# I really need your help with some calculations

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Posted (edited)

This time I am contacting you because I need the help of talented and experienced people for a math problem related to astronomy. This will not be a discussion and my purpose is not to criticize or question your skills and knowledge. I really need your help and all kind of help is appreciated. And if you are a member of this website, you already qualify to help me and will earn the corresponding respect. It is also allowed to make mistakes at first and correct them later to obtain an even better solution (we are all humans after all). Now this is the problem:

Imagine we are standing on Earth and staring at the star Proxima Centauri which has a size of about 1.02 ± 0.08 milliarcsec in the sky. This is about 0,00027778 degrees (correct me if I am wrong).

Now, imagine if the earth rotates and Proxima Centauri is situated just below us behind Earth so we can no longer see it. What I wanted to calculate is the area Proxima Centauri would be shining on on the opposite site of earth corresponding to the point we would see if earth was not in the way. So I took the tangent equation and using the angle alpha of 0,000277778 degrees and the diameter of earth which is about 12.742 kilometres, I obtained the solution that the diameter of the area beeing hit by radiation coming from Proxima Centauri which would become a dot after crossing the diameter of earth and hit us is of about 61,77 meters (once again correct me if I am wrong).

The problem is that this diameter was calculated having in mind that the observer standing on earth is just a dot, so the angle is solely based on the size of Proxima Centauri. But what I now need to know is how this angle would change if the observer was for example an object of the size of 1 meter or 2 meters or 10 meters. And how would increasing the size of the observer increase the diameter of the area on the opposite side of Earth that is beeing hit by radiation coming from Proxima Centauri? Of course, I am only interested in the area of Earth beeing hit that is clearly in the way between Proxima Centauri and the observer. I am aware that Proxima Centauri is always hitting about half the surface of Earth with its radiation, if there are no other objects in the way, but only a small portion of this area would indeed be in the way between Proxima Centauri and the observer.

Please, to make the calculations easier and to first obtain a certain notion of what I am dealing with, consider the surface on the other side as if it was "flat", because this calculation would become incredibly complex if we would incorporate the curvature of Earth, and this might not be important as long as the observer and the area beeing hit by the radiation stay small. Of course, once we are talking about kilometres, Earth´s curvature will start playing an important role and if the observer has the size of Earth him/herself, the total area would be about half the area of Earth.

Any formulas are welcome, but please try to explain things so easy that even I can understand them.

And if you have any questions or do not understand a part of the text, just let me know.

Thank you in advance for your help!

Edited by tmdarkmatter
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13 minutes ago, tmdarkmatter said:

This will not be a discussion

!

Moderator Note

If it's here, and in a mainstream science section, it WILL be a discussion or it will be closed. I'm not sure what form you think feedback on this matter will take, but here at SFN, it's going to be discussed. If you have a viable alternative, we can discuss it. Oooh, the irony!

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I am sorry, what I mean is that I need help and that I am not interested in "defending my point". I just need a solution to this problem.

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54 minutes ago, tmdarkmatter said:

The problem is that this diameter was calculated having in mind that the observer standing on earth is just a dot, so the angle is solely based on the size of Proxima Centauri. But what I now need to know is how this angle would change if the observer was for example an object of the size of 1 meter or 2 meters or 10 meters.

This makes no sense. An angle, by definition, assumes a vertex, which is a point.

Quote

Now, imagine if the earth rotates and Proxima Centauri is situated just below us behind Earth so we can no longer see it. What I wanted to calculate is the area Proxima Centauri would be shining on on the opposite site of earth corresponding to the point we would see if earth was not in the way.

The light from Proxima Centauri hitting the earth can be treated as parallel over such a short distance as the diameter of the earth.

The angular size of the star is not the same as the divergence of light from it.

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2 hours ago, swansont said:

This makes no sense. An angle, by definition, assumes a vertex, which is a point.

That´s the problem. I need to go from a point to an area or distance.

2 hours ago, swansont said:

The light from Proxima Centauri hitting the earth can be treated as parallel over such a short distance as the diameter of the earth.

The angular size of the star is not the same as the divergence of light from it.

Maybe I can also ask you the question: How big must an object be so at a distance of the diameter of earth it is able to block the light coming from Proxima Centauri so it creates a shadow that is just big enough to cover an object of the size of 1 metre, 2 metres, 10 metres?

If you consider the trajectory of the light from Proxima Centauri as parallel, that would mean that a coin would block the light and that is not the case.

