Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hours…

Question

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)/xradh^(-1), where x is _______.

MARKS Content Team · Accepted Answer

Correct answer: 3. T_1=1 hour ⇒ω_1==2π rad/hour T_2=8 hours ⇒ω_2=(π)/4 rad/hour R_1=2×10^3 km As T^2∝R^3 ⇒R2R1^3=T2T1^2⇒R_2R_1=18^(2/3)=4⇒R=8×10^3 km V_1=ω_1R_1=4π×10^3 km/h V_1=ω_2R_2=2π×10^3 km/h Relative ω=(V_1-V_2)/(R_2-R_1)=(2π×10^3)/(6×10^3)=(π)/3 rad/hour x=3

Solution

Relevant Questions