Question

If the value of $G$ (universal gravitational constant) is $6.67\times {10}^{-11}$ unit, then determine the value in CGS system.

Answer 1

Prerequisite: System of units.
Step 1: Find the SI unit of universal gravitation constant
The universal gravitational constant is the gravitational force acting between two bodies of unit mass, kept at a unit distance from each other.
$$\begin{gathered} G=\frac{F{r}^2}{m_1{m}_2} \\ =\frac{force\times {\left(length\right)}^2}{mass\times mass} \\ =\frac{N\times m^2}{k{g}^2} \end{gathered}$$

So, the SI unit of $G$ is $\frac{N{m}^2}{k{g}^2}$.
Step 2: Convert the SI unit of $G$ into C.G.S unit

$$\begin{gathered} 1\frac{N{m}^2}{k{g}^2}=1\times \frac{kg(m/s^2)\times m^2}{k{g}^2} \\ =1\times \frac{m^3}{kg{s}^2} \\ =1\times \frac{{(100cm)}^3}{(1000g)\times s^2} \\ =1\times \frac{{{10}^{6}cm}^3}{{10}^{3}g{s}^2} \\ ={10}^3\frac{{cm}^3}{g{s}^2} \end{gathered}$$

So, the value of 1 SI unit of $G$ in CGS units is ${10}^{3}c{m}^3{g}^{-1}{s}^{-2}$.
Step 3: Calculate the value of $6.67\times {10}^{11}$ SI units of $G$ into C.G.S units
As calculated from the previous step, we have:
$1\frac{N{m}^2}{k{g}^2}={10}^3\frac{{cm}^3}{g{s}^2}$
$$\begin{gathered} \Rightarrow 6.67\times {10}^{11}\frac{N{m}^2}{k{g}^2}=6.67\times {10}^{-11}\times {10}^3\frac{{cm}^3}{g{s}^2} \\ =6.67\times {10}^{-8}\frac{{cm}^3}{g{s}^2} \end{gathered}$$
Thus, the value of $G$ in the C.G.S system is $\mathbf{6}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\frac{{\mathbf{c}\mathbf{m}}^{\mathbf{3}}}{\mathbf{}\mathbf{g}\mathbf{}{\mathbf{s}}^{\mathbf{2}}}$ (or simply $c{m}^3{g}^{-1}{s}^{-2}$).