRYRcZslYHzhslT%5FyEO%2DiVyAzRaC%2D5EHXdtEfBZd3UNKID33MhW8GP7QE57AS99pI96yuNuOvSBIb3hYku3k5LkY1ZhZEJc6yQ%5FXK9ZCErU7GLGeyLHTD773bRrzCT8T%2DuFGYTO8KtYChk7Fa%2DeUUsqxBmU0aeSMuwyHG3BmMfh0JgBzP08r1Q3ISiJNrOdnpBh%5FDbLykhGf6jJjCrkN7U9JsNu81umrbCl28yt8YgyTx2EEyXJW3D%5FzCDL4l8uim%5FnfgWEW0qJHpJfwHdGZaY&amp;lang=sage" rel="nofollow noreferrer">script</a> that I used to do the energy and speed calculations.</p>
<p>And here's the <a href="https://sagecell.sagemath.org/?z=eJydWP1y4jgS%5F99P0ePk1jbjOAbC7i1XrhuSOLXZnUAKqNmdYyhKGAHKGssjyYSE4oHuOe7FrloWX2Fmr%2DpSM46s%5FtSvW93t2LYNXaoEo0sKS5ooLmBCFIGp4Av4hQv2yjMJUy6AZxS4gAUXFBQRM6pgzCeMSgsAQNCUKLakoDiQDPhYUrEkivEMEpopKnwg2QTUnGYJhTzlCtegBHnSVrd6WAb1Wygky2Zw%5FZ9%5FvzIqICnEkspA039nag6CsGzMnyHhKTqWo5nAKumCKUUzGL%5FA4wPUuqinFtZqQfhTUK1atm1bFlvkXCgQ1AdBvxZUKmnp806LLFGcpxIMSyqKUUKSOS3pTFFxRM8FnxSJsqxCpBCBPVcql83LSyknwVOeBhmRJJjx5SXJ2eXcoDmaspQGJGe2RXI2WlIhEaYI7GoQ2pY1JpKOksUEd2zbemj9Fo%5Fix1%5Fih%2Dhz3LM617%2DOblv9VtTuWN34btT73OvHD9H9TffO0lyj%5FufHOPoU3%5FQ73Z71Kb4Z9VvXH%2DOoptc3nW43anfasX772LqOP%5FZQ1U3v0%2Diu031o9bUVjdOH3fHdBVlJ9kojlim3WvM8a0KnMKUqmY8wXdwyIfxdrPOUZNQHqYhQ%2DIvn%2DKS519RRKk83RSs3nYeHVvs2ctaljo1j3cTtftyNnHWpbePogz5%2DbLXjyFlr1RvH6vVb3f6of%5F%2DAm9qQ3uw87vd4rrfix1Hv%5Fl%5FlFsUttHvoxrvz817nbp0sJpv1Fv3Nu%5FPzuHNXMp7lAk%2DeLCaeyfavEO2SJ8i5VG4hUl%5FfnWjtTLlYEOU0wVF0pZyNDxhzGa0dluUFElwnWUwcH13wNqXSfSIIGkhKRDJ3hd16vIdPcbd332k3v8iK%2D6VX8WyduAHq9oKZ4EXuVksdbLpT8y6Cg%5FQqccef8ihT%2D3ciMpbNmoAWtkJMwtqsNz6MC7ylTIJMBMsVUrESrA%5F0bmyvvHiLN367%5F5Tel%5FMv571O7AYVvYw78aHnW4cXqLfNM%5FrWR7vdAZ3Scfe%2DZ3tvqMd6yqioQmRald7ThSyCxQ6iQbU5tA4DutPxD6PU%2DRs4UIGrsNSasozKMtKaL5B5ypTedUsO4%2DmXzA6eOMtcTRs0mtXG0DPICDo9xYZ53fgu7sbtmxjuuq2HGFrtW7jpdLq39%2D1WP%2D5p0GSl97IY8xQWlGCwvh9440eXTqnQJfZOkAWFppaYbrNW44OwWPoC05USJFHlFcaHDzOqX6mM7kgqqbmvBkn8dQqBIQ5kSXId3%5FEGzYvqUCeLxJKOLMMSeNwT%5FHm729xGkz8buHIufVAzpS%2DTRn8w9M1%5FzfBWwy7%2DbLp3f79rXKQyIHlOs4k7tdeCPw%2Dqww2spykn2vggHHqbgxw7g7gEB%5F1huo%5F9AEua8oSpl33G8Wf0r9RSeNq1Ah1DjbXmcLjPWb6zX7ZZbbRZxyzZ8qiZ%2DgZPvTn0ggwLSspe6cT1joL5Bi3LOoPrgqUTuKYHvVNHe0xf3S27iWvO5UjNFETwyvI9saSFPqgQIsjoSrmG09guj52H5flKGzpQuwDlPigEwsjtw4F8ZCzdHC4gDz24hPohnAFdKQRgkIfwHlRYkT4gq9KL4R4sMxQYvHT6vInKAfjbo6BXh%2DgZdKwzuBXkGQjOHmO%2D8uEZxwye69hTSLjIqJBA4AcYayyfmaCjMV%2D5xIexD5XKn88G0iWarpjBwK0grsjjeaU7jxDpqlKfuIPlgJV3hCFSbuhX%5Fbpf80NvaDQaifffFbnyG%5F5P%5Fo%5F%2D1TdEZLFwjZgyRC2pw4JuLQfNq6EPy8FVc3icVI%2DW9YFh6yWJ0sddEJa5FpT9PLJrYfXHi19JdlG1y%2DaOW7XawRbNI7sRTmzfAjMTRHY1tH0zOcrIrv%5F8sw8N%5FQiv6lfV%2Dt81c05SqhSN7LMpafhwFl5NfThrhPgck1qtQTWbmQDLIqXlcCqIejQtL87AiZOU5Yoljg%2DOroW4uOaTF4i%5FFkRx4Qx9kIZ%5FpF5yGjmCTBh3PNS3eon6otCqdZKUbzDhSpqlnocaPqRkTNORPnKI7FM0ZhyDCRF%5FGn5SKD4qcrymu9q6bYEZV%2DCibpywtT3vpBmWZwOiIKVEqnIen8KRlP22GZYWtlPZyTgGkaEFUgmWu57hOXzl%2DeEb3b0daIYInA8OvDevpVETadMZSpl9RzDUg5ZR3g8T%5F%2D9IGeqRVGlsThco5CDgDkKKC6CppOCkbDZXjmn8cEMzJcoBgYsZRHjxAP95uwuac5ap%2DsTlYmbijI9KNWj45TdHZBzBrmHtat6TD%5FtBGNk89JlmxYIKoqiLl86c298eBUcS7yDYppn%2DP6P1YR%5FUak7GqoO8OOB9giiCUH%2De7XMZx8fwWPK0Mf%5Fl%5FIBZv%5FcJI%5FE9daP%5Fpaq8LtbRAbl60%2DPZdFsVjvePTnlKwp%2DswOk1pRm2Oe%2DbLDqeOpnnBXXZZVYsvH0hFiSbURf3hifSu2K8zar8IKe2%2DZR8y6rpo8m2YGvESke8Y%5F5TdHeWd1ZR%2DNSuTtMjZHUhOVamFY31QFGfuEdzxF8o2ifTN7QhIDjG1icu8%5FGUU54p7Z69upALkqb2W0g02jgNHF2qnMtBs7m3tRu7v%2DPBbmD%5FiER58lmxH%2DUHU3vNmvVNE9ZyY1snjshjR3SqvnFl58x4zFcQwWMw5kU2YdlMDw%2DeKUl%5FfC5v9dbv1cve3zKIKUekQh8qAz1SVFChNxw0a8NtBOwzQojtA88JzqhRGNQaWwP9OYX3fwBZMXk6U%2Dgqh1%5F47gVqHYTDQTj0tcuDKq49Ux8PTDUaDXv3Aac73huPdwOS9nQnOBOUZttPxsdAzvmze9QwdSGP9NPHP6%5FgX4cYzyKHCzXnM0Hyue7pPEP3TZH5L%5Fh7gxE=&amp;lang=sage" rel="nofollow noreferrer">3D orbit script</a>. See <a href="https://astronomy.stackexchange.com/a/49823/16685">here</a> for details of an earlier version of that script.</p>
<p>See
<a href="https://astronomy.stackexchange.com/a/49267/16685">https://astronomy.stackexchange.com/a/49267/16685</a> for a distance plotting script.</p>
<hr />
<p>It appears that the Horizons ID# for 'Oumuamua has changed since I first posted this answer. I suspect that there's a bug due to the apostrophe in the name <code>'Oumuamua</code>. Please see the body data page link near the top of this answer to get the current ID mumber. Or just use <code>Oumuamua*</code> instead of the ID number. That <em>should</em> work correctly in any of my Horizons scripts.</p>

