<p>An approach would be to use the Bondi-Hoyle (B-H) accretion rate and then assume that some large fraction of the gravitational potential energy was radiated from the innermost stable circular orbit (ISCO) as the material spiralled into the black hole via an accretion disc.</p>

<p>To calculate the B-H accretion rate we need how fast the black hole traverses the Oort cloud and the density of matter there.
<span class="math-container">$$\dot{M} \simeq \lambda\frac{4\pi G^2 M^2 \rho}{v^3},$$</span>
where <span class="math-container">$M$</span> is the black hole mass, and <span class="math-container">$\lambda$</span> is an accretion efficiency parameter that is judged to be <span class="math-container">$\sim 0.01$</span> on the basis of observed compact objects (see <a href="https://arxiv.org/abs/1301.1341" rel="noreferrer">Fender et al. 2013</a>).
The luminosity will be
<span class="math-container">$$ L \simeq \eta \frac{GM \dot{M}}{R_{\rm ISCO}},$$</span>
where <span class="math-container">$R_{\rm ISCO}= 6GM/c^2$</span> for a Schwarzschild black hole. Thus
<span class="math-container">$$ L \sim \eta \lambda \frac{4\pi G^2 M^2 \rho c^2 }{6v^3}.$$</span></p>

<p>Getting all my information from <a href="https://en.wikipedia.org/wiki/Oort_cloud" rel="noreferrer">wikipedia</a>, I assume there are 5 Earth masses in the Oort cloud in a spherical shell from 20,000 to 50,000 au. This gives an average density <span class="math-container">$\rho \sim 2\times 10^{-23}$</span> kg/m<span class="math-container">$^3$</span>.</p>

<p>This is actually <em>lower</em> by a factor of a thousand or so than the density of the interstellar medium. Thus accretion from the ISM is far more important on average. Putting the numbers in for a 1 solar mass black hole and <span class="math-container">$\lambda=0.01$</span>, <span class="math-container">$\eta =1$</span>:
<span class="math-container">$$L = 3\times 10^{23}\left(\frac{\rho}{10^{-20}\ {\rm kg m}^{-3}}\right)\left(\frac{v}{10\ {\rm km s}^{-1}}\right)^{-3}\ {\rm W.}$$</span></p>

<p>This is quite an appreciable luminosity and much would emerge at UV and X-ray wavelengths. It would be readily detectable. The flux at a distance of 50,000 au would be <span class="math-container">$5\times 10^{-10}$</span> W/m<span class="math-container">$^2$</span>, which is <strong>4 orders of magnitude</strong> above the faint flux limit of the <a href="https://heasarc.gsfc.nasa.gov/docs/rosat/rass.html" rel="noreferrer">ROSAT all sky X-ray survey</a>.</p>

CC BY-SA 4.0
2018-10-09T00:58:32.033Z
So, if I'm doing my math right, that's about 1.25 W/m^2 at 1 AU. I realize a  lot of it's outside the visible spectrum, but it sounds like you'd be able to see with the naked eye it if you got close enough. (Though you might want to stay back on account of the X-rays and use a telescope.)
Mark Foskey
CC BY-SA 4.0
2018-10-08T06:38:24.433Z
2018-10-08T07:51:07.427Z
2018-10-08T07:51:07.427Z
ProfRob
<p>Imagine a stellar-mass black hole passing through the Oort cloud, perhaps a quarter  to half a light-year from the Sun. How black would the black hole be? That is, any matter that fell into the black hole would release energy on the way in, and we know there are comets out there, and presumably smaller objects down to microscopic dust. But we also know they are very few and far between. Has anyone done a serious analysis of whether there would be enough infalling matter for the resulting radiation to be detectable?</p>

<p>As two commenters already recognized, this really is about radiation from infalling matter, and not the broader question of how we could detect such a black hole. Answers to that broader question in the comments are interesting, though.</p>

CC BY-SA 4.0
2018-10-07T14:10:06.283Z
I imagine if it had similar mass to the sun and passed that close, there would be some serious (or at least detectable) perturbations in the planetary orbits. _This doesn't address your question about detecting radiation, so it's a comment, not an answer._
Mike Harris
CC BY-SA 4.0
2018-10-07T18:14:55.900Z
I would think that the WISE survey would observe it by gravitational lensing as well, considerably before it reached 1/2 light year.   That also doesn't answer the question.
userLTK
CC BY-SA 4.0
2018-10-08T00:36:44.323Z
Hmm, @MikeHarris, at half a light year the effect on the planets would be 64x that of Alpha Centauri, based on a crude 1/r^2 argument. I think any effect would be pretty subtle.
Mark Foskey
<black-hole><oort-cloud>
Could a black hole pass quiescently through the Oort cloud?
Mark Foskey
# Could a black hole pass quiescently through the Oort cloud?

