91Ó°ÊÓ

Universal Gravitational constant $\mathrm{G}=6.67×{10}^{−11}\text{N}â‹\dots {\text{m}}^2/{\text{kg}}^2$.

Planck’s constant $\mathrm{h}=6.626×{10}^{−34}\text{J}â‹\dots \text{s}$

Velocity of light $\mathrm{C}=3×{10}^8\text{″¾}/\text{s}$.

The units for universal gravitational constant $\mathbf{G}$are:

$\begin{array}{c}\frac{\mathbf{\text{N}}\mathbf{â‹\dots }{\mathbf{\text{m}}}^{\mathbf{2}}}{\mathbf{}{\mathbf{\text{kg}}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{\text{kg}}\mathbf{â‹\dots }\mathbf{\text{m}}\mathbf{/}{\mathbf{\text{s}}}^{\mathbf{2}}\mathbf{â‹\dots }{\mathbf{\text{m}}}^{\mathbf{2}}}{\mathbf{}{\mathbf{\text{kg}}}^{\mathbf{2}}}\\ \mathbf{=}\frac{{\mathbf{\text{m}}}^{\mathbf{3}}}{\mathbf{}\mathbf{\text{kg}}\mathbf{â‹\dots }{\mathbf{\text{s}}}^{\mathbf{2}}}\end{array}$

So, by dimensional analysis, it is$\left[\mathrm{G}\right]={\left[\mathrm{L}\right]}^3{\left[\mathrm{M}\right]}^{−1}{\left[\mathrm{T}\right]}^{−2}$.

The units for Planck const is shown below:

$\begin{array}{l}\text{J}â‹\dots \text{s}=(\text{N}â‹\dots \text{m})â‹\dots \text{s}\\ \text{J}â‹\dots \text{s}=((\text{kg}â‹\dots \text{m}/{\text{s}}^2)â‹\dots \text{m})â‹\dots \text{s}\\ \text{J}â‹\dots \text{s}=\text{kg}â‹\dots {\text{m}}^2/\text{s}\end{array}$

So, by dimensional analysis, it is $\left[\mathrm{h}\right]={\left[\mathrm{L}\right]}^2{\left[\mathrm{M}\right]}^1{\left[\mathrm{T}\right]}^{−1}$.

The dimensional analysis of speed of light is $\left[\mathrm{c}\right]={\left[\mathrm{L}\right]}^1{\left[\mathrm{T}\right]}^{−1}$.

The dimensional analysis of the fundamental length is $\left[l\right]={\left[L\right]}^1$.

So, obtain:

$\begin{array}{l}\left[G\right]={\left[h\right]}^x{\left[C\right]}^y{\left[I\right]}^z\\ \left[G\right]={\left({\left[L\right]}^2{\left[M\right]}^1{\left[T\right]}^{−1}\right)}^x{\left({\left[L\right]}^1{\left[T\right]}^{−1}\right)}^y{\left[L\right]}^Z\\ \left[G\right]={\left[L\right]}^{2X+y+Z}{\left[M\right]}^x{\left[T\right]}^{-x-y}\\ \left[G\right]={\left[L\right]}^3{\left[M\right]}^{-1}{\left[T\right]}^{-2}\end{array}$

The universal gravitational constant $G$ in terms of$h$ is $G=h^{−1}{c}^3{l}^2$.

Rearrange above equation for $I$.

$l=\sqrt{\frac{Gh}{c^3}}$

Substitute $6.67×{10}^{−11}\text{N}â‹\dots {\text{m}}^2/{\text{kg}}^2$ for role="math" localid="1660036687213" $G,6.626×{10}^{−34}\text{J}â‹\dots \text{s}$ for $h$, and $3×{10}^8\text{″¾}/\text{s}$ for $c$ as:

$\begin{array}{l}\mathrm{l}=\sqrt{\frac{(6.67×{10}^{−11}\text{ N}â‹\dots {\text{m}}^2/{\text{kg}}^2)(6.626×{10}^{−34}\text{ }\text{J}â‹\dots \text{s})}{{(3×{10}^8\text{″¾}/\text{s})}^3}}\\ \mathrm{l}=4×{10}^{−35}\text{ }\text{m}\end{array}$

Thus, the fundamental length should be around$4×{10}^{−35}\text{m}$.