# Question 623a6

Sep 5, 2016

$\text{Of the order of } {10}^{28}$ marbles of $1 m m$ diameter.

#### Explanation:

We know that marble is a small spherical toy. These come in various size. Most commonly, these are about $1 c m$ in diameter, but may range from about $1 m m$ to over $8 c m$.
It is also clear that number of marbles of smaller diameter required to fill a sphere will be greater than the number of marbles of larger diameter.

Since equatorial radius of earth is greater than the polar radius, therefore we use
Average Radius of earth as $= 6.371 \times {10}^{8} c m$
Volume of sphere the size of earth$= \frac{4}{3} \pi {\left(6.371 \times {10}^{8}\right)}^{3} c {m}^{3}$
Radius of smallest marble$= 0.5 \times {10}^{-} 1 c m$
Volume of one smallest marble$= \frac{4}{3} \pi {\left(0.5 \times {10}^{-} 1\right)}^{3} c {m}^{3}$
Number of such marbles which will fill the sphere the size of earth$= \frac{\frac{4}{3} \pi {\left(6.371 \times {10}^{8}\right)}^{3}}{\frac{4}{3} \pi {\left(0.5 \times {10}^{-} 1\right)}^{3}}$
$= 1.2742 \times {10}^{28}$

This calculation assumes: the marbles are liquid and we could fit the number calculated above into the larger sphere of the size of earth. Unfortunately, marbles are not liquid. Therefore, there would be inevitable gaps between the marbles. These gaps are related to packing efficiency, which refers to the ratio of the combined volume of the marbles to the total volume of space that these occupy.

In three-dimensional Euclidean space, the densest packing of equal-volume spheres can be achieved by close-packed structures. It can be shown that for equal-volume spheres the densest packing uses approximately 74%# of the volume.

Actual number of $1 m m$ diameter marbles which will fill the sphere the size of earth after taking in to account above packing efficiency $= 1.2742 \times {10}^{28} \times 0.74$
$= 9.42908 \times {10}^{27}$
$\text{Of the order of } {10}^{28}$.