Well, we CAN see the ISS because it is, compared to most satellites, very large at 240'x356' and much closer than most satellites at just 249 miles. While satellites are usually under 40 feet and are much further out.

Here is Martin Lewis' image:

We can use this Angular Size Calculator to figure out how big something would appear in our field of view.

So the ISS comes in at 62.5 arcseconds and we can just barely see it with the naked eye (little more than a dot). Being large it also reflections a LOT more light, 85440 square feet (well, it's not solid so maybe 1/4 that) -- so that's maybe 21360 sq feet versus 1600 sq. feet -- or about 7% as much light IF it were at the same distance -- but they aren't. Most satellites are 5 times or more further away. So that makes them even dimmer (to the square of the distance)!

Your 40' satellite at 2000 km is only 1.26 arcseconds -- a tiny tiny fraction of the ISS in total area.

Too small, too faint. Here is what a powerful 14" telescope can see of a normal satellite when it is close to Earth, this one was about to reenter the atmosphere (Thierry Legault Emmanual Rietsch):

A $400 Orion 8945 SkyQuest XT8 Classic Dobsonian (8") Telescope -- which gives you about 400x useful magnification max is going to just barely be able to see that and only for a VERY short time as it flits across the viewfinder. You just aren't likely to see most satellites from amateur ground-based telescopes with the exception of the ISS (because it is large and fairly close) and a few others.

You need very expensive equipment to capture these.

If you want to try to catch one here is a good resource listing the brightest.

### The Math

This friendly Flat Earther wanted to share his '

*math*' on why we should expect to see so many satellites...

Here is the image offered:

So let's see how our friend did on his math...

Earth diameter:

**7917.5 miles**

Earth diameter pixels in image (measured from image):

**281**

Miles PER Pixel in that image: 7917.5/281 =

# Pixels for Big Satellites in your image (measured from image):

**28.1762**# Pixels for Big Satellites in your image (measured from image):

**14**
A Satellite in your image would be 14*7917.5/281 =

Actual Big Satellite ~ Bus (see below): 35 feet

35 feet is 35/5280 =

So even a 35 foot SQUARE satellite would only take up:

(35^2)/((7917.5/281)*5280)^2)*100 =

http://talkingpointsmemo.com/idealab/satellites-earth-orbit

Satellites mostly orbit between 99 miles and 22237 miles.

So we take the volume of a sphere 1 Earth radius (3958.75 miles) + 22237 miles and subtract from that the smaller sphere of 3958.75+99 miles and we get:

Volume of sphere = (4/3)πr³

[volume sphere r=(3958.75+22237)] - [volume sphere r=(3958.75+99)]

[7.52976×10^13] - [2.79862×10^11] =

For our satellite we will assume it is a full 35' cube =

0.00000029127 / 75017738000000 * 100 =

Something that small is reflecting that much less light, plus the light falls off with the SQUARE of the distance -- and you are pitting it against vastly closer and thus brighter objects but your camera can't even capture the highlights in the clouds when taking a picture of your friend in the shade.

**394.4662 MILES LONG**Actual Big Satellite ~ Bus (see below): 35 feet

35 feet is 35/5280 =

**0.0066 of a mile**So even a 35 foot SQUARE satellite would only take up:

(35^2)/((7917.5/281)*5280)^2)*100 =

**0.0000055% of a pixel**http://talkingpointsmemo.com/idealab/satellites-earth-orbit

https://www.reference.com/vehicles/long-school-bus-feet-3c674c9adc10c1bd"Size varies. Communication satellitescan be as big as a small school busand weigh up to 6 tons, the Federal Communications Commission says. Most weigh a few tons or less. Some that are used briefly are 4 inch cubes and weigh about 2 pounds."

And just to show you how bad it gets, let's consider the VOLUME of that space."An average 64-passenger school bus is35 feet long"

Satellites mostly orbit between 99 miles and 22237 miles.

So we take the volume of a sphere 1 Earth radius (3958.75 miles) + 22237 miles and subtract from that the smaller sphere of 3958.75+99 miles and we get:

Volume of sphere = (4/3)πr³

[volume sphere r=(3958.75+22237)] - [volume sphere r=(3958.75+99)]

[7.52976×10^13] - [2.79862×10^11] =

**75017738000000 CUBIC MILES**For our satellite we will assume it is a full 35' cube =

**0.00000029127 cubic miles**or0.00000029127 / 75017738000000 * 100 =

**0.000000000000000000388 PERCENT of the space**.Something that small is reflecting that much less light, plus the light falls off with the SQUARE of the distance -- and you are pitting it against vastly closer and thus brighter objects but your camera can't even capture the highlights in the clouds when taking a picture of your friend in the shade.

To be fair, some regions, such as the geosynchronous orbits, are more crowded but those are the ones that are about 22237 miles away. The angular size of a distant object is given by α = 2*arctan(g/r/2) so that means that even a 100' satellite, 22237 miles away would be a mere 0.176 arcseconds in size - good luck resolving that.

So yikes, off by about 20 orders of magnitude which is just comically wrong. That is SOME SERIOUS rounding function you got there Flat Earth!

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