CC BY-SA 4.0
2024-08-26T09:59:49.170Z
Great answer! In the first diagram, would it be possible to show which position of Earth matches the moment 'Oumuamua was (apparently) quite close to Earth's orbit?
jcaron
CC BY-SA 4.0
2024-08-26T12:09:29.967Z
@jcaron I've added another diagram that attempts to make that more clear, but it's hard to show it correctly in 2D. Please try the 3D orbit script, it should run in any Web browser, even on a smartphone.
PM 2Ring
CC BY-SA 4.0
2024-08-26T12:29:03.690Z
Thanks, that was exactly the type of graph I had in mind. One can see that the blue dots of each trajectory are “relatively” close — at least Earth wasn’t on the other side of the Sun at that time. Also visible on the other graph you added, at 24.2 million km. How much closer would it have needed to be?
jcaron
CC BY-SA 4.0
2024-08-26T12:40:03.530Z
@jcaron 1 au is ~150 million km, so 24.2 million km isn't that close. To get significant deflection, it would have to get *very* close indeed. I haven't done the precise calculation, but I expect that it'd have to skim the atmosphere, at least.
PM 2Ring
CC BY-SA 4.0
2024-09-04T02:41:27.990Z
Today the 'Oumuamua ID# is 50434520. This is really strange...
PM 2Ring
CC BY-SA 4.0
2024-08-24T19:21:23.397Z
2024-11-21T05:02:08.917Z
2024-11-21T05:02:08.917Z
PM 2Ring
<p>I'm wondering, if it had entered on a trajectory that had caused it to fly by Jupiter, for example, in just the right way, could it have been captured by our Solar System? Or, was it simply moving too fast for that to have ever been a possibility?</p>
<p><a href="https://en.wikipedia.org/wiki/%CA%BBOumuamua" rel="noreferrer">https://en.wikipedia.org/wiki/%CA%BBOumuamua</a></p>

