1）在地表：F萬(wàn)=GMm/R²=mg
==> GM=gR²
在B衛(wèi)星軌道上：GMm/(R+h)²=m(2π/T)²(R+h)
==> T²=4π²(R+h)³/GM=4π²(R+h)³/gR²
==> T=2π(R+h)/R *√[(R+h)/g]
2) A是地球的同步衛(wèi)星,即角速度為wA=w0
又B的角速度wB=2π/T=R/(R+h) *√[(g/(R+h)]
則再一次相遇時(shí)間t=2π/(wB-wA)=2π/{R/(R+h) *√[(g/(R+h)]-w0}