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. `3pi xx 10^(4)`
B. zero
C. `2pi xx 10^(4)`
D. `pi xx 10^(4)`

Answer

D

Explanation

From the Kepler's law,
`(T_(1)^(2))/(T_(2)^(2)) = (r_(1)^(3))/(r_(2)^(3))`
`rArr (1)/(64) = (r_(1)^(3))/(r_(2)^(3)) rArr r_(2) = 4xx 10^(4)km`
`:. V_(1) = (2pi r_(1))/(T_(1))` and `v_(2) = (2pir_(2))/(T_(2))`
`rArr v_(1) = 2xx 10^(4) pi kmh^(-1)`
and `v_(2) = pi xx 10^(4) kmh^(-1)`
Thus, speed of B relative to A is `v_(21)`.
i.e. `v_(21) = pi xx 10^(4) kmh^(-1)`.