Suppose unknowingly you wrote the universal gravitational constant value as `G = 6.67 xx 10^(11)` Instead of the correct value `G = 6.67 xx 10^(-11)`, what is the acceleration due to this new acceleration due to gravity, what will be your weight W'?

Explanation

Let `G_(SI)` be the gravitational constant in the Sl system and `G_("cgs")` in the egs system. Then
`G_(SI) = 6.6 xx 10^(-11) Nm^(2) kg^(-2) , G_("cgs") = ?`
`n_(2) = n_(1) [(M_(1))/(M_(2))]^(a) [(L_(1))/(L_(2))]^(b) [(T_(1))/(T_(2))]^(c)`
`G_("cgs") = G_(SI) [(M_(1))/(M_(2))]^(a) [(L_(1))/(L_(2))]^(b) [(T_(1))/(T_(2))]^(c)`
`M_(1) 1 kg L_(1) = 1m T_(1) = 1s`
`M_(2) 1 g L_(2) = 1 cm T_(2) = 1s`
The dimensional formula for G is `M^(-1) L^(3)T^(-2)`
a = -1, b = 3 and c = -2
`G_("cgs") = 6.6 xx 10^(-11) [(1kg)/(1g)]^(-1) [(1m)/(1cm)]^(3) [(1s)/(1s)]^(-2)`
`=6.6 xx 10^(-11) xx 10^(-3) xx 10^(6) xx 1`
`G_("cgs") = 6.6 xx 10^(-8)` dyne `cm^(2) g^(-2)`