<p>One constraint is the <a href="https://en.wikipedia.org/wiki/Arago_spot" rel="nofollow">recommendation</a> $F=\frac{d^2}{b\lambda}\geq 1$, in this case with $d=12700 \mbox{ km}$ about the diameter of Earth, $\lambda=600 \mbox{ nm}$ some wavelength of visible light, and $b$ the distance between circular obstacle and observer.
The distance between Earth and the observer should hence be</p>

<p>$b\leq \frac{d^2}{\lambda}=\frac{(12700\cdot10^3\mbox{ m})^2}{600\cdot 10^{-9}\mbox{ m}}=481.67\cdot 10^{18}\mbox{ m}$</p>

<p>Another constraint is the <a href="https://en.wikipedia.org/wiki/Arago_spot#The_circular_object.27s_surface_roughness" rel="nofollow">surface roughness of the circular object</a>:
$\Delta r &lt; \sqrt{r^2 + \lambda\frac{gb}{g+b}}-r$, with $r=6350\mbox{ km}$ the radius of the circular obstacle (here Earth), $g$ the distance between the point light source and the circular obstacle, and $b$ the distance between the circular obstacle and the screen.</p>

<p>To simplify calculations, say $g\gg b$. Then approximately
$\Delta r &lt; \sqrt{r^2 + \lambda\frac{gb}{g}}-r = \sqrt{r^2 + \lambda b}-r$.</p>

<p>After adding $r$ and squaring you get
$(\Delta r +r)^2&lt;r^2+\lambda b$.
This simplifies to
$(\Delta r)^2+2r\Delta r&lt;\lambda b$.
Assume $\Delta r\ll r$, and neglect the second order $(\Delta r)^2$ to get
$2r\Delta r&lt;\lambda b$. Divide by $\lambda$ to get an approximate constraint for $b$ as</p>

<p>$b&gt;\frac{2r\Delta r}{\lambda}$.
With $2r=12700 \mbox{ km}$ about the diameter of Earth, $\lambda=600 \mbox{ nm}$ some wavelength of visible light, we get</p>

<p>$b&gt;\frac{12700\cdot 10^3\mbox{ m}\cdot\Delta r}{600\cdot 10^{-9}\mbox{ m}}
= 21.1667\cdot 10^{12}\Delta r$.</p>

<p>The two constraints allow for reasonable values of $\Delta r$.
Assume a surface roughness of Earth of e.g. $\Delta r = 1\mbox{ km}$.
Then a valid range of distances of observers would be between
$0.00224$ and $50912$ <a href="https://en.wikipedia.org/wiki/Light-year" rel="nofollow">lightyears</a> of $9.4607\cdot 10^{15}\mbox{ m}$ from Earth.</p>

<p>In <a href="https://en.wikipedia.org/wiki/Astronomical_unit" rel="nofollow">astronomcal units of</a> $149597870700\mbox{ m}$ the closest distance of an observer would be $141.49 \mbox{ au}$ from Earth.</p>

<p>Due to Earth's oblateness, however, you would get a <a href="https://en.wikipedia.org/wiki/Point_spread_function" rel="nofollow">point spread function</a> significantly different from a dot for this "short" distance from Earth. It might be possible to correct this by an appropriate telescope optics.</p>

<p>The effect of <a href="https://en.wikipedia.org/wiki/Gravitational_lens#Explanation_in_terms_of_space.E2.80.93time_curvature" rel="nofollow">gravitational lensing</a> is
$\theta=\frac{4GM}{rc^2}=2.969\cdot 10^{-27}\frac{\mbox{m}}{\mbox{kg}}\frac{M}{r}$, after applying the <a href="https://en.wikipedia.org/wiki/Gravitational_constant" rel="nofollow">constant of gravitation</a> $G$ and the <a href="https://en.wikipedia.org/wiki/Speed_of_light" rel="nofollow">speed of light</a> $c$. With the mass $M=5.97237\cdot 10^{24}\mbox{ kg}$ and a radius of $r=6350000\mbox{ m}$ of <a href="https://en.wikipedia.org/wiki/Earth" rel="nofollow">Earth</a>, we get an angle of
$\theta=2.793\cdot 10^{-9}$ by gravitational lensing at the surface of Earth.</p>

<p>This would focus parallel rays of light to a point near a distance of
$b=\frac{r}{\tan \theta}=130.27\cdot 10^{15}\mbox{ m}$, or $13.77$ lightyears, hence well beyond the minimum distance where an Arago spot could form. But, of course, the innermost peak of the point spread function would be closer to a circular disc at this larger distance with relevant gravitational lensing.</p>

