Question

An artificial satellite is describing an equatorial orbit at 1600 km above the surface earth. Calculate its orbital speed and the period of revolution. If satellite is travelling in the same direction as the rotation of the earth (i.e., from west to east), calculate the interval between two successive time at wich it will appear verically overhead to an observer at a fixed point on the equator. Radius of earth = 6400 km.

Answer 1

We know that period of the satellite is
`T = (2pi)/(sqrt(GM))r^(3//2) = (2pi)/(sqrt(gR^(2)))r^(3//2)`
Where `r = 6400+1600=8000xx10^(3)m`,
`g = 9.8 m/sec^(2)` and `R = 6400xx10^(3)m`
Substituting values we get
`T=2xx3.14[((8000xx10^(3))^(3))/(9.8xx(6400xx10^(3))^(2))]^(1//2)`
`= 7096`
Futher, orbital speed,
`v = sqrt((GM)/(r )) = sqrt((gR^(2))/(r ))`
or `v = sqrt(((9.8)/(8000xx10^(3))))xx(6400xx10^(3))`
`= 7083.5 m//s`
Let t be the time interval between two successive moments at which the satellite is overhead to an observer at fixed position on the equator. As both satelite and earth are moving in same direction with angular speed `omega_(S)` and `omega_(E )` respectively, we can write the time of separation as
`t = (2pi)/(omega_(S)-omega_(E ))`
Here `omega_(S)=(2pi)/(7096)` and `omega_(E )=(2pi)/(86400)`
Thus we have `t = (86400xx7096)/(86400-7096)`
`= 7731 s`