# How far could you see from a planet

1. Jun 18, 2014

### StrangeCoin

How far could you see, on average, from a planet that was surrounded with infinite number of identical stars like the Sun, randomly (uniformly) distributed with their average distance of 10 light years apart? In other words, at what point of distance will such distribution tend to "close in" and block the view to any further away stars. Thanks.

2. Jun 18, 2014

### micromass

Staff Emeritus
Countably infinite or uncountably infinite?

Do we treat stars as points?

3. Jun 18, 2014

### StrangeCoin

I didn't know the first one existed. If I have to pick, then whatever is more infinite, I think the second one.

Not sure, if necessary I guess. What's the alternative and how they produce different results?

4. Jun 18, 2014

### micromass

Staff Emeritus
Then I can't answer it since I have no idea how to do probability with uncountably many objects in the way you mean. I'm not even sure if the 10 light year condition can ever be satisfied.

Picking countable infinity has problem too since the answer is then automatically infinity. Unless you don't treat stars a points but as little spheres.

Also, I'm not sure what your 10 light years condition actually means. Do you mean (naively) that the average of all distances between ALL individual stars is 10 light years? Then you probably won't have a uniform distribution.

About a uniform distribution. Do you assume space to be infinite and that the stars can literally pick any place in space with the same probability? I'm afraid this is impossible.

5. Jun 18, 2014

### Stephen Tashi

6. Jun 18, 2014

### StrangeCoin

You know what I mean, please re-define it in whatever way is the most appropriate. As for the distribution, maybe if instead say there is on average 10 stars per any 100 LY cube volume of space?

Whichever is more "realistic", spheres we need I'd say.

Yes, infinite universe with infinite stars, it's just a hypothetical scenario. Surely the stars will "close in" and block the view completely to further away stars before the distance reaches infinity, so infinity shouldn't bother us really.

7. Jun 20, 2014

### Dr J L

I'm no scientist, but It seems to me you would have to take into account perception, of distance before you can determine starting point. Because each angle will be a different distance. The probabilities will be different due to perception.

8. Jun 20, 2014

### Matterwave

So, I'm going to make the assumption that you mean that in every 1000 (10^3) cubic light years you have roughly 1 star the size of the Sun.

We can make some estimates based on this (of course a better answer requires more details).

A star the size of the sun of radius R located at distance D obscures an area of the sky that is basically going to be proportional to $\frac{\pi R^2}{4\pi D^2}=\frac{1}{4}\frac{R^2}{D^2}$. This is how many steradians of the sky you will be obscuring. The sky has a total of 4pi steradians.

A lower bound for the distance can be found if the stars are not allowed to "overlap" in the sky. In this case, you need that:

$$\sum_i \frac{1}{4}\frac{R^2}{D_i^2}=4\pi$$

In a spherical shell of thickness $dD$ (assumed small), located a distance $D$ from you, there are roughly $\frac{4\pi D^2 dD}{1000 ly^3}$ stars in this shell.

So we should integrate out to find:

$$4\pi=\int_0^x \frac{1}{4}\frac{R^2}{D^2}\frac{4\pi D^2}{1000 ly^3} dD$$

This gives:

$$4\pi=\frac{R^2 x}{1000 ly^3}$$

Substituting the radius of the Sun, we find:

$$x=\frac{(4\pi)(1000 ly^3)}{R^2}\approx 2.3\times 10^{18}ly$$

This is way larger than our current observable universe (50 million times larger).

9. Jun 20, 2014

### StrangeCoin

Doctor but not scientist, a doctor of love I guess. Doc, I think it's misleading to look at it from the observer point of view. It would perhaps be more insightful if we look at it from the point of view of a photon that needs to travel all the way to our little planet without crashing into a star on it's way.

Or perhaps better yet, we could look at it from above, from birds-view perspective. Imagine a wooden board with randomly placed nails and you have a small ball that needs go through the nails from one side of the board to another. Field of view and perspective doesn't really matter, the chance for the ball to pass the nails depends only on the distribution of the nails and the distance the ball needs to travel, plus the size of the nails and the size of the ball of course.

10. Jun 20, 2014

### Dr J L

I'm just here to learn. But learning start from perception within any mathematical equation on earth. If you were say on the moon or any where that does not have gravitational, pull then the perception would be different. Perception starts at any stand point, doesn't matter if on the ground or on mars or on the moon. The variables will always be different unless in a double blind study. Plus when you are in water the perception will be different because the gravitational pull is different.

