<p>The enhancement of a cross-section due to gravitational focusing is given by
<span class="math-container">$$ \sigma_{\rm eff} = \pi a_J^2 \left(1 + \frac{2GM_{\odot}}{a_J\ v^2}\right),$$</span>
where  <span class="math-container">$a_J$</span> is the semi-major axis of Jupiter's orbit (assumed circular), <span class="math-container">$v$</span> is the relative velocity (at infinity) and I have ignored the mass of Jupiter.</p>

<p>Thus, using <span class="math-container">$v=30$</span> km/s (as specified in the question) the term in brackets is the gravitational focusing enhancement of the cross-section and equals 1.38.</p>

<p>We now assume an isotropic velocity distribution and a homogeneous density of such objects of <span class="math-container">$n$</span> per unit volume. The flux of such particles is <span class="math-container">$f =n v \sigma_{\rm eff}$</span>.</p>

<p>If <span class="math-container">$f = 0.5$</span> yr<span class="math-container">$^{-1}$</span>, then <span class="math-container">$n \sim 5 \times 10^{12}$</span> per cubic parsec.</p>

CC BY-SA 4.0
2019-09-17T21:31:39.217Z
Thank you for backing me up with math! I wonder how much error I had in my napkin calculations.
alamar
CC BY-SA 4.0
2019-09-17T16:13:01.210Z
2019-09-17T16:13:01.210Z
ProfRob
<p>I had to write my own code to calculate this, since I could not find any tools to solve two-body problem.</p>

<p>From my simulation it seems that Jupiter orbit cross section is <code>6.5 AU</code>, assuming radius of Jupiter orbit as <code>5 AU</code>. Seems like a tiny difference, but surface-wise it's <code>1.7x</code> increase. This assuming <code>20 km/s</code>, it's smaller for faster scouts.</p>

<p>This means one scout passes through a <code>6×10^24 square meters</code> per year. Thus we can expect 30 million of these passing through every square light year.</p>

<p>During that year, scout will travel at least <code>600 million km</code>. This means there's maybe <code>2×10^24</code> scouts in our galaxy.</p>

<p>Each scout probably weighs <code>2×10^9 kg</code>. This means there are <code>4×10^33 kg</code> worth of them in our galaxy. That's 2,000 suns.</p>

CC BY-SA 4.0
2019-09-16T21:02:51.087Z
2019-09-17T13:40:22.333Z
2019-09-17T13:40:22.333Z
alamar
<p>Now we know two of them, so maybe it's not an outlier.</p>

<p>We have two points so we can draw a line. How many similar macroscopic objects should be zipping through space at ~30 km/sec so that we would see one of these passing inside the orbit of Jupiter once every two years? I don't really know what units this should be expressed with. Maybe it's <code>1 / (km^2 * year)</code>  - the frequency of such a body coming through any random square kilometer of empty space in a year of time.</p>

<p>What's the "capture cross section" of intra-Jupiter orbit? I imagine that since the Sun affects trajectories of bodies coming near it, drawing them closer as they pass by, intra-Jupiter orbit will see much more bodies than a circle of same area in empty space. How large is this effect for ʻOumuamua-like bodies? I expect this is to be a unitless coefficient. Is it neligible? Is it 5x? 10x? 1000x?</p>

<p>Sorry if my question is not clear enough.</p>

CC BY-SA 4.0
2019-09-18T17:49:22.633Z
List of Hyperbolic comets: https://en.wikipedia.org/wiki/List_of_hyperbolic_comets (those with eccentricity >1)
Wayfaring Stranger
CC BY-SA 4.0
2019-09-19T09:46:12.200Z
Most of those are *barely* hyperbolic because of gravitation perturbances inside Solar system. They, in turn, may become scouts if they ever arrive to other systems.
alamar
<solar-system><interstellar><modeling>
Estimate prevalence of ʻOumuamua-style interstellar obects inside Jupiter orbit now that C/2019 Q4 (Borisov) is found?
alamar
# Estimate prevalence of ʻOumuamua-style interstellar obects inside Jupiter orbit now that C/2019 Q4 (Borisov) is found?

(Asked by: username_0 on 2019-09-13)

Now we know two of them, so maybe it's not an outlier.

We have two points so we can draw a line. How many similar macroscopic objects should be zipping through space at ~30 km/sec so that we would see one of these passing inside the orbit of Jupiter once every two years? I don't really know what units this should be expressed with. Maybe it's `1 / (km^2 * year)` - the frequency of such a body coming through any random square kilometer of empty space in a year of time.

What's the "capture cross section" of intra-Jupiter orbit? I imagine that since the Sun affects trajectories of bodies coming near it, drawing them closer as they pass by, intra-Jupiter orbit will see much more bodies than a circle of same area in empty space. How large is this effect for ʻOumuamua-like bodies? I expect this is to be a unitless coefficient. Is it neligible? Is it 5x? 10x? 1000x?

Sorry if my question is not clear enough.

**Tags:** solar-system, interstellar, modeling

> **Comments:**
> **username_1**: List of Hyperbolic comets: https://en.wikipedia.org/wiki/List_of_hyperbolic_comets (those with eccentricity >1)  
> **username_0**: Most of those are *barely* hyperbolic because of gravitation perturbances inside Solar system. They, in turn, may become scouts if they ever arrive to other systems.  

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## 2 answers

### Answer by username_2 on 2019-09-17. Score: 3 | Accepted answer

The enhancement of a cross-section due to gravitational focusing is given by
$$ \sigma\_{\rm eff} = \pi a\_J^2 \left(1 + \frac{2GM\_{\odot}}{a\_J\ v^2}\right),$$
where $a\_J$ is the semi-major axis of Jupiter's orbit (assumed circular), $v$ is the relative velocity (at infinity) and I have ignored the mass of Jupiter.

Thus, using $v=30$ km/s (as specified in the question) the term in brackets is the gravitational focusing enhancement of the cross-section and equals 1.38.

We now assume an isotropic velocity distribution and a homogeneous density of such objects of $n$ per unit volume. The flux of such particles is $f =n v \sigma\_{\rm eff}$.

If $f = 0.5$ yr$^{-1}$, then $n \sim 5 \times 10^{12}$ per cubic parsec.

> **Comments:**
> **username_0**: Thank you for backing me up with math! I wonder how much error I had in my napkin calculations.  

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### Answer by username_0 on 2019-09-16. Score: 1

I had to write my own code to calculate this, since I could not find any tools to solve two-body problem.

From my simulation it seems that Jupiter orbit cross section is `6.5 AU`, assuming radius of Jupiter orbit as `5 AU`. Seems like a tiny difference, but surface-wise it's `1.7x` increase. This assuming `20 km/s`, it's smaller for faster scouts.

This means one scout passes through a `6×10^24 square meters` per year. Thus we can expect 30 million of these passing through every square light year.

During that year, scout will travel at least `600 million km`. This means there's maybe `2×10^24` scouts in our galaxy.

Each scout probably weighs `2×10^9 kg`. This means there are `4×10^33 kg` worth of them in our galaxy. That's 2,000 suns.


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