Two stallites A and B revolve round the same planet in coplanar circular orbits lying in the same plane. Their periods of revolutions are 1h and 8h, respectively. The radius of the orbit of A is `10^(4)`km. The speed of B relative to A when they are closed in `kmh^(-1)` is

Options

A. `10^(4)pi`
B. `2xx10^(4)pi`
C. `10^(4)pi//2`
D. `4xx10^(4)pi`

Answer

a

Explanation

`(r_(2))/(r_(1))=((T_(1))/(T_(1)))^(2//3)=((8)/(1))^(2//3)=4`
`r_(2)=4xxr_(1)`
`=4xx10^(4)km`
`v_(1)=r_(1)omega_(1)=10^(4)xx(2pi)/(T_(1))`
`v_(1)=(2pixx10^(4))/(1)=2pixx10^(4)` ltBrgt `v_(2)=v_(2)omega_(2)`
`=4xx10^(4)xx(2pi)/(T_(2))`
`=(4xx10^(4)xx2pi)/(8)=pixx10^(4)"km/h"`
`V_(AB)=V_(B)-V_(A)`
`=2pixx10^(4)-pixx10^(4)=pixx10^(4)(2-1)`
`pixx10^(4)`.