Question 12

Geostationary satellite: period T₁ = 24 h at radius r₁. By Kepler's third law, T ∝ r^(3/2): T₂/T₁ = (r₂/r₁)^(3/2) = (1/1.21)^(3/2) = (1/1.1)³ = 1/1.331 So T₂ = 24/1.331 hours, i.e. ω₂ = 1.331 ω₁ in magnitude. The second satellite orbits in the opposite sense, so the relative angular velocity as seen from the geostationary satellite is: ω_rel = ω₁ + ω₂ = ω₁(1 + 1.331) = 2.331 ω₁ Relative period: T_rel = 2π/ω_rel = T₁/2.331 = 24/2.331 hours Hence p = 2.331.

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