0 Members and 1 Guest are viewing this topic.

All but a few stars appear as mere pinpoints in even the largest telescopes. They are much too far away to derive their diameters from measuring their angular diameters and distances.

"Vanish" means you can't see it. We see things because the emit light. Received photon = visible, no received photon = vanished.Just because your eye, or even a telescope, can't measure the diameter of something, doesn't mean it has vanished....sorry, you cannot view external links. To see them, please REGISTER or LOGIN QuoteAll but a few stars appear as mere pinpoints in even the largest telescopes. They are much too far away to derive their diameters from measuring their angular diameters and distances.

To extend why I asked the question is because if we was to observe from a star that was far way looking in at our galaxy and the sun was to small to see, it had past position x, the observer would see a black hole?

does the suns diameter r2*2/2?

Is twice the distance of radius of a mass from an observer , diameter of the mass divided by 2? Is that the scale?

Because r2*2/2=2r=d you are asking is the sun's diameter its diameter. This sort of question puts people off answering.

Quote from: Thebox on 27/10/2015 18:14:08Is twice the distance of radius of a mass from an observer , diameter of the mass divided by 2? Is that the scale?What do you mean by this? What scale?Why have you just introduced mass into a question about visibility.Can you be more explicit about what you are asking?

This sounds more like a question of trigonometry...The apparent size of the sun (A, measured as an angle) will be A = 2*sin^{–1}(r_{sun}/r_{obs})where A = apparent sizesin^{–1} is the inverse sine or arcsine functionr_{sun} = radius of the sun in kmr_{obs} = distance of observer to the sunwhen A = 0, the sun has vanished (the only problem is there is no value of r_{obs} for which the A is 0 (as long as r_{sun} is nonzero; A will only approach 0 as r_{obs} approaches infinity)

Quote from: chiralSPO on 27/10/2015 18:52:06This sounds more like a question of trigonometry...The apparent size of the sun (A, measured as an angle) will be A = 2*sin^{–1}(r_{sun}/r_{obs})where A = apparent sizesin^{–1} is the inverse sine or arcsine functionr_{sun} = radius of the sun in kmr_{obs} = distance of observer to the sunwhen A = 0, the sun has vanished (the only problem is there is no value of r_{obs} for which the A is 0 (as long as r_{sun} is nonzero; A will only approach 0 as r_{obs} approaches infinity)That can't be true, r_{obs}=y & r_{sun}=x then +dy=-dx?

No. As I defined it, r_{sun} is a constant (the actual radius of the sun), so your dx =d(r_{sun}) = 0.

Quote from: chiralSPO on 27/10/2015 19:48:54No. As I defined it, r_{sun} is a constant (the actual radius of the sun), so your dx =d(r_{sun}) = 0.Obviously the sun does not change size, I am talking visual observation, an object relative shrinks the more distant away from it, so relatively the radius of the sun shrinks relative to distance to an observer. At a point Z point X viewed along a Y-axis will ''vanish'' to observation, so distance must have a scaling system, that is why we have magnification is it not?Your avoiding the actual question and coming back with something not the same as the question, my answer requires a distance of r_{obs} that r_{sun} ''vanishes''. At what distance from the earth does 1392000km of diameter of the sun ''vanish'' if the sun was moving away?

Naked Eye Angular resolution: about 1 arcminute, approximately 0.02° or 0.0003 milliradians,[1] which corresponds to 0.3 m at a 1 km distance.

Quote from: WikipediaNaked Eye Angular resolution: about 1 arcminute, approximately 0.02° or 0.0003 milliradians,[1] which corresponds to 0.3 m at a 1 km distance.On an astronomical eye-chart, the "Sun Line" would become unreadable around the orbit of Uranus.But the Sun is not a black letter on a white eye-chart, or even a white letter against the black backdrop of space.Astronomers measure the visibility of a star as "Apparent Magnitude". Magnitude 7 is the faintest anyone can see with the naked eye, under excellent conditions.Astronomers compare the brightness of different stars by "Absolute Magnitude", which is the Apparent Magnitude at a distance of 10 parsecs, or 32 light-years. The Absolute Magnitude of the Sun is 5, which is still very visible.To reduce the Sun to magnitude 7 (ie invisible to the naked eye), you would need to be at a distance of about 80 light-years. (If I've done the calculations correctly!)See:...sorry, you cannot view external links. To see them, please REGISTER or LOGIN

Thank you, so in reverse if we was 81 light years away from the Sun, we would observe nothing of the sun?

Quote from: Thebox on 27/10/2015 21:03:38Thank you, so in reverse if we was 81 light years away from the Sun, we would observe nothing of the sun?Evan means we could not see the sun by naked eye if it was 81 ly away. With a telescope, we could easily observe the sun from this distance...

....I am presuming now that all the stars we see in the night sky by eye are much further away than 80ly.....

Quote from: Thebox on 27/10/2015 23:12:55....I am presuming now that all the stars we see in the night sky by eye are much further away than 80ly.....Why would you assume something that is incorrect when it is easy to check on the net?

[so If my assumption is incorrect, then why please?

Your avoiding the actual question and coming back with something not the same as the question,

I am presuming now that all the stars we see in the night sky by eye are much further away than 80ly

if the sun was too small to see, ...the observer would see a black hole?

What do you mean by "bigger"?

I am a bit confused you have to ask what I mean by bigger, if you are 20 stone and I am 10 stone, you are bigger than me, if you are 6ft and I am 4ft you are bigger than me.

if star had a smaller radius than our suns radius and was beyond lets say 70 ly away, we would not see it by eye, if the star was bigger and further away we would still see it?

Quote from: Thebox on 28/10/2015 11:34:24I am a bit confused you have to ask what I mean by bigger, if you are 20 stone and I am 10 stone, you are bigger than me, if you are 6ft and I am 4ft you are bigger than me. You have just provided 2 definitions of bigger. If you had asked whether one man was bigger than another Evan would quite reasonably ask what you meant by bigger.This is a similar problem with stars as you have talked about both mass and radius.Quote from: Thebox on 28/10/2015 11:34:24 if star had a smaller radius than our suns radius and was beyond lets say 70 ly away, we would not see it by eye, if the star was bigger and further away we would still see it?Ok, this is better because you are being more specific. However, if you read evan's reply you will see he mentions brightness. It is this, measured as magnitude rather than diameter or mass which determines whether a star, planet, comet is visible.You might find this chart helpful ...sorry, you cannot view external links. To see them, please REGISTER or LOGIN

... I do want to be able to communicate properly and to be understood, the problem with messages is they can be read several ways. I consider I do have some great insight into several things, I do know though if nobody can understand them I will never be heard.

I do have to question now why science considers light magnitude before radius of a star.

I do not understand how science can say something is x big when there is no background for comparison. How do we know the distance stars are not just small and close?

Image you have 2 light bulbs one larger than the other but both of same wattage. At 2km they will both be tiny points of light and you won't be able to tell them apart. The same with stars, at distances of 2ly you can't differentiate size with the naked eye.Now imagine the smaller bulb is twice the wattage of the larger one. In this case you will be able to see the smaller bulb from further away than the larger.

How do we know the distance stars are not just small and close?

So although they look like they are both at 2km, they are not. A sort of illusion created by scaling and distance ?each one of the A's in this diagram are a different size.

Agreed, all very obvious really.