A geostationary satellite is at distance r₁ from Earth's center. A second satellite orbits in the eq

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Question

A geostationary satellite is at distance r₁ from Earth's center. A second satellite orbits in the equatorial plane in the opposite direction at distance r₂ from Earth's center, with r₁ = 1.21r₂. The time period of the second satellite as measured from the geostationary satellite is 24/p hours. The value of p is

Accepted Answer

Correct Answer: p = 2.33Step-by-step Solution:Step 1: Find T₂ (period of second satellite) using Kepler's Third Law.T² ∝ r³ → T₁/T₂ = (r₁/r₂)^(3/2) = (1.21)^(3/2) = (1.1)³ = 1.331T₁ = 24 hours (geostationary). T₂ = 24/1.331 ≈ 18.03 hours.Step 2: Find the apparent period as seen from geostationary satellite.The geostationary satellite rotates at angular velocity: ω₁ = 2π/T₁ = 2π/24 rad/h (eastward).The second satellite orbits in opposite direction: ω₂ = 2π/T₂ (westward, so effectively −2π/T₂ in the eastward-positive convention).Relative angular velocity: ω_rel = ω₁ − (−ω₂) = ω₁ + ω₂ (they move in opposite directions, so their angular velocity relative to each other is the sum).ω_rel = 2π/T₁ + 2π/T₂ = 2π(1/T₁ + 1/T₂)Apparent period: t = 2π/ω_rel = 1/(1/T₁ + 1/T₂) = T₁T₂/(T₁ + T₂)t = (24 × 18.03)/(24 + 18.03) = 432.72/42.03 ≈ 10.30 hours.But the problem says t = 24/p hours → p = 24/t = 24/10.30 ≈ 2.33. ✓Exact calculation:T₂ = T₁/(1.21)^(3/2) = 24/(1.1)³ = 24/1.331t = T₁T₂/(T₁+T₂) = T₁·(T₁/1.331)/(T₁ + T₁/1.331) = T₁/(1.331 + 1) = 24/2.331 ≈ 24/2.33So p = 1 + (1.21)^(3/2) = 1 + 1.331 = 2.331 ≈ 2.33Key Concepts:Kepler's Third Law: T² ∝ r³, so T ∝ r^(3/2). For two satellites: T₁/T₂ = (r₁/r₂)^(3/2).Relative angular velocity for opposite-direction orbits: When two objects move in opposite directions on circular paths, their relative angular velocity is ω₁ + ω₂ (the angular velocities add because they rotate in opposite senses). This is different from same-direction orbits, where ω_rel = |ω₁ − ω₂|.Apparent (synodic) period: The period as seen from one frame is t = 2π/ω_rel. For opposite orbits: 1/t = 1/T₁ + 1/T₂ (analogous to how two clocks running in opposite directions "lap" each other).Geostationary orbit: Period T = 24 hours, fixed with respect to Earth's surface. Any satellite with shorter period orbits faster; any with longer period orbits slower.

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