Sunday, 7 June 2015

space telescope - Could the Earth be used to cast an arago/poisson spot on something

One constraint is the recommendation $F=frac{d^2}{blambda}geq 1$, 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



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



Another constraint is the surface roughness of the circular object:
$Delta r < sqrt{r^2 + lambdafrac{gb}{g+b}}-r$, with $r=6350mbox{ km}$ 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 $ggg b$. Then approximately
$Delta r < sqrt{r^2 + lambdafrac{gb}{g}}-r = sqrt{r^2 + lambda b}-r$.



After adding $r$ and squaring you get
$(Delta r +r)^2<r^2+lambda b$.
This simplifies to
$(Delta r)^2+2rDelta r<lambda b$.
Assume $Delta rll r$, and neglect the second order $(Delta r)^2$ to get
$2rDelta r<lambda b$. Divide by $lambda$ to get an approximate constraint for $b$ as



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



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



The two constraints allow for reasonable values of $Delta r$.
Assume a surface roughness of Earth of e.g. $Delta r = 1mbox{ km}$.
Then a valid range of distances of observers would be between
$0.00224$ and $50912$ lightyears of $9.4607cdot 10^{15}mbox{ m}$ from Earth.



In astronomcal units of $149597870700mbox{ m}$ 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
$theta=frac{4GM}{rc^2}=2.969cdot 10^{-27}frac{mbox{m}}{mbox{kg}}frac{M}{r}$, after applying the constant of gravitation $G$ and the speed of light $c$. With the mass $M=5.97237cdot 10^{24}mbox{ kg}$ and a radius of $r=6350000mbox{ m}$ of Earth, we get an angle of
$theta=2.793cdot 10^{-9}$ by gravitational lensing at the surface of Earth.



This would focus parallel rays of light to a point near a distance of
$b=frac{r}{tan theta}=130.27cdot 10^{15}mbox{ m}$, 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.

No comments:

Post a Comment