## Fermi Friday: Baseball Blackhole

The question of interest this week is:

If all baseballs ever used in an MLB game were compressed into a blackhole, how large would the blackhole be?

To estimate the blackhole radius $$r_{BH}$$, we'll need to compute the number of baseballs used in MLB games, translate that number to a total mass, then compute the blackhole radius of the corresponding mass.

### Counting Baseballs

We begin with a lower-bound on the number of baseballs used in all MLB games. If we define $$n_u$$ as the number of pitches that a ball is used for, then the problem translates to counting pitches. The pitches per game can be computed in terms of the number of pitches per inning $$n_p$$ and the number of innings per game $$n_i$$ as $$n_p n_i$$. The number of regular season games per year can be given in terms of the number of games per team $$n_{g}$$, the number of teams $$n_{t}$$, and the number of postseason games $$n_{ps}$$ as $$n_g n_t / 2 + n_{ps}$$. Scaling by the number of seasons $$n_s$$, then the complete expression for total baseballs is then given by:

$$n_B = \frac{n_s n_p n_i \left(n_g n_t + 2 n_{ps} \right)}{2 n_u}$$

To estimate the blackhole radius $$r_{BH}$$ of an arbitrary mass $$m$$, we use the well-known, elegant equation for the Schwarzschild event-horizon radius:

$$r_{s} = \frac{2 G m}{c^2}$$

### Solution

Putting the above expressions together, we arrive at the baseball black hole radius:

$$r_{b} = \frac{G m_B n_s n_p n_i \left(n_g n_t + 2 n_{ps} \right)}{n_u c^2}$$

Plugging in some numbers:

• Mass of a baseball: $$m_B = 1.5 \times 10^{-1}$$ kg
• Number of pitches per baseball: $$n_u = 2 \times 10^0$$
• Number of seasons: $$n_s = 1.0 \times 10^2$$
• Number of pitches per inning: $$n_p = 2 \times 10^1$$
• Number of innings per game: $$n_i = 1 \times 10^1$$
• Number of games per team: $$n_g = 1.6 \times 10^2$$
• Number of teams: $$n_t = 3 \times 10^1$$
• Number of post-season games: $$n_{ps} = 5 \times 10^1$$
• Gravitational constant: $$G = 6.7 \times 10^{-11}\ \text{m}^3 \text{kg}^{-1}\text{s}^{-2}$$
• Speed of Light: $$c = 3 \times 10^8\ \text{ms}^{-1}$$

The estimate is hidden below. Press the button below to see solution!

The radius is so small, that I had to compute a few other numbers to contextualize it. The following numbers are the amount of time required, at the current rate of game play, for the baseball black hole to reach a certain size.

• Proton Radius ($$10^{-15}$$ m) would require 18 million more years of baseball
• Baseball Radius ($$10^{-2}$$ m) would require $$1.3 \times 10^{21}$$ more years of baseball. Considerably longer than the age of the universe ($$1.4 \times 10^{10}$$ yr)