Why don't we take the universal gravitational constant $G$ to be equal to 1 in $F= \frac{Gm_{1}m_{2}}{r^2}$? In the derivation of Newton's Second Law, we get to an equation $F=kma$. Since this equation is essentially defining force, Newton could have taken the value of $k$ to be anything. For the sake of convenience, he took it to be unity and hence we get $F=ma$. Going by the same argument, in Newton's law of gravitation we get the equation  $F= \frac{Gm_{1}m_{2}}{r^2}$. Why cannot we take the value of $G$ to be unity for sake of convenience. Why does it have such a complicated value of $6.626× 10^{-11}$. How is it different from Newton's second law?
 A: In fact, in the Planck system of units G is taken to be 1. The trouble is that Newton was working with units of mass, distance, and time, that had already been chosen (feet, pounds, seconds), and they were chosen without respect to making the gravitational constant simple. The Planck system of units is very handy for cosmology, but the implied units for mass, force, distance, and time, are very awkward for day to day use. For example, the Planck unit of length is $1.616255(18)\times10^{-35}$ m.
A: Ok let's try it. Say mass $m_1 = 1000$ kg is at distance $1$ metre from mass $m_2 = 0.5$ kg. How will mass $m_2$ move?
With $G=1$ the force on it is $F = m_1 m_2/r^2 = 500$ newtons. So the acceleration will be $a = F/m_2 = 1000$ m/s$^2$.
Now we do the experiment, and we observe that the acceleration is $6.67 \times 10^{-8}$ m/s$^2$. Clearly the above theoretical calculation is wrong!
The point is that the experiment here tells us the value of $G$. It is not an independent parameter. Upon observing $6.67 \times 10^{-8}$ m/s$^2$ in this experiment, we deduce that the value of $G$ must be $6.67 \times 10^{-11}$  m$^3$ / kg s$^2$.
An early experiment of this type was the one done by Cavendish. He did not measure the acceleration directly (it is too small) but rather he inferred the value of the force $F$ from the movement of a torsion balance.
