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 te 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

By kepler's `3^(rd)` law,
`(T^(2))/(R^(3))=` constant `therefore(T_(1)^(2))/(R_(1)^(3))=(T_(2)^(3))/(R_(2)^(3))` or `(1)/((10^(4))^(3))=(64)/(R_(2)^(3))`
or `R_(2)=4xx10^(4)km`
distance travelled in one revolution,
`S_(1)=2piR_(1)=2pixx10^(4)` and `S_(2)=2piR_(2)=2pixx4xx10^(4)`
`v_(1)=(S_(1))/(t_(1))=(2pixx10^(4))/(1)=2pixx10^(4)kmph` and `v_(2)=(S_(2))/(t_(2))=(2pixx4xx10^(4))/(8)=pixx10^(4)kmph`
`therefore` relative velocity `=v_(1)-v_(2)=2pixx10^(4)-pixx10^(4)=pixx10^(4)kmph`.