Photos of Space from Space.

[Zargon]

Hello. As a kid I used to wake up early to watch the Geminis take off from Capr Canavral.
was very excitting as a kid. BUT I never saw a picture of space. When I got older I got into
Electronics. I learned a formula. Path Loss in DB=37+20LogDin Miles+20LogF in Mhz...
Then someone said that don't work for light.. They said the Level of light 1000 miles above the Earth is the Same level
in DB as is as seen from the Planets serface... What they said was the level of brightness of any given star as seen with the human eye in space is the same when viewed on earth... Do you believe that? So my question is>> I have Never seen a photo from space OF space Besides Hubble.. I have never seen a photo of a star from any astronaut with a camera in space.
When you see the guys working on hubble you Never see stars in the background.. Even photos of the earth from the moon you don't see stars..
I apologize for my ignorance but could someone please answer my 45 year question?

[makc]

this is simple - stars do not exist no, really, this is because hubble is one of the few tools dispatched to space to take pictures of stars. evrything you mention are pictures of brighter objects (such as planets or astronauts).

same is here, on earth. take a camera, go out at night, and take a picture of moon and stars together. I bet you can't do it in a single exposure without moon looking like solid white disk.

[BMAONE23]

Most of the Images taken from space by astronauts are concerning subject matter that doesn't relate to the stars. When you're "up there" your focus isn't on the stars. There is this humongous blue marble below you that is so mind-blowingly beautiful, the stars seem to pale in comparison. And this paling effect is also apparent in images taken in space. When the astronauts are taking images in space, the subject matter (Earth, ISS, Shuttle, Hubble, (from the) Lunar Surface, etc...) is so intrinsically bright that the camera's iris is forced to operate faster to avoid pixel burnout (photonic over infusion). The main subjects are usually so bright that, in order to avoid overexposure of the image subject, the stars literally Pale in comparison and do not get imaged.

[BMAONE23]

The APOD for 09-05-20 "Above Earth Fixing Hubble" is a fair example of an image from space that is awash in light with the earth in the background. Though, if you look in the lower right corner of the image you will find 3 stars leaving slight trails

[Chris Peterson]

BUT I never saw a picture of space. When I got older I got into
Electronics. I learned a formula. Path Loss in DB=37+20LogDin Miles+20LogF in Mhz...
Then someone said that don't work for light.. They said the Level of light 1000 miles above the Earth is the Same level
in DB as is as seen from the Planets serface... What they said was the level of brightness of any given star as seen with the human eye in space is the same when viewed on earth...

No, that's not true. But I don't think your formula is applicable to optical wavelengths.

The degree to which the atmosphere attenuates light is called atmospheric extinction. At sea level, for visible wavelengths, it is about 0.28 magnitudes per air mass. One air mass is the column of air between the ground and space. In other words, a star that you observe at the zenith will appear 0.28 magnitudes brighter above the atmosphere. That's brighter, but not by much. The attenuation becomes a lot greater as you look at stars at lower altitudes, however. Obviously, you are looking through more air when you do that. At 45° the attenuation is 0.4 magnitudes. That's still fairly subtle. But you lose an entire magnitude compared to the zenith by the time you're down to 13° above the horizon: the extinction there is 1.28 magnitudes. At the horizon, you're attenuating stars by about 11 magnitudes, which is why you don't see many stars that low.

Realistically, however, what you see staring straight up into the sky from a dark location on Earth isn't going to look very different from what you'd see in space. And it's more than an astronaut will see, looking through his shielded glass visor.

The reason you don't normally see stars in images from space is that they are exposed for sunlit objects, and most stars are far too dim to show up in such exposures.

[bystander]

The APOD for 09-05-20 "Above Earth Fixing Hubble" ...

Other posts have been split and merged with Fixing HST (APOD 2009 May 20).

[Zargon]

Thanks for your reply. It was very informing..

.28 magnatudes per air mass is something I knew not.

That answers the question of photos of space from space...

As far as the path loss formula.
That is correct, I believe, for electromagnetic radiation..
I don't know if light is electromagnetic radiation.
Photons: but do they spin off emf fields?
Interesting.

So maybe the DB Formula don't work for light.
I don't have a degree in physics.

Loss in DB=37+20LogD in Miles + 20LogF in Mhz.
Maybe not for light?

[Zargon]

I thought a little more about that.
The light has Already gone Many "D's" in distance.

The light level at the point of inception with the earths surface
is down only .28 magnitudes from level at the point of interception with the earths upper atmosphere.

And that is due to Atmospheric absorbition.
.28 magnitudes per air mass.

I would assume the .28 magnitudes is on a clear day at 25 degree C and at sea level.

Something like that.

Have a good day..

[Chris Peterson]

As far as the path loss formula.
That is correct, I believe, for electromagnetic radiation..
I don't know if light is electromagnetic radiation.

Light is certainly electromagnetic radiation. But I don't think the formula is valid for any EMR. It is probably derived empirically for a certain narrow wavelength range, part of the radio band. It doesn't make sense that a single formula could describe attenuation at all wavelengths for any single medium. That is determined by a complex mix of factors. The atmosphere, for example, is nearly opaque at some wavelengths, and nearly transparent at others.

[apodman]

I don't think the formula is valid for any EMR. ... The atmosphere, for example, is nearly opaque at some wavelengths, and nearly transparent at others.

As you go from visible wavelengths toward shorter wavelengths, the extinction becomes more pronounced. It still works with the number of air masses as an exponent, but with a larger factor. Extinction of UVA (the ultraviolet "light" closest to visible) is only a little more pronounced than visible light, so you can get a suntan from prolonged exposure in the early morning and late afternoon. Extinction of UVB (farther from visible wavelengths than UVA) is more pronounced, so you get your worst sunburn within a couple of hours before or after high noon and are pretty safe (or at least need less sunblock) before mid-morning and after mid-afternoon. Extinction of UVC (farther yet from visible wavelengths) is extremely pronounced, so you don't get many UVC rays even at mid-day (and a good thing, too, 'cause they would kill you). Unlike atmospheric extinction of visible wavelengths, UV is extincted mostly by ozone in the atmosphere (which means it's hard to get a tan through city smog). If you never threw out your old stack of Sky & Telescope, read "The Astrophysics of Suntanning" in the June 1988 issue.