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12 minutes ago, tmdarkmatter said:

How big must an object be so at a distance of the diameter of earth it is able to block the light coming from Proxima Centauri so it creates a shadow that is just big enough to cover an object of the size of 1 metre, 2 metres, 10 metres?

1 metre, 2 metres, 10 metres, assuming the shadow is perpendicular to the light ray.

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But if the light coming from the star was perpendicular, the star would have the size of just a few millimetres and that is not the case. The further away we are, the smaller we see the star. At a further distance, the angle of the visible light coming from the star becomes smaller.

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3 minutes ago, tmdarkmatter said:

But if the light coming from the star was perpendicular, the star would have the size of just a few millimetres and that is not the case. The further away we are, the smaller we see the star. At a further distance, the angle of the visible light coming from the star becomes smaller.

At the very far distance, the angle between light rays from the star becomes 0, that is the rays are just parallel to each other. Draw the picture and you will see what happens to the shadow when the rays are parallel.

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If you can cover the moon with your thumb, how is this image possible?

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4 minutes ago, tmdarkmatter said:

If you can cover the moon with your thumb, how is this image possible?

They are both far away, such that lens compression has brought them together front to back. The naked eye view will do the same. What you are seeing is the relative  sizes brought together, then the brain does the rest to make sense of it. I think that's how it interprets the scene.

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In that case how do you explain the existence of solar eclipses if the diameter of the moon is 3,474.8 kilometres and the diameter of the sun 1,392,700 kilometres? And even worse, how do you explain that the solar eclipe is only visible on a line of just a few hundred kilometres of width on Earth considering the size of the moon?

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1 hour ago, tmdarkmatter said:

In that case how do you explain the existence of solar eclipses if the diameter of the moon is 3,474.8 kilometres and the diameter of the sun 1,392,700 kilometres? And even worse, how do you explain that the solar eclipe is only visible on a line of just a few hundred kilometres of width on Earth considering the size of the moon?

Assuming you are serious, here is a link from NASA on how eclipses work.  https://eclipse2017.nasa.gov/how-eclipses-work

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2 minutes ago, Bufofrog said:

Assuming you are serious, here is a link from NASA on how eclipses work.

I think you misunderstood my questions. I am not questioning how eclipses work. I am just trying to find a solution to my problem with my calculations. But as we can see in your images, the sun is much bigger than the moon and the moon is much bigger than its shadow on earth. Therefore, the size of an object at a distance of the diameter of earth should be much bigger than our thumb to cover Proxima Centauri, but how big should it be?

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3 hours ago, tmdarkmatter said:

But if the light coming from the star was perpendicular, the star would have the size of just a few millimetres and that is not the case. The further away we are, the smaller we see the star. At a further distance, the angle of the visible light coming from the star becomes smaller.

The size of the “light beam” is approximately the size of the source; even bigger, since over 4 LY you can’t ignore the divergence. Alpha Proxima is bigger than a few mm. The beam illuminates the whole earth, i.e. you could observe the star from any point on the surface unless earth itself is in the way.

3 hours ago, Genady said:

1 metre, 2 metres, 10 metres, assuming the shadow is perpendicular to the light ray.

Diffraction means you need something bigger.

2 hours ago, tmdarkmatter said:

In that case how do you explain the existence of solar eclipses if the diameter of the moon is 3,474.8 kilometres and the diameter of the sun 1,392,700 kilometres? And even worse, how do you explain that the solar eclipe is only visible on a line of just a few hundred kilometres of width on Earth considering the size of the moon?

Diffraction, and that the sun is bigger than the moon, and that the moon is some distance away. The light is not parallel. If you were closer to the moon, the dark spot would be bigger.

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1 minute ago, swansont said:

The beam illuminates the whole earth, i.e. you could observe the star from any point on the surface unless earth itself is in the way.

Yes, of course, this is obvious. But if I am standing on earth, I only see a dot in the sky. But this dot means a certain circle on the opposite side of earth, a circle of a certain size that is necessary to completely block that beam. Imagine if I am standing on a mountain and asking somebody to climb on another mountain about 100 kilometres away, I can see at the horizon and that Alpha Centauri will be visible just a few meters above that mountain from my perspective. So I am asking him to block Alpha Centauri for me. What is he going to use? A coin? A big sign? A house?

But even worse, I want him to create a shadow to Alpha Centauri that has the size of 1 square meter in my position. How do I calculate that?

