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 ?

Options

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)`

Explanation

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)`