This is from Anthony Zee's "QFT in a Nutshell", chapter I.6.
In a D = (3 + n)-dimensional world, the potential energy goes as:
<br />
V(r) \propto \int{d^{3 + n} k \, \frac{e^{i \mathbf{k} \cdot \mathbf{x}}}{k^2}} \propto \frac{1}{r^{1 + n}}<br />
If the gravitational force is proportional to the masses of the interacting particles regardless of the number of spatial dimensions, we may write:
<br />
V(r) = G_{3 + n + 1} \, \frac{m_1 \, m_2}{r^{1 + n}}<br />
In natural units (\hbar = c = 1), the dimensions of mass, length, and time satisfy \mathrm{T} = \mathrm{L} = \mathrm{M}^{-1}, and every physical quantity has only a mass dimension. For example:
<br />
\left[ V(r) \right] = [E] = \mathrm{M} \, \mathrm{L}^2 \, \mathrm{T}^{-2} \stackrel{\mathrm{n.u.}}{\rightarrow} \mathrm{M}<br />
Then, the "Universal Gravitational Constant" in 3 + n + 1 space-time dimensions has a mass dimension:
<br />
\left[ G_{3 + n + 1}\right] = \frac{[E] \, [r]^{1 + n}}{[m]^2} = \mathrm{M}^{-2 - n}<br />
Indeed, this constant defines a Planck mass in 3 + n + 1 space-time dimensions:
<br />
G_{3 + n + 1} = \frac{1}{M_{\mathrm{Pl}, 2 + n}}<br />
Indeed, in (3 + 1)-space-time, M_{\mathrm{Pl},4} = 1/\sqrt{G_{4}} \sim 10^{19} \, \mathrm{GeV}. This is a HUGE energy scale (energy and mass have the same dimension in natural units), a fact we interpret as the reason why gravity is so weak.
But, suppose the extra n spatial dimensions are "curled up" to linear dimensions of the size R. This need not be a microscopic length scale, but is small enough that it had not been detected by gravitational experiments performed on such a small length scale. For distances r \gg R, the gravitational flux spreads only in the 3 unfolded linear dimensions, and the gravitational potential energy goes as:
<br />
V(r) = \frac{m_1 \, m_2}{M^{2 + n}_{\mathrm{Pl}, 3 + n +1}} \, \frac{1}{R^n} \, \frac{1}{r}<br />
If we interpret:
<br />
\frac{1}{M^2_{\mathrm{Pl},4}} \equiv \frac{1}{M^{2 + n}_{\mathrm{Pl}, 3 + n +1} R^{n}}<br />
<br />
\frac{M_{\mathrm{Pl},4}}{M_{\mathrm{Pl},3 + n + 1}} = \left( M_{\mathrm{Pl},3 + n + 1} R \right)^\frac{n}{2}<br />
Now, the Planck mass in 3 + n + 1 space-time dimensions may as well be of the order of magnitude of ordinary particle masses (let us choose a range of values 10 eV - 100 GeV). This would put the dimensioless ratio of masses:
<br />
\frac{M_{\mathrm{Pl}, 4}}{M_{\mathrm{Pl}, 3 + n + 1}} = 10^{17} - 10^{27}<br />
Then, the distance R, should be:
<br />
R = \frac{1}{M_{\mathrm{Pl},3 + n + 1}} \, \left( \frac{M_{\mathrm{Pl}, 4}}{M_{\mathrm{Pl}, 3 + n + 1}} \right)^\frac{2}{n} = \frac{1}{M_{\mathrm{Pl}, 4}} \, \left( \frac{M_{\mathrm{Pl}, 4}}{M_{\mathrm{Pl}, 3 + n + 1}} \right)^{1 + \frac{2}{n}}<br />
This is a monotonically rising function of the dimensionless ratio (the smaller the mass of the Planck particle, the bigger the ratio). For a popular choice n = 6, we would have:
<br />
R = (10^{19} - 10^{31}) \, M^{-1}_{\mathrm{Pl},4} = (10^{-16} - 10^{-4}) \, \mathrm{m}<br />
