Obtain an expression of acceleration produced by gravity of earth. OR Obtain general equation of gravitation force at distance r from the centre of earth and derive the equation of acceleration due to gravity on the surface of earth. 

Explanation

`implies` Imagined a earth to be a sphere made of a large number of concentric spherical shells with the smallest one at the centre and the largest one at its surface.
`implies` A point outside the earth is outside all the shells.
`implies` Thus, all the shells exert a gravitational force at the point outside just as if their masses are concentrated at their common centre.
`implies` For a point inside the earth, the situation is different. This is illustrated in figure.
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`implies` Consider the earth to be made up of concentric shells.
`implies` A point mass m is situated at a distance r `(r lt R_E)`
`implies` from the centre. The point P lies outside the sphere of radius r. For the shells of radius greater than r, the point P lies inside. Hence, no gravitational force exert on mass m kept at P
`implies` If mass of particle is m at P, mass of sphere m, with radius r, then the force on the mass m at P has a magnitude of force
`implies F =(GmM_(r))/r^2 " "...(1)`
`implies` We assume that the entire earth is of uniform density.
Hence its mass is `M_E = (4/3 piR_E^3)rho`
`:. 4/3 pirho = M_E/ R_E^3 " "...(2)`
`implies` But `M_r = 4/3 pir^(3) rho " "...(3)`
`implies` Putting value of equation (3) in equation (1)
`implies :. F = (Gm)/(r^2)((4)/(3) pir^3rho)`
`implies F = Gm (4/3pi rho)r " " ... (4)`
`implies` Value of equation (2) in equation (4),
Equation indicates for F r, r `F = ((GM_Em)/(R_E^3)).R_E`
`:. F = (GM_Em)/(R_E^2) " "....(6)`
`implies ` The acceleration experienced by the mass m is denoted by g. According to Newton.s second law,
`F = mg " "... (7)`
`implies` Equating equation (6) and (7),
`:. mg = (GM_(E)m)/(R_E^2)`
`:. g = (GM_E)/(R_E^2)" "....(8)`
`implies` Acceleration g is readily measurable.
`G = 6.61 x 10^(-11) "Nm"^(2) //kg^(2), M_E = 6 xx 10^(24) kg`
`R_E = 6400 km,` then value g can be obtained. `g = 9.77050 m//s^(2) = 9.8 m//s^(2)`.
`implies` The measurement of G is combined with knowledge of g and `R_E` enables one to estimate `M_E`. This is the reason why there is a popular statement regarding Cavendish : "Cavendish weighed the earth".