Generalizing the Newton-Coulomb Analogy

For mechanics students, they know the law of universal gravitation, which would hereafter be referred to as Newton's law (universal gravitational constant pulled out to clarify comparison):

${\displaystyle F = G \frac{m_1 m_2}{r^2} }$

For electricity and magnetism students, they know Coulomb's law (constants pulled out to clarify comparison):

${\displaystyle F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} }$

Both formulas look similar. In fact, in this blog post, we will extend the analogy.

The analogies are: ${\displaystyle G=\frac{1}{4\pi\epsilon_{0}}}$ and ${\displaystyle m=q}$.

Dielectric Constant is to what in Mechanics?

To do so, we will set the constants equal to each other and solve for ${\displaystyle \epsilon_0}$ in terms of ${\displaystyle G}$:

${\displaystyle G=\frac{1}{4\pi\epsilon_{0}}}$

${\displaystyle \epsilon_{0}=\frac{1}{4G\pi}}$

Now we evaluate the RHS, which is approximately: 1.192×10⁹ kilogram seconds squared per meter cubed. I will dub this the digravitational constant.

Conversion of formulas:

Electricity and Magnetism	Mechanics	Additional Analogy Used	Notes
${\displaystyle \vec{E}=\frac{\vec{F_{E}}}{q}}$	${\displaystyle \vec{G}=\frac{\vec{F_{G}}}{m}}$	N/A	N/A
${\displaystyle E_{x}=\frac{dV}{dx}}$	${\displaystyle G_{x}=\frac{dU}{dx}}$	${\displaystyle U=V}$	N/A
${\displaystyle \Delta V=-\int_{ }^{ }\vec{E} \cdot d\vec{r}}$	${\displaystyle \Delta U=-\int_{ }^{ }\vec{G} \cdot d\vec{r}}$	${\displaystyle U=V}$	N/A
${\displaystyle V=\frac{1}{4\pi\epsilon_{0}}\sum_{i}^{ }\frac{q_{i}}{r_{i}}}$	${\displaystyle U=G\sum_{i}^{ }\frac{m_{i}}{r_{i}}}$	${\displaystyle U=V}$	N/A
${\displaystyle U_{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q_{1}q_{2}}{r}}$	${\displaystyle U_{G}=\frac{Gm_{1}m_{2}}{r}}$	N/A	Using ${\displaystyle U_{E}=qV}$ gives us the product of mass and energy.
${\displaystyle \Delta V=\frac{Q}{C}}$	${\displaystyle \Delta U=\frac{m}{?}}$	${\displaystyle U=V}$	${\displaystyle ?}$ represents mass per energy
${\displaystyle C_{P}=\sum_{ }^{ }C_{i}}$	${\displaystyle ?_{P}=\sum_{ }^{ }?_{i}}$	N/A	${\displaystyle ?}$ represents mass per energy
${\displaystyle \frac{1}{C_{S}}=\sum_{ }^{ }\frac{1}{C_{I}}}$	${\displaystyle \frac{1}{?_{S}}=\sum_{ }^{ }\frac{1}{?_{I}}}$	N/A	${\displaystyle ?}$ represents mass per energy
${\displaystyle I=\frac{dQ}{dt}}$	${\displaystyle ?=\frac{dm}{dt}}$	N/A	${\displaystyle ?}$ represents mass per time
${\displaystyle U_{C}=\frac{1}{2}C\left(\Delta V\right)^{2}}$	${\displaystyle ?=\frac{1}{2}?\left(\Delta U\right)^{2}}$	N/A	Using the formula gives us the product of mass and energy.