00:01
So we can say that d, the distance apart of the license plate letters divided by the height of the satellite above earth, it should be closely related to 1 .22 times the wavelength of the light using divided by the diameter of the dish.
00:24
Okay, so i just did some quick estimation.
00:28
So let's say a license plate is 2 centimeters.
00:31
So that's 2 times 10 to the negative second.
00:36
And the height of a low -ish, right between low and medium -height satellite, is going to be 2 ,000 kilometers, which is 2 times 10 to the 6 meters.
00:49
I'm going to set that equal to 1 .22.
00:51
And i'm just going to use the average wavelength of light, which is around 550 times 10 to the negative 9th, and then all over d.
01:04
And so this is all going to, we're going to solve for what.
01:06
The diameter of this dish would have to be.
01:10
Okay, so plug that into a calculator.
01:12
We're going to give that d is equal to around 67 .1 meters in diameter.
01:27
So that's relatively large.
01:30
So maybe like with a lower hanging satellite, it could be possible.
01:33
It would make that diameter smaller and smaller, the lower that could get it in orbit.
01:40
But discerning something that's two centimeters apart and getting the letters right...