Gravitational Constant with ENM Units?

Question

To give some context, there's a conspiracy 'theory' that I saw called Electric Universe that says that gravity is not a fundamental force and instead is a "incoherent dielectric acceleration". Of course this is not supported by any evidence and they don't offer any real model, but it can fool laypeople. I'm a student of physics and I haven't gotten to ENM yet so I don't feel very comfortable with most things having to do with electricity and magnetism, however even I could figure out that there was something wrong here. This question isn't necessarily about Electric Universe nonsense, that's clearly false for a number of reasons.

The thing that was interesting to me was a paper(not published in any journals that I know of), cited by proponents of this conspiracy, written by Morton F. Spears, an electrical engineering graduate from MIT that served in the army which helped him gain skills in radar and nuclear physics. The paper is titled: "An Electrostatic Solution for the Gravity Force and the Value of G". The paper claims to claim that gravity is electrostatic and from what even I can tell, is quite wrong. The value for G he gets is $6.68541 × 10^{−11} \text{ CVm/kg}^2$. This value is very far off from other measured values, not to mention the units are strange. There are many reasons for his paper being incorrect that I have found from a quick look and there are many that I'm sure others will find. The units that he has for his value of G are what I'm unsure about and I would appreciate some insight/clarification.

The way that he gets to these units is by the following: \begin{align} \phantom{4}\\ &A = \frac{C}{s}\\ &V = \frac{W}{A} = \frac{m^2kg^2}{s^3A} = \frac{m^2kg^2}{s^2C}\\ &\phantom{4}\\ \text{From }&\text{here we can say that,}\\ &\phantom{4}\\ &CVm = \frac{m^3kg}{s^2}\\ &\phantom{4}\\ \text{Hence,}&\\ &\phantom{4}\\ &\frac{CVm}{kg^2} = \frac{m^3}{kgs^2}. &\phantom{4}\\ \end{align}

Is this dimensional analysis valid? If so what does it mean? If someone could shed some light on the validity and some interpretations of it, that would be great!

Answer 1

Gravity is not an electrostatic phenomenon and the paper is doomed to be incorrect.

However, the dimensional analysis is fine. I think maybe what you are confused by is that "Coulomb is a unit of charge" and "Voltage is a unit involving charge", so how can the final result of multiplying charge by voltage not involve charge at all?

Well, note that the magnitude of theforce on point between two charges $q_1$ and $q_2$ is $$ |F| = \frac{k q_1 q_2}{r^2} $$ so the combination $k q_1 q_2 = |F| r^2$ does not depend on any electrical units at all. In fact, $k q_1 q_2$ has units of ${\rm kg}\cdot {\rm m^3}\cdot {\rm s^{-2}}$.

Now consider the potential due to a point charge $$ V = \frac{kq}{r} $$ If we multiply $V$ by a charge $Q$, then the combination $k Q q$ will appear, which does not depend on any electrical units. Indeed, $QV$ is equal to the energy of a point charge $Q$ placed at potential $V$, and of course energy does not depend on electrical units, since there are non-electrical types of energy.

Similarly, $Gm_1 m_2 /r$ has units of energy. So, at least dimensionally, $$ QV $$ and $$ \frac{Gm_1 m_2}{r} $$ have the same units, and an equation like $$ G = \frac{QV r}{m_1 m_2} $$ is dimensionally consistent. It is physically nonsense because it is not motivated by any physical ideas and is not correct. The interpretation is simply that not every dimensionally consistent equation is correct; an equation being dimensionally consistent is necessary but not sufficient for it to be relevant for physics.

Answer 2

It is certainly true that $1 \space C\cdot V = 1\space kg\cdot m^2\cdot s^{-2}$ and thus one could choose to express $G$ in units of $C\cdot V\cdot m \cdot kg^{-2}$. Does this mean anything significant? No.

The dimensions of the above expression are $[Mass]\times[Length]^2\times[Time]^{-2}$. Notice how there is no factor with a charge dimension. This is hardly surprising since $1\space V$ has dimensions $[Mass]\times[Length]^2\times[Time]^{-2}\times [Charge]^{-1}$ and $1\space C$ predictably has dimensions $[Charge]$, so the factors of charge cancel each other. A Volt is just a Joule per Coulomb after all.

One can easily find new expressions for other units in a similar way. From $1\space A = 1\space C\cdot s^{-1}$ we get $1\space s = 1\space C\cdot A^{-1}$. I could truthfully claim that I walked to the bus stop in 180 Coulomb per Ampère, but that doesn't imply that "time is actually electricity" or anything of the sort.