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I know that we are able to "see back into time" by looking very far away through telescopes such as the Hubble telescope, but my question is, wouldn't you run into at least a couple of stars that were blocking your vision when you were trying to look even farther?

Imagine a big Boulder in front of another smaller Boulder and if we are looking from the side of the larger Boulder, we wouldn't be able to see the smaller one because the bigger one was in the way. Hopefully that simplifies my question. If it is as simple as that, wouldn't we miss stars and not be able to view them?

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  • $\begingroup$ This sounds half-way to Olbers's Paradox $\endgroup$
    – Nick T
    Commented Jun 27, 2017 at 17:40

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OK, Imagine the stars at distance x block an area of the sky. At a distance of 2x there should be four times as many stars, but they would seem four times smaller in terms of area covered. Thus, the blocking grows linearly until a significant part of the sky is blocked. The stars within 20 light years (excluding the Sun), blocks approximately $4.3 \cdot 10^{-16}$ of the sky. Even if we scale that up to the radius of the universe, 45 billion light years, only about $2 \cdot 10^{-6}$ of the sky is covered.

Even that tiny number is an overestimate, as we live in an extremely star-dense part of the universe, our galaxy.

Space is simply too big.

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  • $\begingroup$ Great answer, but where do you have the $3\times10^{-19}$ from? $\endgroup$
    – pela
    Commented Feb 14, 2016 at 14:41
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    $\begingroup$ @pela It is calculated from the catalogue of stars within 21 light years johnstonsarchive.net/astro/nearstar.html , using an average radius. It may not be exact or representative in any way, and therefore deviate with at least one order of magnitude. But even then the conclusion holds. $\endgroup$
    – SE - stop firing the good guys
    Commented Feb 14, 2016 at 14:56
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    $\begingroup$ @RobJeffries One can apparently not be careful enough with data. Manually checking the result of my table look-up, I found an error. The RMS stellar radius is 0.55 of the Sun and the representative distance 15 ly. The correct fraction is then $4.3 \cdot 10^{-16}$. Sorry about the previous error. $\endgroup$
    – SE - stop firing the good guys
    Commented Feb 14, 2016 at 17:26

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