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Assuming the radius of the earth to be `6.4xx10^(6)` m calculate the time period T of a satellite for equatorial orbit at `1.4xx10^(3)` km above the surface of the earth and the speed of the satellite in this orbit ?A. 6831 and 7200 `ms^(-1)`B. 6850 s and 7151 `ms^(-1)`C. 68321 s and 7190 `ms^(-1)`D. 6850 and 7400 `ms^(-1)` |
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Answer» Given radius of earth `R_(e )=6.4xx10^(3) m` height `h=1.4xx10^(3)km` Mass of the earth `M_(e )=5.98xx10^(24) kg` Gravitational constant `G=6.67xx10^(11) Nm^(2)(kg)^(2)` `r=R_(e )+h` `=6.4xx10^(6)+1.4xx10^(6)m=7.8xx10^(6)` m Time period `T=[(4 pi^(2)r^(12))/(Gm_(e ))]^(1//2)` `=[(4xx(3.14)^(2))xx(7.8xx10^(6))/(6.67xx10^(-11)xx5.98xx10^(24))]=6853.4` s speed of satellite `v=sqrt(GM_(e ))/(r )` `=[(6.67xx10^(11)xx5.98xx10^(24))/(7.8xx10^(6))]^(1//2) =7151 ms^(-1)` |
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