<p>TLDR; No this is not possible because 1I/'Oumuamua is not coming back. Looking at the orbital elements for 'Oumuamua at e.g the  <a href="https://ssd.jpl.nasa.gov/tools/sbdb_lookup.html#/?sstr=1I" rel="nofollow noreferrer">JPL SBDB entry for 'Oumuamua</a> shows that the eccentricity <span class="math-container">$e$</span> is <span class="math-container">$\sim1.2$</span> i.e. it's on a hyperbolic orbit: <a href="https://i.sstatic.net/FvpkW6Vo.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/FvpkW6Vo.gif" alt="orbit diagram from Prof. Richard Pogge's Astronomy 161:
An Introduction to Solar System Astronomy notes" /></a>.</p>
<p>This means that it is on an orbit with sufficient energy that is greater than the escape velocity of the Solar System, primarily set by the mass of the Sun, and that the orbit isn't closed and so it won't come back. (See also the Orbit Viewer applet on the SBDB page above)</p>

CC BY-SA 4.0
2024-08-09T16:11:33.470Z
I wasn't suggesting it's coming back inside the solar system. I'm asking if it's returning to a prior location outside the solar system, or returning to an earlier trajectory (again, outside teh solar system).
Robert Frost
CC BY-SA 4.0
2024-08-09T16:32:21.843Z
The orbit's not closed, there's no way it can come back to the same patch of sky it was when inbound. It's going further and further away, getting fainter and fainter and because it's not purely radial, the sky position (RA and Dec) is always going to change
astrosnapper
CC BY-SA 4.0
2024-08-11T10:28:46.987Z
Thanks. What is the property "purely radial" and what object do you claim has the property of being purely radial?
Robert Frost
CC BY-SA 4.0
2024-08-12T04:56:27.270Z
The additional condition $\sqrt{\frac{1}{2}} \, v_{C}<v$ is necessary for the blue line. Also an additional line $v<\sqrt{\frac{1}{2}} \, v_{C}$ is need for falling.  $\sqrt{2} \, v_{C}$ may be instead of $v_{E}$ for a better understanding.
Imyaf
CC BY-SA 4.0
2024-08-12T17:55:25.433Z
@RobertFrost I was merely meaning that if Oumuamua was heading away from the Sun on a straight line out of the Solar System, once it got far enough away from Earth, you're always going to see the same "back" of the object and it would always appear to be in the same place on the sky (imagining you had a big enough telescope to detect it). Since the orbit isn't closed, it will never "turn around'" and return to any sky position as when it was first detected on the one and only inbound pass through the solar system. I may have misunderstood your question; if so, can you reword or restate?
astrosnapper
CC BY-SA 4.0
2024-08-09T15:35:05.793Z
2024-08-09T15:35:05.793Z
astrosnapper
<p>Oumuamua is apparently a cigar-shaped asteroid that recently passed through the solar system on an unusual orbit, exhibiting unexpected acceleration.</p>
<p>Oumuamua is known to have fairly minimal powers of thrust other than gravitational free fall, insufficient in themselves to quickly turn around and accelerate the body, but they could be sufficient to enable it to influence itself to harness the gravitational effects of surrounding planetary bodies.  The extragravitational thrust is of unknown cause. The most popular explanation is that Oumuamua has some cometary properties and that it is caused by cometary outgassing in proximity to the Sun, although no cometary tail is apparent and it is widely thought to be rock - with only minimal outgassing possible.</p>
<p>Based on what we know of its trajectory, is there some location in the sky where Oumuamua returns close to its former location, or where it returns to a location which could plausibly be on common trajectory with some other extra-solar-systemic body?</p>

CC BY-SA 4.0
2024-08-09T14:51:33.717Z
Wasn't part of the issue that we had thought Oumuamua was cigar shaped when it was really probably more like a hockey puck or frisbee, and we were seeing it edge-on?
Darth Pseudonym
CC BY-SA 4.0
2024-08-09T16:09:58.693Z
@DarthPseudonym interesting, I hadn't heard that.
Robert Frost
CC BY-SA 4.0
2024-08-09T17:59:05.413Z
Yeah there was a paper in 2019 that showed the light curve we saw could be generated by a spindle or a disc, but simulations showed there were more ways a disc could be spinning to generate the observation, while a spindle would need to be in a pretty specific orientation to Earth. The theories for the acceleration that don't require it to be a powered object all revolve around offgassing either hydrogen or nitrogen, which would be really hard for us to detect.
Darth Pseudonym
<asteroids><navigation><oumuamua>
Where in the sky does Oumuamua return closest to a former location?
Robert Frost
# Where in the sky does Oumuamua return closest to a former location?

