Two satellites `S_(1) and S_(2)` are revolving round a planet in coplanar circular orbits of radii `r_(1)` and `r_(2)` in the same directin, respectively.Their repective periods of revolution are 1 h and 8gh.The radius of orbit of satellitye `S_(1)` is equal to `10^(4)` km.What will be their relative speed (in km//h)when they are closes?

Explanation

By Kepler's `3^(rd)` law, `T^(2)/R^(3)=` constant `:. T_(1)^(2)/R_(1)^(3)=T_(2)^(2)/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)=2pi R_(1)=2 pi xx10^(4)` and `S_(2)=2pi R_(2)=2pixx4xx10^(4)`
`v_(1)=S_(1)/t_(1)=(2pixx10^(4))/1=2pixx10^(4)` km/h and `v_(2)= S_(2)/t_(2)=(2pixx4xx10^(4))/8=pixx10^(4)` km/h
`:.` Relative velocity `=v_(1)-v_(2)=2pixx10^(4)-pixx10^(4)=pixx10^(4)` km/h