Two satellites A and B revolve round a planet in coplanar orbits in the same direction, rheir periods are 1 hour and 8 hours respectively. The orbital radius of A is `10^(4)` km the speed of B relative to A when they are closest is :

Options

A. `10^(4)pi km//h`
B. `2xx10^(4)pi km//h`
C. `(v_(1))/(r_(1))=(10^(4)pi)/(2)km//h`
D. `4xx10^(4)pi km//h`.

Answer

A

Explanation

As shown in Fig. A and B are closest in this position
`therefore v_(1)=(2pi r_(1))/(T_(1))=(2pi xx 10^(4))/(1)km//hr`
As`r_(1)=10^(4)km`

By Kepler.s law `(r_(2))/(r_(1))=((T_(2))/(T_(1)))^(2//3)=((8)/(1))^(2//3)`
or `r_(2)=4xx10^(4)km`
`v_(2)=(2pi r_(2))/(T_(2))=(2pi xx4xx10^(4))/(8)`
`=pi xx 10^(4) km//hr`.
Correct choice is (a).