Gravity is from spacetime geodesic acceleration:
$\ddot{r} = -{\Gamma^r}_{tt} (c \cdot \dot{t})^2 + ...$
For $\dot{t} = 1$ and $\dot{x}^i = 0$ otherwise
$\ddot{r} = c^2 \frac{R (R - r)}{2 r^3}$
Let $2 \frac{G}{c^2} M = R$, for mass M in kg and Schwarzschild radius R in m
For gravitational constant $G =$ $\frac{m^3}{kg \cdot s^2}$
$\ddot{r}_{Schwarzschild} = -G \frac{M}{r^2} (1 - \frac{G}{c^2} \frac{2 M}{r})$
Notice that the ratio of Schwarzschild gravity to Newton gravity is:
For the Earth:
$M =$ $kg$
$R = \frac{2 G M}{c^2} =$ {{earth_Schwarzschild_radius_in_m = 2 * gravitational_constant_in_m3_per_kg_s2 * earth_mass_in_kg / (speed_of_light_in_m_per_s * speed_of_light_in_m_per_s)}} $m$.
At the surface $r =$ $m$
so $\frac{R}{r} = \frac{2 G M}{c^2 r} =$ {{earth_Schwarzschild_to_radius_ratio = earth_Schwarzschild_radius_in_m / earth_radius_in_m}}
and $1 - \frac{R}{r} = $ {{1 - earth_Schwarzschild_to_radius_ratio}}
For very small $\frac{G}{c^2} \frac{2 M}{r}$, we can approximate
$\ddot{r}_{Newton} = -\frac{G M}{r^2}$
which is Newton gravity.
At the Earth's surface this is {{gravitational_constant_in_m3_per_kg_s2 * earth_mass_in_kg / (earth_radius_in_m * earth_radius_in_m)}} $\frac{m}{s^2}$
Gravity is:
$\ddot{r} = -{\Gamma^r}_{tt} (c \cdot \dot{t})^2 + ...$
For $\dot{t} = 1$ and $\dot{x}^i = 0$ otherwise.
$\ddot{r} = G (\frac{2 M_{,r}}{r} - \frac{M}{r^2}) (1 - \frac{G}{c^2} \frac{2M}{r})$
Let's assume M is much smaller, such that $\frac{M}{r^2} \approx 0$
$\ddot{r} = \frac{2 G M_{,r}}{r} = $ {{2 * gravitational_constant_in_m3_per_kg_s2}} $\frac{m^3}{kg \cdot s^2} \cdot \frac{1}{r} \cdot (\frac{\partial M}{\partial r})$
While we're here, let's look at charge:
Replace M with Q
(But can you simply replace M with Q?
Relativistic gravitation is caused from $G_{ab} = T_{ab}$.
The 'swarm-of-mass' stress-energy attributed to matter gravitational effects is due to the density -- i.e. the intrinsic rest-mass of all particles within a region of space.
However the stress-energy attributed to electromagnetism gravitational effects is only due to the electric and magnetic fields, which are derived from calculations of the charge of the particles in nearby space.
Should there exist a 'swarm-of-mass' stress-energy based on particle charge, in addition to mass?
Should there exist a stress-energy based on gravitational effects at a particular location in spacetime, as there is due to electromagnetic fields?)
Coulomb's constant: $k_e =$ $\cdot \frac{kg \cdot m^3}{C^2 \cdot s^2}$
so $\sqrt{k_e \cdot G} =$ {{sqrt_ke_G_in_m3_per_C_s2 = Math.sqrt(Coulomb_constant_in_kg_m3_per_C2_s2 * gravitational_constant_in_m3_per_kg_s2)}} $\cdot \frac{m^3}{C \cdot s^2}$
and $\sqrt{\frac{k_e \cdot G}{c^4}} =$ {{sqrt_ke_G_in_m3_per_C_s2 / (speed_of_light_in_m_per_s * speed_of_light_in_m_per_s)}} $\cdot \frac{m}{C}$
Notice that in the same way $M = \int_V \rho dV$, so is $Q = \int_V q dV$
Where $[\rho] = \frac{kg}{m^3}$ and $[q] = \frac{C}{m^3}$
Side thoughts... charge vs. mass of particles, and their relativistic effects of each...
Winding a wire into a ball.
