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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. |
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Answer» SOLUTION :`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. ` (##KPK_AIO_PHY_XI_P1_C08_E01_012_Q01.png" width="80%"> `implies` Consider the earth to be made up of concentric shells. `implies`A point mass m is situated at a distance r `(r ltR_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^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 : "Cavendishweighed the earth". |
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