I have the question " Newton's Gravitational Law is given as. $F = G (mM/r)$, Where $m$ and $M$ are masses, $r$ is the distance between the masses and $F$ is the attraction force, and $G$ is a constant.
Derive the dimension of $G$ and work out the standard units for $G$?"
So I have rearranged the equation to make $G$ the subject so I get $G = F (mM/r)$.
I then put in the units where $F = [\text{kg}\cdot\text{m}\cdot\text{s}^{-2}]$, $m$ and $M = [\text{kg}]$ and $r = [\text{m}]$.
The final answer that I get is $G = [\text{kg}^2\cdot\text{m}^2\cdot\text{s}^{-2}]$ is This correct ? And if not could you explain where I went wrong?
By the physics law (p.s. you are missing an $r$!)
$$F = G \frac{mM}{r^2}$$
You get
$$G = \frac{F r^2}{mM}$$
Hence the dimensions are well given:
$$[G] = \frac{[N]\cdot [m]^2}{[kg]^2}$$
Now, the Newton is not an elementary unit, and you can obtain it by the simplest force:
$$F = ma$$
Hence
$$[F] = \text{kg}\cdot \text{m}/\text{s}^2$$
SUbstituting above:
$$[G] = \frac{[kg]\cdot [m]}{[s]^2}\cdot \frac{[m]^2}{[kg]^2}$$
Hence
$$[G] = \frac{[m]^3}{[s]^2 \cdot [kg]}$$