CC BY-SA 4.0
2024-08-24T13:04:14.620Z
https://astronomy.stackexchange.com/questions/58245/can-it-be-predicted-if-an-interstellar-object-will-get-bound-to-the-solar-system
Mithoron
CC BY-SA 4.0
2024-08-24T14:09:30.393Z
The question that Mithron links to implies an answer:  There would have to be a strong interaction between the object and a planet:  essentially a reverse slingshot with the object losing, rather than gaining, orbital energy to a planet.  It's quite possible, but space is big and so such events are unlikely.
James K
<solar-system><extra-terrestrial>
In theory, could an object like 'Oumuamua have been captured by a three-body interaction with the sun and planets?
phil1008
# In theory, could an object like 'Oumuamua have been captured by a three-body interaction with the sun and planets?

(Asked by: username_0 on 2024-08-24)

I'm wondering, if it had entered on a trajectory that had caused it to fly by Jupiter, for example, in just the right way, could it have been captured by our Solar System? Or, was it simply moving too fast for that to have ever been a possibility?

<https://en.wikipedia.org/wiki/%CA%BBOumuamua>

**Tags:** solar-system, extra-terrestrial

> **Comments:**
> **username_1**: https://astronomy.stackexchange.com/questions/58245/can-it-be-predicted-if-an-interstellar-object-will-get-bound-to-the-solar-system  
> **username_2**: The question that Mithron links to implies an answer:  There would have to be a strong interaction between the object and a planet:  essentially a reverse slingshot with the object losing, rather than gaining, orbital energy to a planet.  It's quite possible, but space is big and so such events are unlikely.  

---

## 1 answers

### Answer by username_3 on 2024-08-24. Score: 25 | Accepted answer

Maybe, but the probability is *very* low.

There are excellent answers at the recent related question, [Can it be predicted if an Interstellar Object will get bound to the solar system by knowing its speed and direction?](https://astronomy.stackexchange.com/questions/58245/) But I'll take this opportunity to give some actual numbers relevant to the trajectory of 'Oumuamua.

