Wave Optics

JEE Main level, numerical answer, medium

81 of 191

Question 81 of 191

Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 11 hour and 88 hours respectively. The radius of the orbit of nearer satellite is 2×103km.2\times {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 πxradh−1\dfrac{\pi }{x}rad{h}^{-1}, where xx is _______.

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Answer: 3

T1=1{T}_{1}=1 hour
⇒ω1==2πrad/hour\Rightarrow {\omega }_{1}==2\pi rad/hour
T2=8{T}_{2}=8 hours
⇒ω2=π4rad/hour\Rightarrow {\omega }_{2}=\dfrac{\pi }{4}rad/hour
R1=2×103km{R}_{1}=2\times {10}^{3}km
As T2∝R3{T}^{2}\propto {R}^{3}
⇒(R2R1)3=(T2T1)2⇒R2R1=(18)2/3=4⇒R=8×103km\Rightarrow {\left(\dfrac{{R}_{2}}{{R}_{1}}\right)}^{3}={\left(\dfrac{{T}_{2}}{{T}_{1}}\right)}^{2}\Rightarrow \dfrac{{R}_{2}}{{R}_{1}}={\left(\dfrac{1}{8}\right)}^{2/3}=4\Rightarrow R=8\times {10}^{3}km
V1=ω1R1=4π×103km/h{V}_{1}={\omega }_{1}{R}_{1}=4\pi \times {10}^{3}km/h
V1=ω2R2=2π×103km/h{V}_{1}={\omega }_{2}{R}_{2}=2\pi \times {10}^{3}km/h
Relative ω=V1−V2R2−R1=2π×1036×103\omega =\dfrac{{V}_{1}-{V}_{2}}{{R}_{2}-{R}_{1}}=\dfrac{2\pi \times {10}^{3}}{6\times {10}^{3}}=π3rad/hour=\dfrac{\pi }{3}rad/hour
x=3x=3

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