A question for those with telescopes. (Earth, stars, light, spacecraft)

[ocpaul20]

In 2017/2018, how large does does a lunar feature have to be to be seen clearly from Earth?

I guess what I am asking is, if we look at the Moon through an optical telescope, what is the smallest feature we can see?

Could we see something that was 10 feet wide for example? or 20ft or 50ft, or 100ft.

I realise it depends on the telescope and Earth weather and atmosphere viewing conditions, but what is ideally possible these days with amateur telescopes?

If I was looking for a telescope, what could I expect from it if I looked at the Moon?

[Michael Way]

Originally Posted by ocpaul20

In 2017/2018, how large does does a lunar feature have to be to be seen clearly from Earth?

I guess what I am asking is, if we look at the Moon through an optical telescope, what is the smallest feature we can see?

Could we see something that was 10 feet wide for example? or 20ft or 50ft, or 100ft.

I realise it depends on the telescope and Earth weather and atmosphere viewing conditions, but what is ideally possible these days with amateur telescopes?

If I was looking for a telescope, what could I expect from it if I looked at the Moon?

''A convenient expression of the resolution of a telescope used visually is the Dawes limit, which says that the smallest angle we can resolve (in arcseconds) equals 116/D, where D is the aperture’s diameter in millimeters. If we could look at the Moon through the Hubble Space Telescope, for which D is 2,400 mm (94.5 inches), we’d be able to discern surface features as small as 0.05 arcsecond. When the Moon is closest to Earth, only 356,000 kilometers (221,000 miles) away, 0.05 arcsecond corresponds to about 85 meters (280 feet). Not only is this insufficient to resolve a person on the Moon, but it’s not even good enough to detect the 10-meter-wide Apollo landers left there in the 1960s and ’70s.

So how big a telescope would it take to see a moonwalker? To get less than 2-meter (6-foot)resolution when the Moon is closest, we’d need a telescope able to resolve angles as small as 0.001 arcsecond. That translates into a diameter exceeding 100 meters (4,000 inches). And this behemoth would have to be in space, too, since atmospheric seeing limits even the biggest ground-based scopes to an angular resolution of at best a half arcsecond or so, corresponding to lunar features no smaller than about 1 km (0.6 mile).''

Could you see astronauts on the Moon? - Sky & Telescope

According to the information above, to see an astronaut on the moon through a telescope you would need to have a 100 meter (4,000 inches) diameter lens, and it would have to be in space above the earth's atmosphere. There is no way then that a ground based telescope can resolve a feature of 10 feet or even 100 ft.

Even the Hubble telescope can't discern lunar features smaller than about 328 ft across.

''One of the most common questions asked by the public when we’re looking at the moon through a telescope is why we can’t we see the American flags or any other sign of Apollo with the Hubble Space Telescope. It IS the most powerful telescope, right? Here’s the rub. The smallest possible thing Hubble can see on the moon is about 328 feet across or the length of a football field. While impressive feat of resolution, no Apollo spacecraft comes anywhere near that size. Every piece of man-made hardware is below the space telescope’s resolution limit.''

https://astrobob.areavoices.com/2012...-the-moon-yes/

[ocpaul20]

Maybe someone can check my calculations please.

===============================
Using NASA Moon fact sheet
Moon distance from Earth (equator, km) 378,000 (NASA)
Apparent diameter (seconds of arc) 1896 (NASA)
Moon equator radius = 1738.1Km(NASA)

diameter= 2xradius = 17,381,000*2 = 34,762,000m
Apparent diameter (seconds of arc) 1896 = 1896/60 = 31.15 arc-minutes

I got this calculation from a Yahoo answers page.
A = angular size in arcminutes of an object (galaxy, nebula, etc.)
D = distance to object
S = linear size of object (in same units as D)

Then
S = A * D * 0.000291
This number, 0.000291, is the tangent of 1 arc-minute.

Plugging in our values...
S = 31.6 * 378,000,000m * 0.000291 = 34,759,368m diameter (near enough NASA's calc)

Using diameter from NASA above
============================
34762000/31.6 = 1,100,063m per arc-minute at the Moon equator
1100063/60 = 18,334m per arc-second = 18.3Km per arc-second at the equator

So, if I have got my calculation correct, it all depends on the effective resolution of the Earth-based telescope.


The Palomar Observatory built in 1948 (1949 first light) has a 200 inch mirror.

The reflective surface is adjusted in real-time, up to 2000 times a second, to correct for atmospheric distortions and refocus starlight into sharp images. PALM-3000 brings the optical power of the Hale Telescope closer to its diffraction limit by producing images typically 10–20 times sharper than seeing-limited instruments.

Astronomers measure the resolving power of a telescope in terms of degrees. Degrees are further divided into arcminutes and arcseconds.

1 degree = 60 arcminutes = 3600 arcseconds

The best telescopes on the ground can rarely see detail that is less than 1 arcsecond wide or differentiate between two stars that are less than 1 arcsecond apart in the sky.

Hubble can see detail down to less than 0.1 arcsecond across — more than 10 times clearer.

From that page it says , the full moon is roughly 0.5 degrees or 1800 arcseconds across in a telescope view.

So, the theoretical Dawes limit is for the Hale Telescope
116/508cm = 116/50800mm = 0.0023 arcseconds

[TimAZ]

The "theoretical" resolution of a telescope is just that. Telescopes in the real world have errors in their optics, misalignments, and on earth they look through the atmosphere above. All of these factors reduce the performance and distort or blur images so the actual resolution is always worse than the theoretical number.

[ocpaul20]

Originally Posted by TimAZ

The "theoretical" resolution of a telescope is just that. Telescopes in the real world have errors in their optics, misalignments, and on earth they look through the atmosphere above. All of these factors reduce the performance and distort or blur images so the actual resolution is always worse than the theoretical number.

I realise this, but if I have got these calculations correct, the theoretical resolution is amazing and the ACTUAL resolution is going to end up pretty good with all the different technologies used. So, 23 thousandths of an arcsecond is theoretically possibly, so what would a practical value be on a good night? Pulling a figure out of the air maybe 1/100 arcsecond perhaps? Looks like that would be 182m per 1/00 arcsecond

The best telescopes on the ground can rarely see detail that is less than 1 arcsecond wide

Assuming that this is the best they can do, we are talking about being able to resolve a structure on the Moon that is 18,000m long. I am sure they can do better than that, given the theoretical limits we have calculated.

Maybe someone on here knows some correct resolution figures for the Hale Telescope so that we can get some accurate figures rather than estimates?

There must have been loads of people who have participated in viewing nights held there who would know this information.