A satellite is revolving around the Earth. If its height is 4 times the height of a geo-stationary satellite, what will become its time period?
Answer: 8 days
Explanation
Kepler's law states:T² = T₀ × (R/R₀)³/²
Given:
R₀ = Rₑ + h₀ (geo-stationary satellite)
R = Rₑ + 4h₀ (new satellite)
T₀ = 24 hours (1 day)
R/R₀ = 5
T² ≈ 1 × 5³/² ≈ 8 days
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