JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 22)
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 $$\times$$ 103 km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $${\pi \over x}rad\,{h^{ - 1}}$$ where x is ____________.
Answer
3
Explanation
_1st_September_Evening_Shift_en_22_1.png)
T1 = 1 hour
$$\Rightarrow$$ $$\omega$$1 = 2$$\pi$$ rad/hour
T2 = 8 hours
$$\Rightarrow$$ $$\omega$$2 = $${\pi \over 4}$$ rad/hour
R1 = 2 $$\times$$ 103 km
As T2 $$\propto$$ R3
$$ \Rightarrow {\left( {{{{R_2}} \over {{R_1}}}} \right)^3} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^2}$$
$$ \Rightarrow {{{R_2}} \over {{R_1}}} = {\left( {{8 \over 1}} \right)^{2/3}} = 4 \Rightarrow {R_2} = 8 \times {10^3}$$ km
_1st_September_Evening_Shift_en_22_2.png)
V1 = $$\omega$$1R1 = 4$$\pi$$ $$\times$$ 103 km/h
V2 = $$\omega$$2R2 = 2$$\pi$$ $$\times$$ 103 km/h
Relative $$\omega$$ = $${{{V_1} - {V_2}} \over {{R_2} - {R_1}}} = {{2\pi \times {{10}^3}} \over {6 \times {{10}^3}}}$$
$$ = {\pi \over 3}$$ rad/hour
x = 3
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