gravity - Is $4 pi G$ the true most fundamental gravitational constant?

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.