Fill in the blanks by suitable conversion of units:
$(a)$ $1\; kg\; m^{2}\; s^{-2} = \ldots\; g\; cm^{2}\; s^{-2}$
$(b)$ $1\; m = \ldots\; ly$
$(c)$ $3.0\; m\; s^{-2} = \ldots\; km\; h^{-2}$
$(d)$ $G = 6.67 \times 10^{-11}\; N\; m^{2}\; kg^{-2} = \ldots\; cm^{3}\; s^{-2}\; g^{-1}$

(N/A) Since $1\; kg = 10^{3}\; g$ and $1\; m^{2} = 10^{4}\; cm^{2}$,then $1\; kg\; m^{2}\; s^{-2} = 10^{3}\; g \times 10^{4}\; cm^{2}\; s^{-2} = 10^{7}\; g\; cm^{2}\; s^{-2}$.
$(b)$ $1\; ly$ (light year) is the distance light travels in one year. $1\; ly = (3 \times 10^{8}\; m/s) \times (365.25 \times 24 \times 3600\; s) \approx 9.46 \times 10^{15}\; m$. Thus,$1\; m = 1 / (9.46 \times 10^{15})\; ly \approx 1.057 \times 10^{-16}\; ly$.
$(c)$ $1\; m = 10^{-3}\; km$ and $1\; s = (1/3600)\; h$,so $1\; s^{-2} = (3600)^{2}\; h^{-2}$. Therefore,$3.0\; m\; s^{-2} = 3.0 \times 10^{-3}\; km \times (3600)^{2}\; h^{-2} = 3.0 \times 10^{-3} \times 12960000\; km\; h^{-2} = 3.888 \times 10^{4}\; km\; h^{-2}$.
$(d)$ $G = 6.67 \times 10^{-11}\; N\; m^{2}\; kg^{-2}$. Since $1\; N = 1\; kg\; m\; s^{-2}$,$G = 6.67 \times 10^{-11}\; (kg\; m\; s^{-2})\; m^{2}\; kg^{-2} = 6.67 \times 10^{-11}\; kg^{-1}\; m^{3}\; s^{-2}$. Converting units: $1\; kg^{-1} = (10^{3}\; g)^{-1} = 10^{-3}\; g^{-1}$ and $1\; m^{3} = (10^{2}\; cm)^{3} = 10^{6}\; cm^{3}$. Thus,$G = 6.67 \times 10^{-11} \times 10^{-3}\; g^{-1} \times 10^{6}\; cm^{3}\; s^{-2} = 6.67 \times 10^{-8}\; cm^{3}\; s^{-2}\; g^{-1}$.