The mean radius of earth is R, its angular speed on its own axis is w and the acceleration due to gravity at earth's surface is g. What will be the radius of the orbit of a geostationary satellite

Options

A. `((R^(2)g)/(omega))^(1//3)`
B. `((R^(2)omega^(2))/(g))^(1//3)`
C. `((R^(2)g)/(omega^(2)))^(1//3)`
D. `((Rg)/(omega^(2)))^(1//3)`

Answer

C

Explanation

For a geostationary satellite, moving in a circular orbit of radius r with a speed `v_(0)`,
`v_(0)=sqrt((GM)/(r ))=sqrt((gR^(2))/(r ))" " because g= (GM)/(R^(2))`
`therefore T=(2pi r)/(v_(0))=(2pi r)/((gR^(2))/(r ))^(1//2)=(2pi r^(3//2))/(sqrt(gR^(2)))`
But `T=(2pi)/(omega)" " therefore (2pi r)/(omega)=(2pi r^(3//2))/(sqrt(gR^(2)))`
`therefore r^(3//2)=(sqrt(gR^(2)))/(omega)`
`therefore r^(3)=(gR^(2))/(omega^(2))" " therefore r = ((gR^(2))/(omega^(2)))^(1//3)`