The average density of the earth is `5500kg*m^(-3)` the gravitational constant is `6.7xx10^(-11)N*m^2*kg^(-2)` and the radius of the earth is 6400 km. Using the given values, find the magnitude of the acceleration due to gravity on the surface of the earth.

Explanation

It is known `rho =(3g)/(4piRG)`
or,`g=(4pi rhoRG)/(3)`
`therefore g=(4xx22xx5500xx6400xx10^3xx6.7xx10^(-11))/(7xx3)`
`=9.88m*s^(-2)`.