A satellite close to the earth is in orbit above the equator with a period of rotation of `1.5` hours. If it is above a point `P` on the equtor at some time, it will be above `P` again after time ...........

Options

A. (ii),(iv)
B. (ii),(iii)
C. (i),(iii)
D. (i),(iv)

Answer

A

Explanation

Let `omega_(0)`, angular velocity of earth, `omega_(0)=(2pi)/(T_(0))=(2pi)/24`
`omega`: angular velocity of satellite, `omega=(2pi)/T=(2pi)/1.5`
For satellite, rotating west to east, `omega_(1)=omega-omega_(0)`
Time periode of rotation relative to earth
`t_(1)=(2pi)/(omega_(1))=(2pi)/((2pi)/1.5-(2pi)/24)=1.6 hr`
For satellite, rotating east to west, `omega_(2)=omega+omega_(0)`
Time periode of rotation relative to earth
`t_(2)=(2pi)/(omega_(2))=(2pi)/((2pi)/1.5+(2pi)/24)=24/17 hr`