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3 minutes ago, tmdarkmatter said:

Yes, of course, this is obvious. But if I am standing on earth, I only see a dot in the sky. But this dot means a certain circle on the opposite side of earth, a circle of a certain size that is necessary to completely block that beam. Imagine if I am standing on a mountain and asking somebody to climb on another mountain about 100 kilometres away, I can see at the horizon and that Alpha Centauri will be visible just a few meters above that mountain from my perspective. So I am asking him to block Alpha Centauri for me. What is he going to use? A coin? A big sign? A house?

If you want to block the whole star from all of earth, you need something the size of the earth.

If you just want to block one person from seeing it, you need something roughly the size of someone’s face, though diffraction will mess with this. Arago’s spot means light will still be there.

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1 minute ago, swansont said:

If you just want to block one person from seeing it, you need something roughly the size of someone’s face.

But the size of the object needed depends on the distance where the object is located. How do you calculate the size according to the distance?

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5 minutes ago, tmdarkmatter said:

Yes, of course, this is obvious. But if I am standing on earth, I only see a dot in the sky. But this dot means a certain circle on the opposite side of earth, a circle of a certain size that is necessary to completely block that beam. Imagine if I am standing on a mountain and asking somebody to climb on another mountain about 100 kilometres away, I can see at the horizon and that Alpha Centauri will be visible just a few meters above that mountain from my perspective. So I am asking him to block Alpha Centauri for me. What is he going to use? A coin? A big sign? A house?

Have you ever noticed that e.g. a tree or house some distance away can block your view of things further away?

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2 minutes ago, swansont said:

If you want to block the whole star from all of earth, you need something the size of the earth.

If I want to block the whole star from all of earth half way to Proxima Centauri, I might need a star about half the size of Proxima Centauri.

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Just now, tmdarkmatter said:

If I want to block the whole star from all of earth half way to Proxima Centauri, I might need a star about half the size of Proxima Centauri.

Yes.

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Just now, swansont said:

Yes.

But we are just saying things and I would need equations to calculate these shadows.

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Posted (edited)

As a rule of thumb (i.e. close enough), you'd block Proxima Centauri from all viewers on Earth at the same time with a disk the diameter of Earth, if the disk is next to Earth, or the diameter of Proxima Centauri if the disk is next to that star.

Anywhere in between would be a ratio.

(Give or take a little gravitational lensing, and assuming everything stays still ... etc ...)

Edited by pzkpfw
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15 minutes ago, tmdarkmatter said:

But we are just saying things and I would need equations to calculate these shadows.

The blocked light forms a cone (see the eclipse link for an example)

s = r*theta

s is the size of the block. theta is the angular size. r is the distance to the screen blocking the light.

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8 hours ago, swansont said:

s = r*theta

Yes, this is getting closer to what I need. We can also write it as theta = s/r indicating that if r increases with an increasing distance, s also has to increase for theta to be the same.

Later today I will do some calculations moving the observer away from the surface of earth. This would provide me a certain surface on earth as a blocker which can be compared to the blocker surface on the opposite side of earth. This way I can compare them and figure out how both values increase if the observer is moving away from the surface.

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Posted (edited)

Ok, I have done these simple calculations and the results are as follows:

If the observer is 1 km over the surface of earth, according to the tangent ecuation, the object on the surface of earth must have the size of 4,8481 mm (the size of a very small coin) to block Proxima Centauri. So if you are staring at a rooftop that is 1 km away and Proxima Centuari is just showing up behind the rooftop, you might estimate that a tiny coin would be enough to block the light coming from Proxima Centauri. But what is interesting is that the blocking object on the other side of earth at a distance of 12,742 km would also increase in size by 4,8481 mm, so the equation is totally linear. At a distance of 100 km above earth, the object on the surface would have a size of 0,48481 meters to block the light coming from Proxima Centauri. But what is interesting is that if I insert the total distance to Proxima Centauri in that equation, the result is an object 907 times the size of Proxima Centauri, meaning that the light we see coming from Proxima Centauri is 907 times bigger than the star itself, because we do not only see the star but also its bright surroundings. (Correct me if I am wrong)

But the ratio between blocking object on the surface and blocking object on the opposite side of earth would begin at zero when the observer is on the surface to a value of almost 1 when the observer is light years away.

And the size of the blocking object would in both cases become bigger than earth itself.

So if we are standing on earth and would fit in this area of 0,1844 square meters, the object to block the light from Alpha Centauri from us on the opposite of earth (maybe I can call it antishadow) would have to be 16.630,78 times bigger than ourselves. This seems to be a constant for all objects of all sizes on the surface in the case of the angle of Alpha Centauri.

Edited by tmdarkmatter

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