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:

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

Blackhole Radius

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)