The earth may be regarded as a spherically shaped uniform core of density `rho_(1)` and radius `(R)/(2)` surrounded by a uniform shell of thickness `(R)/(2)` and density `rho_(2)`. Find the ratio of `(rho_(1))/(rho_(2))` if the value of acceleration due to gravity is the same at surface as at depth `(R)/(2)` from the surface.

Answer

`(7)/(3)`

Explanation

Total mass of earth
`M=(4)/(3)pi((R)/(2))^(3)rho_(1)+(4)/(3)pi(R^(3)-(R^(3))/(8))rho_(2)`
`M=(4piR^(3))/(24)(rho_(1)+7rho_(2))`
Acceleration due to gravity at earth's surface `=(GM)/(R^(2))=(4piGR)/(24)(rho_(1)+7rho_(2))`
Acceleration due to gravity at depth `(R)/(2)` from the surface `=(G[(4)/(3)pi((R)/(2))^(2)rho_(1)])/(((R)/(2))^(2))=(4piGrho_(1)R)/(6)`
Now according to question
`(4piGr(rho+7rho_(2)))/(24)=(4piGrrho_(1))/(rho_(2))=(7)/(3)`