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From Newton's law of gravitation we know that

$$F=G\frac{m_1m_2}{r^2}$$

where $G$ is gravitational constant. We can also see that it has dimensions

$$[G]=\frac{[L]^3}{[M][T]^2}$$

and we have a good numerical estimate of its value ($G\simeq 6.67\times 10^{-11} N(m/kg)^2$).

I've read (for instance in the paper I'm currently mostly interested in, Kerr/CFT correspondence) that in higher dimensions, there are 4 dimensional gravitational constant, and in the search for explanation of it, I've also found that in string theory we have gravitational constants of higher dimensions.

But I haven't seen any formula that would give me mean to calculate it in higher dimensions.

Is there such a thing, or is the definition of it theory dependent?

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There is, in fact, a standard way of defining the gravitational constant in higher dimensions. We reason as follows:

We might try to generalize the formula for the gravitational force given by Newton's Law of Gravitation as you have written above, but this doesn't lead to natural generalization because it's unclear how the power of $r$ should be generalized to higher dimensions. On the other hand, we can rewrite this law in the form of a Poisson equation: \begin{align} \Delta \Phi = 4\pi G \rho. \end{align} Where $\Delta$ is the three-dimensional laplacian, $\Phi$ is the gravitational potential and $\rho$ is the mass density. Now, to generalize to higher dimensions, we simply assert that the Poisson equation still describes Newtonian gravitation in higher dimensions. This can be motivated by taking general relativity in higher dimensions and then taking the weak field limit. Then the gravitational constant is defined by the higher-dimensional Poisson equation; \begin{align} \Delta^{(D-1)}\Phi^{(D)} = 4\pi G^{(D)}\rho^{(D)} \end{align} My notation here is that $D$ is the number of spacetime dimensions, so $G^{(4)}$ is the standard gravitational constant. In particular, I write $\Delta^{(D-1)}$ to emphasize that there are only spatial derivatives in the laplacian; \begin{align} \Delta^{(D-1)} = \partial_1^2 + \cdots + \partial_{D-1}^2. \end{align}

The Laplacian has units of one over length squared in all dimensions, and $\Phi^{(D)}$ has units of energy over mass in all dimensions, so the units of the left hand side are dimension-invariant. This means that the units of the right hand side must also be dimension-invariant; \begin{align} [G^{(4)}\rho^{(4)}] = [G^{(D)}\rho^{(D)}] \end{align} But the units of $\rho^{(D)}$ are mass per unit spatial volume, $M/L^{D-1}$ in any spactime dimension $D$, so we get the following relationship between the $D$-dimensional gravitational constant and the $4$-dimensional gravitational constant: \begin{align} [G^{(D)}] = [G^{(4)}]\frac{[\rho^{(4)}]}{[\rho^{(D)}]} =[G^{(4)}]\frac{ML^{D-1}}{ML^3} = [G^{(4)}]L^{D-4}. \end{align} So, for example, the gravitational constant in $5$ spacetime dimensions has units of length times that the of gravitational constant in $4$ spacetime dimensions.

There is a nice discussion of this with more detail in section 3.8 of

A First Course in String Theory, Zwiebach (2nd Ed.)

Zwiebach also has a discussion of how the numerical vale of $G$ changes when you add extra compact spatial dimensions. For example, he demonstrates that with one extra compact dimension of length $\ell_C$, the five-dimensional gravitational constant becomes \begin{align} G^{(5)} = \ell_C G^{(4)} \end{align} Generally speaking, the value of the gravitational constant in higher dimensions depends on the sizes of these extra (compact) dimensions. If the extra dimensions are non-compact, I'm not exactly sure how one would proceed because one needs an extra characteristic length scale for each extra dimension.

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This is a really nice answer, thank you :) –  dingo_d Nov 14 '13 at 19:34
    
@dingo_d Sure thing. –  joshphysics Nov 14 '13 at 19:37
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