CC BY-SA 3.0
2016-01-21T15:10:45.040Z
+1 certainly for the detail and effort - I think I've now followed all of it closely enough... (?) -- if I understand, then the answer would be somewhere around 141.5 AU from Earth, with an essentially negligible reduction of that from gravitational lensing, and it would be spread to an extent but you estimate that could be corrected for with optics. sos ~141 AU would make any adjustments for looking at a different part of the sky quite impractical to do quickly. Though... the universe is big and who knows what we could find to look at with this much of an increase in telescope power?
Code Jockey
CC BY-SA 3.0
2016-01-21T16:04:57.843Z
I wonder if using a slightly smaller planet - perhaps Mars or even Mercury(?) might be slightly more feasible, though they'd be more "rough" than Earth, I think Mercury has a much longer day, and thus I'd guess less oblation due to centrifugal "force" (not sure how much it might be deformed due to proximity with the sun?) ---- is the last part of your answer saying that gravitational lensing would at least somewhat correct for the oblate shape of the Earth?
Code Jockey
CC BY-SA 3.0
2016-01-22T00:52:06.200Z
You could use our Moon. It has no dense atmosphere. But the limiting factor will still be the surface roughness: http://www.nasa.gov/images/content/604359main_WAC_CSHADE_O000N1800_1000.jpg
Gerald
CC BY-SA 3.0
2016-01-20T16:27:35.707Z
2016-01-20T16:27:35.707Z
Gerald
<p>There's a <a href="http://www.gizmag.com/aragoscope-lensless-telescope/35761/" rel="nofollow">potential design for a space-based telescope</a> that uses a large opaque circular disk in place of a transparent lens to focus light for analysis. It uses the light that bends around the disk that focuses on the Poisson or Arago spot, but requires a circular object. It could theoretically produce much clearer images (1000x) than telescopes we have today, albeit with a narrower field of vision.</p>

<p>I understand that the Earth is not perfectly round, but I believe I've also heard it's very "smooth" despite mountains and such, so I wonder if instead of a spot (very small circle) on some point behind it, it might cast an oval or other oblong shape of some size that, using lenses or computer reconstruction, might produce an image that is usable, and presumably even higher resolution (millions, billions, trillions, or even more times as detailed?).</p>

<p>If that's not possible, what are the issues? Does the atmosphere get in the way? If we could imagine a sufficiently circular disk roughly the size of Earth, about how far would the image processing receiver have to be from the disk opposite the studied object in order to capture the convergence of bent light? Would the gravity of the Earth affect the distance at all, due to gravitational lensing?</p>

CC BY-SA 4.0
2019-09-12T15:58:36.763Z
other astronomical arago-related question: [Arago Spot in the Shadows of Celestial Bodies](https://astronomy.stackexchange.com/q/28277/7982) and also [see this ansswer](https://space.stackexchange.com/a/38689/12102) to *Is there any wavelength at which the Sun is both transparent to and quiet of electromagnetic radiation?*
uhoh
<space-telescope><gravitational-lensing>
Could the Earth be used to cast an arago/poisson spot on something
Code Jockey
# Could the Earth be used to cast an arago/poisson spot on something

(Asked by: username_0 on 2016-01-18)

There's a [potential design for a space-based telescope](http://www.gizmag.com/aragoscope-lensless-telescope/35761/) that uses a large opaque circular disk in place of a transparent lens to focus light for analysis. It uses the light that bends around the disk that focuses on the Poisson or Arago spot, but requires a circular object. It could theoretically produce much clearer images (1000x) than telescopes we have today, albeit with a narrower field of vision.

I understand that the Earth is not perfectly round, but I believe I've also heard it's very "smooth" despite mountains and such, so I wonder if instead of a spot (very small circle) on some point behind it, it might cast an oval or other oblong shape of some size that, using lenses or computer reconstruction, might produce an image that is usable, and presumably even higher resolution (millions, billions, trillions, or even more times as detailed?).

If that's not possible, what are the issues? Does the atmosphere get in the way? If we could imagine a sufficiently circular disk roughly the size of Earth, about how far would the image processing receiver have to be from the disk opposite the studied object in order to capture the convergence of bent light? Would the gravity of the Earth affect the distance at all, due to gravitational lensing?

