Could the Earth be used to cast an arago/poisson spot on something There's a potential design for a space-based telescope 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?
One constraint is the recommendation $F=\frac{d^2}{b\lambda}\geq 1$, 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:
$\Delta r &lt; \sqrt{r^2 + \lambda\frac{gb}{g+b}}-r$, 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.

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

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

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

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

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 of $9.4607\cdot 10^{15}\mbox{ m}$ from Earth.

In astronomcal units of $149597870700\mbox{ m}$ 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 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 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 $G$ and the 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, 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.