If the Earth were made of lead of relative density `11.4`, then find the value of acceleration due to gravity on the surface of Earth ? Radius of the Earth is `6400 km` and `G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`.

Explanation

Density of Earth `rho =` relative density
`xx` density of water
`= 11.4 xx 10^(3) kg m^(-3)`
Acceleration due to gravity on the surface of Earth is
`g = (GM)/(R^(2)) = (G)/(R^(2)) xx (4)/(3) pi R^(3) rho = (4 piGR rho)/(3)`
`= (4 xx (3.14) xx (6.67 xx 10^(-11)) xx (6400 xx 10^(3)) xx (11.4 xx 10^(3)))/(3)`
`= 20.38 ms^(-2)`