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Prove that the critical orbital speed of a satellite in a near-Earth is 2Rsqrt(pi_(p)G//3), where p and R are the average density and radius of the Earth and G is the universal gravitional constant. |
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Answer» Solution :Suppose a satellite is in a circular orbit of radius `r = R + h` around the Earth, where R is the radius of the Earth and h the ALTITUDE of the orbit. The critical orbital speed of the satellite is `v_(c)=SQRT((GM)/(r))""...(1)` Where M is the mass of the Earth and G the universal gravitional constant. For a low-altitude or near-Earth orbit `h LT lt R`, so that ignoring h , `r = R + h ~= R ` and `u_(c)~=sqrt((GM)/(R))""...(2)` If we consider the Earth to be a SPHERE of average density p, then `rho=("mass")/("volume")=(M)/((4)/(3)piR^(3))` `therefore M = (4)/(3)pirhoR^(3)""...(3)` Substituting this value of M in Eq. (2), we get, `u_(c)=sqrt((G((4)/(3)pirhoR^(3)))/(R))=sqrt((rpirhoGR^(2))/(3))=2Rsqrt((pirhoG)/(3))` Which is the required EXPRESSION. |
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