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If the Earth were a homogenous sphere of wood of density 800 kg/m^(3). What would be (i)the acceleration due to gravity on the Earth's surface ? (ii) the critical velocity of a satellite orbiting close to its surface ? |
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Answer» <P> Solution :Data : `p = 800 kg//m^(3) , G = 6.67 xx 10^(-11) N.m^(2)//Kg^(2)R_(E) = 6.4 xx 10^(6)` m(i) ` g = GM_(E)//R_(E)^(2)` ASSUMING the Earth to be a HOMOGENOUS sphere of density p. ` M_(E) = 4/3 piR_(E)^(3)p` Then , the surface gravity on a wooden Earth . ` g = (G (4/3 piR_(E)^(3)p))/(R_(E)^(2)) = 4/3piGR_(E)p` ` 4/3 (3.142) (6.67 xx 10^(-110 ) (6.4 xx 10^(6)) (800) ` ` = 1.431 m//s^(2)` (ii) The critical VELOCITY close the the surface of the wooden Earth. ` v_(c0= sqrt(gR_(E)) = sqrt(1.431 xx 6.4xx10^(6))` `3.026 xx 10^(3) m//s` |
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