The acceleration due to gravity g and density of the earth `rho` are related by which of the following relations? (where G is the gravitational constant and `R_(E)` is the radius of the earth)

Options

A. `((g)/(G)) ( 4 pi)/( 3) R_(e)^(3) = rho`
B. `(((g)/(G)))/((4 pi)/( 3) R_(e))= rho`
C. `(g)/(G) ((4 pi)/93) R_(e)^(2) = rho`
D. `(((g)/(G)))/(((4 pi)/( 3) R_(e)^(3)))= rho`

Answer

B

Explanation

Let us take that, earth to be a sphere of radius `R_(e)` and mean density is given as `rho` then mass of the earth is given by ,
`M = ( 4 pi)/( 3) R_(e)^(3) rho`
According tot he law of gravitation the force on a body of mass m, near the surface of earth is given by , `F = (GM m)/( R_(e)^(2))` acting along the line joining the centre of the earth and the centre of the body .
Now, if this force produce an acceleration, g in the body of mass m . Then
`(GM m)/(R_(e)^(2)) = mg rArr g = (GM)/( r_(e)^(2))`
`rArr g = (G)/( R_(e)^(2)) xx (4)/( 3) pi R_(e)^(3) rho rArr ((g)/(G)) = (4)/(3) pi R_(e) rho rArr (((g)/(G)))/(((4)/(3) pi R_(e)))`