A standard baseball has a circumference of approximately 23 cm. If a baseball had the same mass pear unit volume (see Tables in section 1–5) as a neutron or a proton, about what would its mass be?

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_{\circ }={10}^{-15}\mathrm{m}$$r_{\circ }={10}^{-15}\mathrm{m}$

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

$\rho =\frac{M}{V}.$

In the above formula, M is the mass of the material, V is the volume, and $\rho$is the density of the material.

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

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

${\rho }_{\circ }=\frac{m_{\circ }}{\left(\frac{\pi d_{\circ }^3}{6}\right)}.$

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

$\begin{array}{c}{\rho }_{\circ }=\frac{{10}^{-27}}{\left(\frac{\pi {\left({10}^{-15}\right)}^3}{6}\right)}\\ =1.91\times {10}^{18}{\mathrm{kgm}}^{-3}.\end{array}$

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

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

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

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

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

$m_{\mathrm{ball}}={\rho }_{\circ }·V_{\mathrm{ball}}.$

Here, ${\rho }_{\circ }$is the density of the proton/neutron.

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

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

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