Observation starts with a thought that creates a mathematical equation. That equation needs to be put to test. Doesn't matter the size of object or the velocity it is the starting point as to where things end up. To start from the observer to the observee to get reasoning and deduction to make a mathematical equation. So basically science has not figured out yet that the universe is infinite and they need a starting point. That point can be on earth or on the moon or in water. I don't want to get off topic, cause I'm here to learn. Basically in the human body same as any planet or star there is random energy, that science has yet to explain.

11. Jun 20, 2014

### StrangeCoin

I think not allowing them to overlap imposes unnatural constraints on the starting premise that the distribution is random. Valuable example in any case.

Thank you for that. It's beyond me though, so could you please tell us something about how it scales, how that distance varies with the stars distribution density, what is their relation?

12. Jun 20, 2014

### Hornbein

13. Jun 20, 2014

### Matterwave

This is true. But putting in a truly random distribution means additionally calculating the probability that a star farther away is obscured in view by a closer star in the distribution. No simple way to implement this come to mind. In addition, stars are NOT randomly distributed, they are clumped together in galaxies and galaxy clusters and super clusters and super structures, etc. So, it's not like the starting premise was a realistic one anyways.

You can see this in the final answer:

$$x=\frac{4\pi\bar{D}^3}{R^2}$$

Where $\bar{D}$ is the average distance between stars. So if the average distance between stars was 2 times closer, then $x$ would be smaller by a factor of 8.

The largeness of $x$ reflects the fact that space is mostly empty space. The average distance between stars is much much larger than the average size of a star (assumed to be the Sun), thus, each star only obstructs a tiny amount of your field of view. This can be readily seen in the night sky. The observable universe is several tens of billions of light years in diameter. In all this space, there are >200 billion galaxies on average each containing more than a hundred billion stars. Of all of these stars and all of this distance, the night sky is still dark.

14. Jun 20, 2014

### StrangeCoin

When you say "space" in that context it seems to me we are not talking about the same thing. I'm really talking about a chance, some probability percentage, rather than anything geometrical.

Code (Text):

x   x        x   x    x     x   xx    x     x
x       x          x     x        x      x
x    x    xx        x  xx          x
x                   x x       x    xx    x
x    x    x               x      x
x           x      xx  x                 x             [o] cyclops eye
x     x x     x  x          x    x
x          x        x      x      x
x x   x x           x  xx            x
x             x x       x    x    x
xx       x       x x        x      x
x       x       x           x         xx  x

This is how I see the problem, in 2D. All the 'x' are shooting little balls, as the distance to x increases, the chance for its balls to reach the eye, decreases. Now, if that chance does not drop to zero before the distance reaches infinity, then that means the distribution was not random to start with. Would that be true, can we make such claim about "random"?

15. Jun 20, 2014

### StrangeCoin

The thing about it is that, perhaps somewhat contrary to intuition, the more they obscure each other, the more they "clump" or "cluster" together, the number of visible stars would actually become less, not more.

So then the maximum "clump" factor would be if all the stars were in-line one right after another, and then only the closest star would be visible. It would be interesting to see results if we could include this "clump" factor into the equation.

I was searching the internet to find something related to this problem, but I miserably failed, likely due to my inability to recognize the problem expressed in some unrelated example. "Poisson distribution" kept showing in the search though, so whatever is that about might hold a solution, or at least point to some other, maybe more practical, way to think about it.

Thanks.

Last edited: Jun 20, 2014
16. Jun 20, 2014

### Matterwave

If you input clumping of the stars, then obviously the answer will depend on which direction you look in. If you tend to look in a direction with a lot of stars, then you won't be able to see as far, if you look in a direction with few stars, then you can look much farther.

17. Jun 20, 2014

### StrangeCoin

We are looking for some 'average' number then. But, "direction with a lot of stars" and "direction with fewer stars" doesn't really make sense in relation to infinity, nor random distribution. Every direction should on average hold the same number of stars, some grouped closer, some grouped further away, but at some point in infinity every line of sight is supposed to go through about equal amount of stars.

18. Jun 20, 2014

### Matterwave

Well, if you can come up with a model for that, let me know.

19. Jun 20, 2014

### StrangeCoin

I don't expect to come up with it, but rather to find it. Someone must have thought of something like this before. If I only knew what to look for, or at least recognize it when I find it.