[Horizons](https://ssd.jpl.nasa.gov/horizons/) has a file for 'Oumuamua. From the [body data page](https://ssd.jpl.nasa.gov/api/horizons.api?format=text&MAKE_EPHEM=NO&COMMAND=%27Oumuamua*%27),

> 1I/'Oumuamua (A/2017 U1)
>
> NOTE: This orbit solution includes non-gravitational perturbations [...] The behavior of these accelerations outside the observed data arc from 2017 October 14 to 2018 January 2 can only be assumed. Predictions outside this time interval, especially prior to October 2017, could be much more uncertain than reported here.

So please take the data in this answer with a grain of salt. ;)

---

We can determine whether a body is bound to the Sun by calculating its [specific orbital energy](https://en.wikipedia.org/wiki/Specific_orbital_energy), which is the sum of the body's kinetic and potential energies (relative to the Sun), per unit mass. The equation is
$$\varepsilon = \frac{v^2}{2} - \frac{\mu}{r}$$
where $\varepsilon$ is the specific orbital energy, $v$ is the body's speed (relative to the Sun), and $r$ is its distance from the centre of the Sun. $\mu$ is the [gravitational parameter](https://en.wikipedia.org/wiki/Standard_gravitational_parameter) of the Sun.

Also,
$$\varepsilon = -\frac{\mu}{2a}$$

where $a$ is the semi-major axis of the orbit; $a$ is negative for hyperbolic orbits.

In an isolated two body system, $\varepsilon$ is constant. If the body's orbit gets altered by interacting with a planet, then $\varepsilon$ can change.

If $\varepsilon$ is negative, the body is in a bound orbit. Otherwise, it's on an escape trajectory. If $\varepsilon$ is zero, the body's speed tends to the limit of zero as its distance approaches infinity. If $\varepsilon$ is positive, the body's speed tends to $v\_{\infty} = \sqrt{2\varepsilon}$, the hyperbolic excess velocity.

Using the data from the Horizons page, we get
$\varepsilon \approx 348.432 \, \rm{(km/s)^2}$ and $v\_{\infty} \approx 26.405 \, \rm{km/s}$. So to capture 'Oumuamua we need to supply a fairly hefty amount of $\Delta v$. By comparison, the mean orbital speed of Earth is $\approx 29.78 \, \rm{km/s}$.

The specific orbital energy equation is closely related to the vis-viva equation,
$$v^2 = \mu\left(\frac2r - \frac1a\right)$$

(I have a derivation of the vis-viva equation at the end of [this answer](https://physics.stackexchange.com/a/676872/123208)).

For a circular orbit, $r=a$, so $v=\sqrt{\mu/r}$. If a body at distance $r$ from the Sun has $v \ge \sqrt{2\mu/r}$, then its $\varepsilon \ge 0$, so it's unbound.

Wikipedia [Gravity assist](https://en.m.wikipedia.org/wiki/Gravity_assist) discusses the principles of the gravitational slingshot. Briefly, a slingshot can change a body's speed by upto twice the planet's speed, if the trajectories are in line. But that's *extremely* unlikely to occur for a body falling towards the Sun from the outer Solar System. It's much more likely for the body to cross the planet's orbit at roughly a right angle, and in that case, the maximum $\Delta v$ is roughly equal to the planet's speed. And in order for the body to be captured, its speed has to be reduced to less than $\sqrt{2}$ of the planet's speed.

The orbital radius of Jupiter is ~$5.2$ au, and its speed is ~$13.07 \, \rm{km/s}$. At that radius the speed of 'Oumuamua is ~$32.2 \, \rm{km/s}$, so an inline close Jupiter encounter *could* reduce its speed sufficiently, but a right-angle encounter wouldn't provide enough $\Delta v$.

At $1$ au, 'Oumuamua speed is up to ~$49.7 \, \rm{km/s}$, but Earth's speed is $29.78 \, \rm{km/s}$, as mentioned earlier. So a right-angle slingshot off Earth *could* work! We get even better results with Venus or Mercury.

Those estimates are based on treating the bodies as points.
After further thoughts and rough calculations, it looks like the asteroid would need to get closer than 1000 km to the planet centre to get sufficient deflection. Obviously, that's not possible. On the other hand, perhaps it can lose enough speed by atmospheric drag.

---

However, the probability of such an encounter is *very* small. Space is big, and a planet is very small compared to the perimeter of its orbit. For example, it takes Earth a year to traverse its orbit, but it covers a distance equal to its diameter in around 7 minutes. So even if an asteroid's trajectory crosses Earth's orbit, the probability that Earth will be in the vicinity at that time is ~(7 minutes) / (1 year).