**Tags:** space-telescope, gravitational-lensing
> **Comments:**
> **username_1**: other astronomical arago-related question: [Arago Spot in the Shadows of Celestial Bodies](https://astronomy.stackexchange.com/q/28277/7982) and also [see this ansswer](https://space.stackexchange.com/a/38689/12102) to *Is there any wavelength at which the Sun is both transparent to and quiet of electromagnetic radiation?*  

---

## 1 answers

### Answer by username_2 on 2016-01-20. Score: 2 | Accepted answer

One constraint is the [recommendation](https://en.wikipedia.org/wiki/Arago_spot) $F=\frac{d^2}{b\lambda}\geq 1$, in this case with $d=12700 \mbox{ km}$ about the diameter of Earth, $\lambda=600 \mbox{ nm}$ some wavelength of visible light, and $b$ the distance between circular obstacle and observer.
The distance between Earth and the observer should hence be

$b\leq \frac{d^2}{\lambda}=\frac{(12700\cdot10^3\mbox{ m})^2}{600\cdot 10^{-9}\mbox{ m}}=481.67\cdot 10^{18}\mbox{ m}$

Another constraint is the [surface roughness of the circular object](https://en.wikipedia.org/wiki/Arago_spot#The_circular_object.27s_surface_roughness):
$\Delta r < \sqrt{r^2 + \lambda\frac{gb}{g+b}}-r$, with $r=6350\mbox{ km}$ the radius of the circular obstacle (here Earth), $g$ the distance between the point light source and the circular obstacle, and $b$ the distance between the circular obstacle and the screen.

To simplify calculations, say $g\gg b$. Then approximately
$\Delta r < \sqrt{r^2 + \lambda\frac{gb}{g}}-r = \sqrt{r^2 + \lambda b}-r$.

After adding $r$ and squaring you get
$(\Delta r +r)^2

$b>\frac{2r\Delta r}{\lambda}$.
With $2r=12700 \mbox{ km}$ about the diameter of Earth, $\lambda=600 \mbox{ nm}$ some wavelength of visible light, we get

$b>\frac{12700\cdot 10^3\mbox{ m}\cdot\Delta r}{600\cdot 10^{-9}\mbox{ m}}
= 21.1667\cdot 10^{12}\Delta r$.

The two constraints allow for reasonable values of $\Delta r$.
Assume a surface roughness of Earth of e.g. $\Delta r = 1\mbox{ km}$.
Then a valid range of distances of observers would be between
$0.00224$ and $50912$ [lightyears](https://en.wikipedia.org/wiki/Light-year) of $9.4607\cdot 10^{15}\mbox{ m}$ from Earth.

In [astronomcal units of](https://en.wikipedia.org/wiki/Astronomical_unit) $149597870700\mbox{ m}$ the closest distance of an observer would be $141.49 \mbox{ au}$ from Earth.

Due to Earth's oblateness, however, you would get a [point spread function](https://en.wikipedia.org/wiki/Point_spread_function) significantly different from a dot for this "short" distance from Earth. It might be possible to correct this by an appropriate telescope optics.

The effect of [gravitational lensing](https://en.wikipedia.org/wiki/Gravitational_lens#Explanation_in_terms_of_space.E2.80.93time_curvature) is
$\theta=\frac{4GM}{rc^2}=2.969\cdot 10^{-27}\frac{\mbox{m}}{\mbox{kg}}\frac{M}{r}$, after applying the [constant of gravitation](https://en.wikipedia.org/wiki/Gravitational_constant) $G$ and the [speed of light](https://en.wikipedia.org/wiki/Speed_of_light) $c$. With the mass $M=5.97237\cdot 10^{24}\mbox{ kg}$ and a radius of $r=6350000\mbox{ m}$ of [Earth](https://en.wikipedia.org/wiki/Earth), we get an angle of
$\theta=2.793\cdot 10^{-9}$ by gravitational lensing at the surface of Earth.

This would focus parallel rays of light to a point near a distance of
$b=\frac{r}{\tan \theta}=130.27\cdot 10^{15}\mbox{ m}$, or $13.77$ lightyears, hence well beyond the minimum distance where an Arago spot could form. But, of course, the innermost peak of the point spread function would be closer to a circular disc at this larger distance with relevant gravitational lensing.

> **Comments:**
> **username_0**: +1 certainly for the detail and effort - I think I've now followed all of it closely enough... (?) -- if I understand, then the answer would be somewhere around 141.5 AU from Earth, with an essentially negligible reduction of that from gravitational lensing, and it would be spread to an extent but you estimate that could be corrected for with optics. sos ~141 AU would make any adjustments for looking at a different part of the sky quite impractical to do quickly. Though... the universe is big and who knows what we could find to look at with this much of an increase in telescope power?  
> **username_0**: I wonder if using a slightly smaller planet - perhaps Mars or even Mercury(?) might be slightly more feasible, though they'd be more "rough" than Earth, I think Mercury has a much longer day, and thus I'd guess less oblation due to centrifugal "force" (not sure how much it might be deformed due to proximity with the sun?) ---- is the last part of your answer saying that gravitational lensing would at least somewhat correct for the oblate shape of the Earth?  
> **username_2**: You could use our Moon. It has no dense atmosphere. But the limiting factor will still be the surface roughness: http://www.nasa.gov/images/content/604359main_WAC_CSHADE_O000N1800_1000.jpg  

---

