Chapter 8, Problem 58E

To determine

Check whether the gravitational force between two protons in nucleus is balanced by their electrical repulsion, calculate the ratio of this electrical force to the gravitational force and determine whether this ratio has any dependence on the distance between the protons.

Answer to Problem 58E

No. The value of $F_{ele}/F_g$ is $1.24\times {10}^{36}$ and is independent of the distance between protons.

Explanation of Solution

The gravitational force between two protons in a nucleus is too weak in comparison with the electoral repulsive force between them. In femtometer $\left(1\text{fm}={10}^{-15}\text{m}\right)$ range, strong nuclear force is the dominant attractive force.

Write the formula to calculate the magnitude of Columb electrical force between two protons.

$F_{ele}=\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r^2}$ (1)

Here,

$F_{ele}$ is the Columb electric force between protons

${\epsilon }_0$ is the permittivity of free space

$e$ is the magnitude charge of electron

$r$ is the distance of separation between protons

Write the formula to calculate the magnitude of gravitational force between two protons.

$F_g=G\frac{m_p^2}{r^2}$ (2)

Here,

$F_g$ is the gravitational force between the protons

$G$ is the gravitational constant

$m_p$ is the mass of proton

Divide equation (1) by equation (2).

$\begin{array}{c}\frac{F_{ele}}{F_g}=\frac{\left(\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{r^2}\right)}{\left(G\frac{m_p^2}{r^2}\right)}\\ =\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{G{m}_p^2}\end{array}$

Substitute $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $1/4\pi {\epsilon }_0$,$1.6\times {10}^{-19}\text{C}$ for $e$,$6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$ and $1.67\times {10}^{-27}\text{kg}$ for $m_p$ in the above equation to find $\frac{F_{ele}}{F_g}$.

$\begin{array}{c}\frac{F_{ele}}{F_g}=\left(9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\frac{{\left(1.6\times {10}^{-19}\text{C}\right)}^2}{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right){\left(1.67\times {10}^{-27}\text{kg}\right)}^2}\\ =1.24\times {10}^{36}\end{array}$

The above ratio shows that $F_{ele}$ is $1.24\times {10}^{36}$ times stronger than $F_g$. This means that $F_{ele}$ cannot be counteracted by $F_g$. The relation $\frac{F_{ele}}{F_g}=\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{G{m}_p^2}$ is independent of term $r$. This means that separation between protons has no any role determining the value of $\frac{F_{ele}}{F_g}$.

Conclusion: Therefore, force of gravity between protons is not balanced by their electrical repulsive force, the value of $F_{ele}/F_g$ is $1.24\times {10}^{36}$ and is independent of the distance between protons.