Two satellites `S_(1)` and `S_(2)` are revolving round a planet in coplanar and concentric circular orbit of radii `R_(1)` and `R_(2)` in the same direction respectively. Their respective periods of revolution are 1 hr and 8 hr. the radius of the orbit of satellite `S_(1)` is equal to `10^(4)`km. Find the relative speed in kmph when they are closest.

Explanation

Let the mass of the plane be M, that `S_1 be m_1 and of S_2 be m_2.` Let the radius of the orbit of `S_1 be R_1(-10^4 km) and of S_2be R_2`.
`Let v_1 and v_2` be the linear speeds of `S_1 and S_2` with respect to the planet. ure shows the situation
AS the square of the time period is proprotioN/Al to the cube to the radius,
`(R_2/R_1)^3=(T_2/T_1)^2((8h)/(1h))^2=64`
or `R_2/R_1=4`
or `R_2=4R_1=4xx10^4km`.
Now the time period of `S_1 is 1h. So,`
`(2piR_1)/v_1=1h`
or `v_1=(2piR_1)/(1h)=2pixx1064kmh^-1`
similarlly `v_2=(2piR_2)/(8h)=pixx10^4kmh^-1`
a. At the closet sepration, they ar movng in the same direction. Hence the speed of `S_2` with respect to `S_1 is |v_1-v_1|=pixx10^3 kmh^-1`.
b. As seen from `S_1 the satellite S_2` is at a distance `R_2-R_1=3xx10^4` km at the closet separation. Also it is moving at `pixx10^4kmh^-1` in a direction perpendicular to the line joining them. Thus the angular speed of `S_2` as observed by `S_1` is
`omega=(pix10^4kmh^-1)/(3x10^4km)=pi/3radh^-1`.