Show that in SI units, the unit of G is Newton `m^(2)kg^(-2).`

Explanation

we know that, Gravitational Force, `F=G(m_(1)m_(2))/(d^(2))`
Rearranging the terms, we get, `G=(F.d^(2))/(m_(1)m_(2))`
Using the respective SI units for the above quantities,
we get, `G=("Newton."m^(2))/(kg^(2))`
`therefore" In SI units, the unit of G is Newton "m^(2)kg^(-2)`
In CGS unit, the unit of G = dyne `cm^(2)g^(-2)`
The value of G is `6.67xx10^(-11)Nm^(2)//kg^(2)`
(This value was first experimentally measured by Henry Cavendish)