But it's unlikely that an asteroid's trajectory *will* cross a planet's orbit. It could be falling towards the Sun from any angle. In the case of 'Oumuamua, its orbital plane is inclined to the ecliptic by ~$122.7417°$. It dropped into the Solar System from the north, whipped around the Sun, and headed back towards the north. Here's a diagram, produced using Horizons. In this diagram, 'Oumuamua is pink, approaching from the left, Jupiter is purple, Earth is blue. The grey plane is the ecliptic, and the pale grey line is the equinox direction. (The timestep is 50 days).

[!['Oumuamua, Jupiter & Earth, in 3D](https://i.sstatic.net/oTuyikMA.png)](https://i.sstatic.net/oTuyikMA.png)

Here are some other plots produced using Horizons. Here's its distance and speed near the approach time, with a 1 day timestep.

[![Asteroid distannce](https://i.sstatic.net/9nEaGyEK.png)](https://i.sstatic.net/9nEaGyEK.png)

[![Asteroid speed](https://i.sstatic.net/MBI89Ngp.png)](https://i.sstatic.net/MBI89Ngp.png)

Here's its specific angular momentum, relative to the Solar System barycentre (SSB). It actually picked up a little bit of orbital energy relative to the SSB by getting a slingshot off the Sun.

[![Specific angular momentum, relative to the SSB](https://i.sstatic.net/2fGefz8M.png)](https://i.sstatic.net/2fGefz8M.png)

Here's its specific orbital energy, relative to the SSB.
[![Specific orbital energy, relative to the SSB](https://i.sstatic.net/TrrTurJj.png)](https://i.sstatic.net/TrrTurJj.png)

And here's its specific orbital energy, relative to the Sun. Both of these calculations use the $\mu=GM$ value for the Sun from [this JPL page](https://ssd.jpl.nasa.gov/astro_par.html).

[![Specific orbital energy, relative to the Sun](https://i.sstatic.net/kEL9o3db.png)](https://i.sstatic.net/kEL9o3db.png)

Here's a plot of its distance from Earth near the time of its closest approach
[![Distance from Earth](https://i.sstatic.net/mGaCcKDs.png)](https://i.sstatic.net/mGaCcKDs.png)

Here's a close-up of the Sun, Earth, and 'Oumuamua from the 3D plot, from 2017-Aug-1 to 2017-Dec-1, with a 5 day timestep. Earth is moving anticlockwise, 'Oumuamua comes in from the upper right. This plot uses the rainbow mode, where the dot colour of the moving bodies is determined by the time, so dots with the same colour show the position at the same time. Note that this plot is rotated relative to the previous plot, in an attempt to make the 3D relationship more clear. But it's easier to see what's going on in the interactive 3D view.

[![Earth & asteroid, 3D closeup](https://i.sstatic.net/ErGtvTZP.png)](https://i.sstatic.net/ErGtvTZP.png)

---

Here's a little Python / Sage [script](https://sagecell.sagemath.org/?z=eJxVUMtunDAU3d%2DvOIJFIWLGDzzDUIlFpEpVFlUXUbpk5BKnuI1hagPq%5FH1lTyOlK%2Dv6PO%5FNIR7Yh6%2DrW7VbNYp7Jrlo8CTKCg%2Df8o84cFUrIQVRjvuw%2DHmanR30K54mu1RwpFd0EKo9tM2p4Q3nRMEM4Xwx%5Fvysr%2DhwOqr4mz%2DuEz5%5FqeD6moVekkvKWjZCKsW5EnvZtEq0po1hj8ZZp3%5FOHvqPDaTRYRcZslYHzhslT%5FyEO%2DiVyAzRaC%2D5EHXdtEfBZd3UNKID33MhW8GP7QE57AS99pI96yuNuOvSBIb3hYku3k5LkY1ZhZEJc6yQ%5FXK9ZCErU7GLGeyLHTD773bRrzCT8T%2DuFGYTO8KtYChk7Fa%2DeUUsqxBmU0aeSMuwyHG3BmMfh0JgBzP08r1Q3ISiJNrOdnpBh%5FDbLykhGf6jJjCrkN7U9JsNu81umrbCl28yt8YgyTx2EEyXJW3D%5FzCDL4l8uim%5FnfgWEW0qJHpJfwHdGZaY&lang=sage) that I used to do the energy and speed calculations.

