Question

An artificial satellite revolves round the earth at a height of `1000 km`. The radius of the earth is `6.38xx10^(3)km`. Mass of the earth `6xx10^(24)kg,G=6.67xx10^(-11)Nm^(2)kg^(-2)`. Find the orbital speed and period of revolution of the satellite.

Answer 1

Here, `h = 3600 km, R = 6400 km = 6.4 xx 10^(6) m`
Radius of orbita of satellite, `r = R + h = 6400 + 3600`
`= 10000 km`
`= 10^(7) m`
Time period satellite, `T = (2pi)/(R ) sqrt(((R + h)^(3))/(g)) = 2pi sqrt(((R + h)^(3))/(g R^(2)))`
`T = 2 xx 3.4 sqrt(((10^(7))^(3))/(10 xx (6.4 xx 10^(6))^(2))) = 9812.5 s`
Orbital velocity of the satellite
`upsilon = (2 pi r)/(T) = (2 xx 3.14 xx 10^(7))/(9812.5) = 6.4 xx 10^(3) ms^(-1)`