Assuming the earth of to be a uniform sphere of radius `6400 kg` and density `5.5 g//c.c.`, find the value of `g` on its surface. `G=6.66xx10^(-11)Nm^(2)kg^(-2)`

Explanation

Density of earth = relative density `xx`
density of water `= 11.5 xx 10^(3) kg//m^(3)`
Radius of earth, `R = 6400 km = 6.4 xx 10^(6) m`
Acceleration due to gravity on the surface of earth
`g = (GM)/(R^(2)) = (G)/(R^(2)) xx (4)/(3) pi R^(3) rho = (4)/(3) pi GR rho`
`= (4)/(3) xx (22)/(7) xx (6.67 xx 10^(-11))`
`xx (6.4 xx 10^(6)) xx (11.5 xx 10^(3))`
`= 20.57 ms^(-2)`