Two satellites `S_(1)` and `S_(2)` are revolving around the earth in coplanar concentric orbits in the opposite sense. At `t=0`, the position of satellites are shown in the diagram. The periods of `S_(1)` and `S_(2)` are `4h` and `24h`, respectively. The radius of orbit of `S_(1)` is `1.28xx10^(4)km`. For this situation, mark the correct statement (s).

Options

A. The angular velocity of `S_(2)` as observed by `S_(1)` at `t=12 h` is `0.486 pi rad s^(-1)`
B. The two satellites are closest to each other for the first time at `t=12 h` and then after every `2h` they are closest to each other
C. The orbital velocity of `S_(1)` is `0.64 pixx10^(4) km`
D. The velocity of `S_(1)` relative to `S_(2)` is continuously changing in magnitude and direction both

Answer

A, B, C, D

Explanation

From Kepler's law `T^(2)propR^(3)`. So, `(T_(1))/(T_(2))=((R_(1))/(R_(2)))^(3/2)`
`R_(2)=((T_(2))/(T_(1)))^(2/3)xxR_(1)=(24/6)^(2/3)xx1.28xx10^(4)km`
`=3.22xx10^(4)km` orbital velocity of `S_(1)` is `v_(1)=(2piR_(1))/(T_(1))=(2pixx1.28xx10^(4))/4=0.64pixx10^(4)km`
Orbital velocity of `S_(2)` is
`v_(2)=(2piR_(2))/(T_(2))=(2pixx1.28xx10^(4))/4=0.64pixx10^(4) km`. At `t=12 h` the two satellites are closest to each other and after every `24h` they come at same position relative to each other. it is clear that direction of `v_(2)` w.r.t `v_(1)` is changing continuously in both magnitude and direction.
Angular velocity of `S_(2)` w.r.t. `S_(1)` at `t=12 h` is `omega=v_(1)+v_(2)//R_(2)-R_(1)=0.469 pi rads^(-1)`. this can be easily proved by writing the basic equations.