Weird universe

Suppose that we live in some crazy open universe, also the concentration of stars is constant and equal to n = 1 0 9 1 M p c 3 n=10^9 \frac{1}{Mpc^3} and the mean radius of the star is R = 1 π 1 0 8 km R=\frac1{\sqrt \pi} 10^8 \text{ km} . Find the average distance 'view' will travel before it 'hits' the star. Answer in M p c Mpc

Example: if you look at tree 9m away from your eyes your look will travel distance of 9m.

1 M p c = 1 0 6 × 206265 × 1.5 × 1 0 8 km *1Mpc=10^6\times 206265 \times 1.5\times 10^8 \text{ km}

This is one of short problems from IOAA 2015.


The answer is 9.57E+13.

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1 solution

Let's separate a whole universe in a thin shell of thickness d r dr and radius r r . Probability of finding star in one of shell is

d P = S e f f e c t i v e ˙ n ˙ 4 π r 2 d r dP=S_{effective}\dot{}n\dot{} 4\pi r^2 dr

where S e f f e c t i v e = π R 2 4 π r 2 S_{effective}=\frac{\pi R^2}{4\pi r^2}

Now the probability of finding star in first or second,.. or n-th shell is equal to sums of all probabilities in a single shells. So to be sure you have found a star your probability must be equal to 1 1 so now suppose that it is going to happen in a shell of radius r k r_k we have equation:

0 1 d P = π R 2 ˙ n ˙ 0 r k d r \int_{0}^{1} dP=\pi R^2\dot{} n\dot{} \int_{0}^{r_k} dr

What leads to:

r k = 1 π R 2 ˙ n r_k=\frac{1}{\pi R^2\dot{} n}

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