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Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hours respectively. The radius of the orbit of nearer satellite is 2 × 10 3 km . The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is π x rad h - 1 , where x is _ _ _ _ _ _ _ .

Correct answer

3

Step-by-step solution

T 1 = 1 hour ⇒ ω 1 = = 2 π   rad / hour T 2 = 8 hours ⇒ ω 2 = π 4   rad / hour R 1 = 2 × 10 3   km As T 2 ∝ R 3 ⇒ R 2 R 1 3 = T 2 T 1 2 ⇒ R 2 R 1 = 1 8 2 / 3 = 4 ⇒ R = 8 × 10 3   km V 1 = ω 1 R 1 = 4 π × 10 3   km / h V 1 = ω 2 R 2 = 2 π × 10 3   km / h Relative ω = V 1 - V 2 R 2 - R 1 = 2 π × 10 3 6 × 10 3 = π 3   rad / hour x = 3

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