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 × 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 π/x rad h⁻¹ where x is ____________. Numerical answer type. Work it out, then open the answer.

Answer: 3. T ∝ r3/2: (r₂)/(r₁) = (8/1)2/3 = 4, so r₂ = 8×10³ km. Speeds: v₁ = (2π(2×10³))/(1) = 4π×10³ km/h, v₂ = (2π(8×10³))/(8) = 2π×10³ km/h, both in the same direction. When closest, the relative velocity is perpendicular to the line joining them: ωrel = (v₁ - v₂)/(r₂ - r₁) = (2π×10³)/(6×10³) = π/3 rad/h So x = 3.

Two satellites revolve around a planet in coplanar circular orbits in… | JEE Main 2021 (1 Sep, Shift 2)

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Satellites - Orbital Velocity & Period

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 $\frac{\pi}{x}\text{ rad h}^{-1}$ where x is ____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

$T \propto r^{3/2}$ : $\frac{r_2}{r_1} = \left(\frac{8}{1}\right)^{2/3} = 4$ , so $r_2 = 8\times10^3$ km.
Speeds: $v_1 = \frac{2\pi(2\times10^3)}{1} = 4\pi\times10^3$ km/h, $v_2 = \frac{2\pi(8\times10^3)}{8} = 2\pi\times10^3$ km/h, both in the same direction.
When closest, the relative velocity is perpendicular to the line joining them: $\omega_{rel} = \frac{v_1 - v_2}{r_2 - r_1} = \frac{2\pi\times10^3}{6\times10^3} = \frac{\pi}{3}$ rad/h
So $x = 3$ .

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