What is the Angular Resolution Formula for the Hubble Space Telescope?

[colourpalette]

My main issue with this problem is the formula and how to derive it.

Homework Statement

The Hubble Space Telescope has a resolution of about 0.05 arc second. What is the smallest object it could see on the surface of the Moon? Give your answer in metres.

Homework Equations

I tried to get my own equation since there was nothing relevant in my textbook, so I did
\theta/360 degrees = distance/2\pir
But it's wrong (why?)

I went online and I found
\theta/206265"=d/D
This gives the right answer but I don't understand where it came from.

The Attempt at a Solution

The answer is 93m, I just don't get the formula

 

[willem2]

1 arcsecond = 1/3600 degrees
180 degrees = \pi radians.

for small angles a \approx tan(a) if a is in radians.

You get 93 m if you apply this, but I have no Idea where the number 206265 comes from

 

[cepheid]

206265 is the number of arcseconds per radian.

 

[cepheid]

I tried to get my own equation since there was nothing relevant in my textbook, so I did
\theta/360 degrees = distance/2\pir
But it's wrong (why?)

It's wrong because 2*pi*r is the expression for the circumference of a circle. But there is no full circle here, so what are you trying to do?

I went online and I found
\theta/206265"=d/D
This gives the right answer but I don't understand where it came from.

remember that the definition of an angle (in radians) is:

theta = s/r​

You can think of an angle as being generated by sweeping out an arc (a portion of a circle) by rotating a line about a fixed point at one of its ends. Here, 's' is the length of the arc that is swept out by the radial line as it rotates by that angle, and 'r' is the radial distance to that arc (in other words it is the length of the line). So, in this case 's' is the size of the surface feature on the moon, and 'r' is the Earth-moon distance. (Alternatively you could use 'd' and 'D' or any other letters you want). You know theta and you know r, so all you have to do is solve this equation for s.

Since the equation is only valid for theta in radians, you have to convert theta to radians by multiplying it by (1/206265) radians/arcsecond.

 

[colourpalette]

It's wrong because 2*pi*r is the expression for the circumference of a circle. But there is no full circle here, so what are you trying to do?

What I was thinking was the part of the angle (or angular resolution) out of 360 degrees of the whole circle would be the same as arc distance out of the circumference of the circle. And I have the diameter, which is kind of close to the arc distance so my answer would be close.
I know where I went wrong now though
It's out of 2\pi, not 360 degrees, thanks for clearing that up!

 

[cepheid]

Ahh yes I see. Your approach was good, actually! You were thinking that the ratio of the angle subtended to one full rotation should be equal to the ratio of the arc length spanned to one full circle. This is another way of arriving at the same result. You just got tripped up by different unit systems.