$l = $ length of wire
$R_{out} = $ radius of wire
$r = $ ball radius
volume of wire cylinder equals volume of ball:
$\pi R_{out}^2 l = \frac{4}{3} \pi r^3$
$l = \frac{4}{3} \frac{r^3}{R_{out}^2}$
$r = (\frac{3}{4} {R_{out}}^2 l)^{\frac{1}{3}}$
change in length with respect to change in radius:
$\frac{\partial l}{\partial r} = 4 (\frac{r}{R_{out}})^2$
Wire signal source:
$\phi(t) = $ signal source.
$\phi(t)$ is unitless.
$\phi'(t)$ is in units of $\frac{1}{s}$.
$\Phi(t) = \int_{t=0}^{t=t'} \phi(t') dt'$ is in units of $s$.
$\Phi'(t) = \phi(t)$.
Wire signal value at any point in space and time:
$f(l,t) = \phi(t - l/v)$ the amplitude function of the current through the wire, which propagates at a velocity of $v$.
for $v = $ mean velocity of electrons through wire.
$\frac{d f}{d l} = -\frac{1}{v} \phi'$ has units $\frac{1}{m}$
Assume the wire has a charge density at length $l$ along the wire of $q(l)$ in units of $\frac{C}{m}$
$q(l,t) = q_0 \cdot A \cdot f(l,t) = q_0 \cdot \pi \cdot R_{in}^2 \cdot f(l,t)$
$R_{in} =$ the inner radius of the wire which is conductive, so $0 < R_{in} < R_{out}$
$Q(l,t) = \int_{l=0}^{l=l'} q(l',t) dl'$
$Q(r,t) = \int_{l=0}^{l = \frac{4}{3} \frac{r^3}{R_{out}^2}} q(l,t) dl$
$Q(r,t) = q_0 \cdot \pi \cdot R_{in}^2 \cdot \int_{l=0}^{l = \frac{4}{3} \frac{r^3}{R_{out}^2}} \phi(t - l/v) dl$
$Q(r,t) = q_0 \cdot \pi \cdot R_{in}^2 \cdot ( -v \cdot \Phi(t - l/v) )|_{l=0}^{l = \frac{4}{3} \frac{r^3}{R_{out}^2}}$
$Q(r,t) = - v \cdot q_0 \cdot \pi \cdot R_{in}^2 \cdot (
\Phi(t - \frac{1}{v} \frac{4}{3} \frac{r^3}{R_{out}^2} )
- \Phi(t)
)$
$Q(r,t)$ is in units of $C$.
$\frac{\partial Q}{\partial r} = \frac{\partial Q}{\partial l} \cdot \frac{d l}{d r} = q(l,t) \cdot \frac{d l}{d r} = q(l,t) \cdot 4 \cdot (\frac{r}{R_{out}})^2$ is in units of $\frac{C}{m}$
AWG 10 wire: https://www.engineeringtoolbox.com/awg-wire-gauge-circular-mils-d_819.html
has a radius of about 1.3 mm, but this includes the sheath, so that is $R_{out}$
copper according to https://www.st-andrews.ac.uk/~www_pa/Scots_Guide/audio/part6/page2.html
copper's atomic weight: $63.54 g$
Avogadro's constant $N_A = 6.02214085774 \cdot 10^{23} \cdot \frac{1}{mol}$
copper's nominal valency = 1, so 1 atom = 1 electron.
so $63.54 g = 6.02214085774 \cdot 10^{23} atoms = 6.02214085774 \cdot 10^{23} \cdot e$
free electrons per gram of copper $= N_A / $atomic weight$ = \frac{6.02214085774 \cdot 10^{23}}{63.54} \cdot \frac{e}{g}$
copper's density: $8.95 \frac{g}{cm^3} = 8.95 \cdot 10^3 \cdot \frac{kg}{m^3}$
copper electron density: $\rho_{copper} = (\frac{6.02214085774 \cdot 10^{23}}{63.54} \cdot \frac{e}{g} \cdot \frac{10^3 g}{1 kg}) \cdot (8.95 \cdot 10^3 \cdot \frac{kg}{m^3})$
$= 8.4825559768292 \cdot 10^{28} \frac{e}{m^3}$
electron charge: $e = 1.602176620898 \cdot 10^{-19} C$
copper charge density: $q_0 = 1.3590552871534 \cdot 10^{10} \frac{C}{m^3}$
mean velocity of moving electrons: $v_{copper} = \frac{I}{A \cdot q_{copper}} = \frac{I}{\pi R_{in}^2 \cdot q_{copper}} \approx 7.5 \cdot 10^{-5} \frac{m}{s}$