# Is $4 \pi G$ the true most fundamental gravitational constant? [closed]

Newton's law of gravitation is:

$$F = G m_1 m_2 \frac{1}{r^2}$$

It looks simple and natural.

But that's only in 3 dimensions. Let's look what happens in $n$ dimensions:

$$n=2 : F = 2 G m_1 m_2 \frac{1}{r}$$ $$n=4 : F = \frac{2}{\pi} G m_1 m_2 \frac{1}{r^3}$$ $$n=5 : F = \frac{3}{2 \pi^2} G m_1 m_2 \frac{1}{r^4}$$ $$n=6 : F = \frac{4}{\pi^2} G m_1 m_2 \frac{1}{r^5}$$

Oh no! Newton's force law becomes cluttered with unintuitive constants! But by defining $G^* = 4 \pi G$ Newton's law of gravitation can be reformulated as such:

$$F = G^* m_1 m_2 \frac{1}{4 \pi r^2}$$

Immediately we recognize that $4 \pi r^2$ is simply the surface area of a sphere of radius $r$.

But that's only in 3 dimensions. Let's look what happens in $n$ dimensions:

$$n=2 : F = G^* m_1 m_2 \frac{1}{2 \pi r}$$ $$n=4 : F = G^* m_1 m_2 \frac{1}{2 \pi^2 r^3}$$ $$n=5 : F = G^* m_1 m_2 \frac{1}{\frac{8}{3} \pi^2 r^4}$$ $$n=6 : F = G^* m_1 m_2 \frac{1}{\pi^3 r^5}$$

$2 \pi r$ is the surface area of a 2 dimensional sphere of radius $r$.

$2 \pi^2 r^3$ is the surface area of a 4 dimensional sphere of radius $r$.

$\frac{8}{3} \pi^2 r^4$ is the surface area of a 5 dimensional sphere of radius $r$.

$\pi^3 r^5$ is the surface area of a 6 dimensional sphere of radius $r$.

Newton's law of gravitation in $n$ dimensions is:

$$F = G^* m_1 m_2 \frac{1}{S_n}$$

Where $S_n$ is simply the surface area of a $n$ dimensional sphere of radius $r$. From this, it seems like $G^*$ would be a nicer definition for the gravitational constant.

## closed as primarily opinion-based by Jim, Kyle Kanos, John Rennie, DavePhD, jinaweeMay 16 '14 at 16:29

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

• You could add to the question that in cosmology we have so many equations with $4\pi G$ or $8\pi G$ in them that we often take one or the other to be the natural unit and equal to 1 – Jim May 16 '14 at 15:18
• Possible duplicate – Ruslan May 16 '14 at 15:25
• See also the difference between electrostatics expressed in SI and Gaussian units. – dmckee May 17 '14 at 1:04

$$R_{ab} - \frac{1}{2}Rg_{ab} = 8\pi G\,T_{ab}$$
and working physicists often get sick of carrying around the factor of 8$\pi$G, and will define $\kappa = 8\pi G$. (or, as Jim says, define the unit of mass such that $G =1$ or 8$\pi$G =1)