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The minimum velocity of projection of a body to send it to infinity from the surface of a planet is (1)/(sqrt(6)) times that is required from the surface of the earth. The radius of the planet is (1)/(36) times the radius of the earth. The planet is surrounded by an atmosphere which contains monoatomic innert gas (gamma = 5//3) of constant density up to a height h(h lt lt radius of the planet). Find the velocity of sound on the surface of the planet. |
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Answer» <P> Solution :Escape velocity from the surface of the planet`v_(p)=sqrt(2g_(p)R_(p))` Given `v_(p)=(v_(e ))/(sqrt(6)) = sqrt((2g_(e )R_(e ))/(6))` `sqrt((g_(e )R_(e ))/(3)) = sqrt(2g_(p)R_(e )//36) rArr g_(p) =6g_(e )` Pressure exerted by the atmospheric column of height h on the surface of the planet `P = rho g_(P) h` USING equation of state `P = (rho RT)/(M)` Hence SPEED of the sound `v = sqrt((yRT)/(M)) = sqrt(yg_(p)h) = sqrt(6y g_(e )h) = sqrt(10 g_(e )h)` |
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