And here's the [3D orbit script](https://sagecell.sagemath.org/?z=eJydWP1y4jgS%5F99P0ePk1jbjOAbC7i1XrhuSOLXZnUAKqNmdYyhKGAHKGssjyYSE4oHuOe7FrloWX2Fmr%2DpSM46s%5FtSvW93t2LYNXaoEo0sKS5ooLmBCFIGp4Av4hQv2yjMJUy6AZxS4gAUXFBQRM6pgzCeMSgsAQNCUKLakoDiQDPhYUrEkivEMEpopKnwg2QTUnGYJhTzlCtegBHnSVrd6WAb1Wygky2Zw%5FZ9%5FvzIqICnEkspA039nag6CsGzMnyHhKTqWo5nAKumCKUUzGL%5FA4wPUuqinFtZqQfhTUK1atm1bFlvkXCgQ1AdBvxZUKmnp806LLFGcpxIMSyqKUUKSOS3pTFFxRM8FnxSJsqxCpBCBPVcql83LSyknwVOeBhmRJJjx5SXJ2eXcoDmaspQGJGe2RXI2WlIhEaYI7GoQ2pY1JpKOksUEd2zbemj9Fo%5Fix1%5Fih%2Dhz3LM617%2DOblv9VtTuWN34btT73OvHD9H9TffO0lyj%5FufHOPoU3%5FQ73Z71Kb4Z9VvXH%2DOoptc3nW43anfasX772LqOP%5FZQ1U3v0%2Diu031o9bUVjdOH3fHdBVlJ9kojlim3WvM8a0KnMKUqmY8wXdwyIfxdrPOUZNQHqYhQ%2DIvn%2DKS519RRKk83RSs3nYeHVvs2ctaljo1j3cTtftyNnHWpbePogz5%2DbLXjyFlr1RvH6vVb3f6of%5F%2DAm9qQ3uw87vd4rrfix1Hv%5Fl%5FlFsUttHvoxrvz817nbp0sJpv1Fv3Nu%5FPzuHNXMp7lAk%2DeLCaeyfavEO2SJ8i5VG4hUl%5FfnWjtTLlYEOU0wVF0pZyNDxhzGa0dluUFElwnWUwcH13wNqXSfSIIGkhKRDJ3hd16vIdPcbd332k3v8iK%2D6VX8WyduAHq9oKZ4EXuVksdbLpT8y6Cg%5FQqccef8ihT%2D3ciMpbNmoAWtkJMwtqsNz6MC7ylTIJMBMsVUrESrA%5F0bmyvvHiLN367%5F5Tel%5FMv571O7AYVvYw78aHnW4cXqLfNM%5FrWR7vdAZ3Scfe%2DZ3tvqMd6yqioQmRald7ThSyCxQ6iQbU5tA4DutPxD6PU%2DRs4UIGrsNSasozKMtKaL5B5ypTedUsO4%2DmXzA6eOMtcTRs0mtXG0DPICDo9xYZ53fgu7sbtmxjuuq2HGFrtW7jpdLq39%2D1WP%2D5p0GSl97IY8xQWlGCwvh9440eXTqnQJfZOkAWFppaYbrNW44OwWPoC05USJFHlFcaHDzOqX6mM7kgqqbmvBkn8dQqBIQ5kSXId3%5FEGzYvqUCeLxJKOLMMSeNwT%5FHm729xGkz8buHIufVAzpS%2DTRn8w9M1%5FzfBWwy7%2DbLp3f79rXKQyIHlOs4k7tdeCPw%2Dqww2spykn2vggHHqbgxw7g7gEB%5F1huo%5F9AEua8oSpl33G8Wf0r9RSeNq1Ah1DjbXmcLjPWb6zX7ZZbbRZxyzZ8qiZ%2DgZPvTn0ggwLSspe6cT1joL5Bi3LOoPrgqUTuKYHvVNHe0xf3S27iWvO5UjNFETwyvI9saSFPqgQIsjoSrmG09guj52H5flKGzpQuwDlPigEwsjtw4F8ZCzdHC4gDz24hPohnAFdKQRgkIfwHlRYkT4gq9KL4R4sMxQYvHT6vInKAfjbo6BXh%2DgZdKwzuBXkGQjOHmO%2D8uEZxwye69hTSLjIqJBA4AcYayyfmaCjMV%2D5xIexD5XKn88G0iWarpjBwK0grsjjeaU7jxDpqlKfuIPlgJV3hCFSbuhX%5Fbpf80NvaDQaifffFbnyG%5F5P%5Fo%5F%2D1TdEZLFwjZgyRC2pw4JuLQfNq6EPy8FVc3icVI%2DW9YFh6yWJ0sddEJa5FpT9PLJrYfXHi19JdlG1y%2DaOW7XawRbNI7sRTmzfAjMTRHY1tH0zOcrIrv%5F8sw8N%5FQiv6lfV%2Dt81c05SqhSN7LMpafhwFl5NfThrhPgck1qtQTWbmQDLIqXlcCqIejQtL87AiZOU5Yoljg%2DOroW4uOaTF4i%5FFkRx4Qx9kIZ%5FpF5yGjmCTBh3PNS3eon6otCqdZKUbzDhSpqlnocaPqRkTNORPnKI7FM0ZhyDCRF%5FGn5SKD4qcrymu9q6bYEZV%2DCibpywtT3vpBmWZwOiIKVEqnIen8KRlP22GZYWtlPZyTgGkaEFUgmWu57hOXzl%2DeEb3b0daIYInA8OvDevpVETadMZSpl9RzDUg5ZR3g8T%5F%2D9IGeqRVGlsThco5CDgDkKKC6CppOCkbDZXjmn8cEMzJcoBgYsZRHjxAP95uwuac5ap%2DsTlYmbijI9KNWj45TdHZBzBrmHtat6TD%5FtBGNk89JlmxYIKoqiLl86c298eBUcS7yDYppn%2DP6P1YR%5FUak7GqoO8OOB9giiCUH%2De7XMZx8fwWPK0Mf%5Fl%5FIBZv%5FcJI%5FE9daP%5Fpaq8LtbRAbl60%2DPZdFsVjvePTnlKwp%2DswOk1pRm2Oe%2DbLDqeOpnnBXXZZVYsvH0hFiSbURf3hifSu2K8zar8IKe2%2DZR8y6rpo8m2YGvESke8Y%5F5TdHeWd1ZR%2DNSuTtMjZHUhOVamFY31QFGfuEdzxF8o2ifTN7QhIDjG1icu8%5FGUU54p7Z69upALkqb2W0g02jgNHF2qnMtBs7m3tRu7v%2DPBbmD%5FiER58lmxH%2DUHU3vNmvVNE9ZyY1snjshjR3SqvnFl58x4zFcQwWMw5kU2YdlMDw%2DeKUl%5FfC5v9dbv1cve3zKIKUekQh8qAz1SVFChNxw0a8NtBOwzQojtA88JzqhRGNQaWwP9OYX3fwBZMXk6U%2Dgqh1%5F47gVqHYTDQTj0tcuDKq49Ux8PTDUaDXv3Aac73huPdwOS9nQnOBOUZttPxsdAzvmze9QwdSGP9NPHP6%5FgX4cYzyKHCzXnM0Hyue7pPEP3TZH5L%5Fh7gxE=&lang=sage). See [here](https://astronomy.stackexchange.com/a/49823/16685) for details of an earlier version of that script.