(Asked by: username_0 on 2024-08-09)

Oumuamua is apparently a cigar-shaped asteroid that recently passed through the solar system on an unusual orbit, exhibiting unexpected acceleration.

Oumuamua is known to have fairly minimal powers of thrust other than gravitational free fall, insufficient in themselves to quickly turn around and accelerate the body, but they could be sufficient to enable it to influence itself to harness the gravitational effects of surrounding planetary bodies. The extragravitational thrust is of unknown cause. The most popular explanation is that Oumuamua has some cometary properties and that it is caused by cometary outgassing in proximity to the Sun, although no cometary tail is apparent and it is widely thought to be rock - with only minimal outgassing possible.

Based on what we know of its trajectory, is there some location in the sky where Oumuamua returns close to its former location, or where it returns to a location which could plausibly be on common trajectory with some other extra-solar-systemic body?

**Tags:** asteroids, navigation, oumuamua

> **Comments:**
> **username_1**: Wasn't part of the issue that we had thought Oumuamua was cigar shaped when it was really probably more like a hockey puck or frisbee, and we were seeing it edge-on?  
> **username_0**: @username_1 interesting, I hadn't heard that.  
> **username_1**: Yeah there was a paper in 2019 that showed the light curve we saw could be generated by a spindle or a disc, but simulations showed there were more ways a disc could be spinning to generate the observation, while a spindle would need to be in a pretty specific orientation to Earth. The theories for the acceleration that don't require it to be a powered object all revolve around offgassing either hydrogen or nitrogen, which would be really hard for us to detect.  

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## 1 answers

### Answer by username_2 on 2024-08-09. Score: 3 | Accepted answer

TLDR; No this is not possible because 1I/'Oumuamua is not coming back. Looking at the orbital elements for 'Oumuamua at e.g the [JPL SBDB entry for 'Oumuamua](https://ssd.jpl.nasa.gov/tools/sbdb_lookup.html#/?sstr=1I) shows that the eccentricity $e$ is $\sim1.2$ i.e. it's on a hyperbolic orbit: [![orbit diagram from Prof. Richard Pogge's Astronomy 161:
An Introduction to Solar System Astronomy notes](https://i.sstatic.net/FvpkW6Vo.gif)](https://i.sstatic.net/FvpkW6Vo.gif).

This means that it is on an orbit with sufficient energy that is greater than the escape velocity of the Solar System, primarily set by the mass of the Sun, and that the orbit isn't closed and so it won't come back. (See also the Orbit Viewer applet on the SBDB page above)

> **Comments:**
> **username_0**: I wasn't suggesting it's coming back inside the solar system. I'm asking if it's returning to a prior location outside the solar system, or returning to an earlier trajectory (again, outside teh solar system).  
> **username_2**: The orbit's not closed, there's no way it can come back to the same patch of sky it was when inbound. It's going further and further away, getting fainter and fainter and because it's not purely radial, the sky position (RA and Dec) is always going to change  
> **username_0**: Thanks. What is the property "purely radial" and what object do you claim has the property of being purely radial?  
> **username_3**: The additional condition $\sqrt{\frac{1}{2}} \, v_{C}<v$ is necessary for the blue line. Also an additional line $v<\sqrt{\frac{1}{2}} \, v_{C}$ is need for falling.  $\sqrt{2} \, v_{C}$ may be instead of $v_{E}$ for a better understanding.  
> **username_2**: @username_0 I was merely meaning that if Oumuamua was heading away from the Sun on a straight line out of the Solar System, once it got far enough away from Earth, you're always going to see the same "back" of the object and it would always appear to be in the same place on the sky (imagining you had a big enough telescope to detect it). Since the orbit isn't closed, it will never "turn around'" and return to any sky position as when it was first detected on the one and only inbound pass through the solar system. I may have misunderstood your question; if so, can you reword or restate?  

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