91Ó°ÊÓ

Protons, neutrons and baseball are perfect spheres.

Circumference of the baseball,$L_{\mathrm{ball}}=23\mathrm{cm}.$

From Table 1–3, mass of the proton or neutron,$m_{\mathrm{p}}={10}^{-27}\mathrm{kg}$

From Table 1–2, diameter of the proton or neutron,$r_{∘}={10}^{-15}\mathrm{m}$$r_{∘}={10}^{-15}\mathrm{m}$

The density of a material is defined as its mass per unit volume. The formula for density is

$\mathrm{ÒÏ}=\frac{M}{V}.$

In the above formula, M is the mass of the material, V is the volume, and $\mathrm{ÒÏ}$is the density of the material.

The formula for the volume of a sphere having diameter d is $V=\frac{\mathrm{Ï€}{d}^3}{6}$.

Using this value in the equation of density of proton/neutron, you will get

${\mathrm{ÒÏ}}_{∘}=\frac{m_{∘}}{\left(\frac{\mathrm{Ï€}{d}_{∘}^3}{6}\right)}.$

Substituting the variables by their values in the above equation, you will get

$\begin{array}{c}{\mathrm{ÒÏ}}_{∘}=\frac{{10}^{-27}}{\left(\frac{\mathrm{Ï€}{\left({10}^{-15}\right)}^3}{6}\right)}\\ =1.91×{10}^{18}{\mathrm{kgm}}^{-3}.\end{array}$

The formula for the circumference of a circle with diameter d is $L=\mathrm{Ï€}d$. Using this, you get the diameter of the baseball:

$\begin{array}{c}{d}_{\mathrm{ball}}=\frac{L_{\mathrm{ball}}}{\mathrm{Ï€}}\\ =\frac{23}{\mathrm{Ï€}}\\ =7.32\mathrm{cm}\\ =0.0732\mathrm{m}\end{array}$

Using the formula for the volume of a sphere, you get the volume of the baseball:

$\begin{array}{c}{V}_{\mathrm{ball}}=\frac{\mathrm{π}{d}_{\mathrm{ball}}^3}{6}\\ =\frac{\mathrm{π}×{\left(0.0732\right)}^3}{6}\\ =2.054×{10}^{-4}{\mathrm{m}}^3\end{array}$

If you assume that the density of the baseball is equal to the density of the proton/neutron, then by the definition of density, the mass of the baseball is

$m_{\mathrm{ball}}={\mathrm{ÒÏ}}_{∘}·V_{\mathrm{ball}}.$

Here, ${\mathrm{ÒÏ}}_{∘}$is the density of the proton/neutron.

Substituting the variables by their values in the above equation, you will get

$\begin{array}{c}{m}_{\mathrm{ball}}=\left(1.91×{10}^{18}\right)·\left(2.054×{10}^{-4}\right)\\ =3.92×{10}^{14}\mathrm{kg}.\end{array}$

Therefore, if the density of the baseball was equal to that of the neutron/proton, then its mass would be $3.92×{10}^{14}\mathrm{kg}$.