Reviewed CSCA Physics question · Hard

一颗人造卫星绕地球做匀速圆周运动，轨道半径为地球半径的2倍。已知地球表面的重力加速度为g，地球半径为R，求卫星的周期T。

T = 2π√(2R/g)
T = 4π√(2R/g)
T = 2π√(R/g)
T = 4π√(R/g)

Show the answer and explanation

Correct answer

B. T = 4π√(2R/g)

Principle or equation

万有引力提供向心力：GMm/r² = m(4π²/T²)r，且GM = gR²。轨道半径r = 2R。

Why this answer is correct

由GM = gR²，代入向心力公式：gR²/(2R)² = 4π²(2R)/T² → gR²/(4R²) = 8π²R/T² → g/4 = 8π²R/T² → T² = 32π²R/g → T = 4π√(2R/g)。

Example

若轨道半径为4R，则T = 2π√((4R)³/(gR²)) = 2π√(64R/g) = 16π√(R/g)，但本题为2R。

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