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It's not an independent check.It's a circular argument.it uses the value for the mass of the Earth"We now multiply Mp by the mass of the earth"But nobody has ever measured that directly. It is calculated from the measured value of the local acceleration due to (the Earth's) gravity, the Earth's radius, and the Universal gravitational constant G.So, you are using G to calculate GSorry, but this method is dead.

"A manufactured mass with an exactly known mass, radius and value for g is all you need. g can be measured just by recording the speed at which the mass attracts an object over a well defined period of time. "Or you can do it the easy way- as Cavendish did.The point remains that you have calculated G by using a given value of G.So you have not really achieved anything." If the adjusted value of G is used as in the method you would arrive at probably the most accurate value for G."No it would not.It would give you exactly the same value of G as you measured.Do you not understand that you are using G to calculate G so, if you assumed that it was 42 it would come out to exactly 42?Also c/(c^2) is a pointlessly complicated way of saying 1/c

Try (c/c^2) * (2 * 10^-2). Elegant and simple.