Question

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

Short answer

The mass of the baseball will be \(3.92 \times {10^{14}}{\rm{ kg}}\).

Step-by-step solution

Assumptions and data identification 

Protons, neutrons and baseball are perfect spheres.

Circumference of the baseball, \({L_{{\rm{ball}}}} = 23{\rm{ cm}}{\rm{.}}\)

Reading data from the table 

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

From Table 1–2, diameter of the proton or neutron, \({r_ \circ } = {10^{ - 15}}{\rm{ m}}\)

Finding the density of proton/neutron

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

\(\begin{aligned}{c}{\rho _ \circ } = \frac{{{{10}^{ - 27}}}}{{\left( {\frac{{\pi {{\left( {{{10}^{ - 15}}} \right)}^3}}}{6}} \right)}}\\ = 1.91 \times {10^{18}}{\rm{ kg }}{{\rm{m}}^{ - 3}}.\end{aligned}\)

Finding the volume of the baseball

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

\(\begin{aligned}{c}{d_{{\rm{ball}}}} = \frac{{{L_{{\rm{ball}}}}}}{\pi }\\ = \frac{{23}}{\pi }\\ = 7.32{\rm{ cm}}\\{\rm{ = 0}}{\rm{.0732 m}}\end{aligned}\)

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

\(\begin{aligned}{c}{V_{{\rm{ball}}}} = \frac{{\pi d_{{\rm{ball}}}^3}}{6}\\ = \frac{{\pi \times {{\left( {0.0732} \right)}^3}}}{6}\\ = 2.054 \times {10^{ - 4}}{\rm{ }}{{\rm{m}}^3}\end{aligned}\)

Finding the mass of the baseball

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

\({m_{{\rm{ball}}}} = {\rho _ \circ } \cdot {V_{{\rm{ball}}}}.\)

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

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

\(\begin{aligned}{c}{m_{{\rm{ball}}}} = \left( {1.91 \times {{10}^{18}}} \right) \cdot \left( {2.054 \times {{10}^{ - 4}}} \right)\\ = 3.92 \times {10^{14}}{\rm{ kg}}{\rm{.}}\end{aligned}\)

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