See
<https://astronomy.stackexchange.com/a/49267/16685> for a distance plotting script.

---

It appears that the Horizons ID# for 'Oumuamua has changed since I first posted this answer. I suspect that there's a bug due to the apostrophe in the name `'Oumuamua`. Please see the body data page link near the top of this answer to get the current ID mumber. Or just use `Oumuamua*` instead of the ID number. That *should* work correctly in any of my Horizons scripts.

> **Comments:**
> **username_4**: Great answer! In the first diagram, would it be possible to show which position of Earth matches the moment 'Oumuamua was (apparently) quite close to Earth's orbit?  
> **username_3**: @username_4 I've added another diagram that attempts to make that more clear, but it's hard to show it correctly in 2D. Please try the 3D orbit script, it should run in any Web browser, even on a smartphone.  
> **username_4**: Thanks, that was exactly the type of graph I had in mind. One can see that the blue dots of each trajectory are “relatively” close — at least Earth wasn’t on the other side of the Sun at that time. Also visible on the other graph you added, at 24.2 million km. How much closer would it have needed to be?  
> **username_3**: @username_4 1 au is ~150 million km, so 24.2 million km isn't that close. To get significant deflection, it would have to get *very* close indeed. I haven't done the precise calculation, but I expect that it'd have to skim the atmosphere, at least.  
> **username_3**: Today the 'Oumuamua ID# is 50434520. This is really strange...  

---

