Problem Statement
A satellite orbits Earth at height h above the surface. Find its orbital velocity and period.
Given
M_E, R_E, h, G.
Concepts & Formulas
Circular orbit: centripetal = gravity. mv²/(R_E+h) = GM_Em/(R_E+h)² → v = √(GM_E/(R_E+h)).
Step-by-Step Solution
Step 1: v = √(GM_E/(R_E+h)).
Step 2: T = 2π(R_E+h)/v = 2π(R_E+h)^{3/2}/√(GM_E).
Step 3: At h=0: v₀ = √(gR_E) ≈ 7.9 km/s. T₀ = 2πR_E/v₀ = 2π√(R_E/g) ≈ 84 min.
Worked Calculation
v = √(GM_E/(R_E+h)). T = 2π(R_E+h)^{3/2}/√(GM_E).
Boxed Answer
v = √(GM_E/(R_E+h)); T = 2π(R_E+h)^{3/2}/√(GM_E)
Physical Interpretation
Kepler’s 3rd law: T² ∝ r³. A low-orbit satellite (h≪R_E) has period ~84 min; geostationary orbit at h≈36,000 km has period 24 h.
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