Two saetllites `S_(1)` and `S_(2)` revolve around a planet in coplaner circular orbit in the same sense. Their periods of revolutions are `1` hour and `8` hours respectively. The radius of orbit of `S_(1)` is `10^(4) km`. When `S_(2)` is closed to `S_(1)`, the speed of `S_(2)` relative to `S_(1)` is `pi xx 10^(n) km//h`. what is the value of `n`?

Answer

`(4)`

Explanation

Here, `T_(1) = 1h, T_(2) = 8h , r_(1) = 10^(4) km, r_(2) = ?`
From kepler's law
`r_(2) = r_(1) ((T_(2))/(T_(1)))^(3//2) = 10^(4) ((8)/(1))^(2//3) = 4 xx 10^(4) km`
Relative speed of `S_(2)` w.r.t. `S_(1) = |upsilon_(2) - upsilon_(1) |`
`= (2pi r_(2))/(T_(2)) - (2pi r_(1))/(T_(1)) = 2pi [(r_(2))/(T_(2)) - (r_(1))/(T_(1))]`
`= 2pi [(4 xx 10^(4))/(8) - (10^(4))/(1)]` ltbr. `= pi xx 10^(4) km//h = pi xx 10^(n) km//h` (Given)
`:. n = 4`