(Asked by: username_0 on 2018-10-07)

Imagine a stellar-mass black hole passing through the Oort cloud, perhaps a quarter to half a light-year from the Sun. How black would the black hole be? That is, any matter that fell into the black hole would release energy on the way in, and we know there are comets out there, and presumably smaller objects down to microscopic dust. But we also know they are very few and far between. Has anyone done a serious analysis of whether there would be enough infalling matter for the resulting radiation to be detectable?

As two commenters already recognized, this really is about radiation from infalling matter, and not the broader question of how we could detect such a black hole. Answers to that broader question in the comments are interesting, though.

**Tags:** black-hole, oort-cloud
> **Comments:**
> **username_1**: I imagine if it had similar mass to the sun and passed that close, there would be some serious (or at least detectable) perturbations in the planetary orbits. _This doesn't address your question about detecting radiation, so it's a comment, not an answer._  
> **username_2**: I would think that the WISE survey would observe it by gravitational lensing as well, considerably before it reached 1/2 light year.   That also doesn't answer the question.  
> **username_0**: Hmm, @MikeHarris, at half a light year the effect on the planets would be 64x that of Alpha Centauri, based on a crude 1/r^2 argument. I think any effect would be pretty subtle.  

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## 1 answers

### Answer by username_3 on 2018-10-08. Score: 9 | Accepted answer

An approach would be to use the Bondi-Hoyle (B-H) accretion rate and then assume that some large fraction of the gravitational potential energy was radiated from the innermost stable circular orbit (ISCO) as the material spiralled into the black hole via an accretion disc.

To calculate the B-H accretion rate we need how fast the black hole traverses the Oort cloud and the density of matter there.
$$\dot{M} \simeq \lambda\frac{4\pi G^2 M^2 \rho}{v^3},$$
where $M$ is the black hole mass, and $\lambda$ is an accretion efficiency parameter that is judged to be $\sim 0.01$ on the basis of observed compact objects (see [Fender et al. 2013](https://arxiv.org/abs/1301.1341)).
The luminosity will be
$$ L \simeq \eta \frac{GM \dot{M}}{R\_{\rm ISCO}},$$
where $R\_{\rm ISCO}= 6GM/c^2$ for a Schwarzschild black hole. Thus
$$ L \sim \eta \lambda \frac{4\pi G^2 M^2 \rho c^2 }{6v^3}.$$

Getting all my information from [wikipedia](https://en.wikipedia.org/wiki/Oort_cloud), I assume there are 5 Earth masses in the Oort cloud in a spherical shell from 20,000 to 50,000 au. This gives an average density $\rho \sim 2\times 10^{-23}$ kg/m$^3$.

This is actually *lower* by a factor of a thousand or so than the density of the interstellar medium. Thus accretion from the ISM is far more important on average. Putting the numbers in for a 1 solar mass black hole and $\lambda=0.01$, $\eta =1$:
$$L = 3\times 10^{23}\left(\frac{\rho}{10^{-20}\ {\rm kg m}^{-3}}\right)\left(\frac{v}{10\ {\rm km s}^{-1}}\right)^{-3}\ {\rm W.}$$

This is quite an appreciable luminosity and much would emerge at UV and X-ray wavelengths. It would be readily detectable. The flux at a distance of 50,000 au would be $5\times 10^{-10}$ W/m$^2$, which is **4 orders of magnitude** above the faint flux limit of the [ROSAT all sky X-ray survey](https://heasarc.gsfc.nasa.gov/docs/rosat/rass.html).

> **Comments:**
> **username_0**: So, if I'm doing my math right, that's about 1.25 W/m^2 at 1 AU. I realize a  lot of it's outside the visible spectrum, but it sounds like you'd be able to see with the naked eye it if you got close enough. (Though you might want to stay back on account of the X-rays and use a telescope.)  

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