地球半径为R，地面上重力加速度为g，在高空绕地球做匀速圆周运动的人造地球卫星，其线速度大小可能是（ ）
①
②
③
④
A．①②③
B．②③
C．①④
D．②③④
答案：B
题目分析：已知地球半径为R，地面上重力加速度为g，我们可以求出第一宇宙速度，第一宇宙速度又叫最大环绕速度，所以在高空绕地球做匀速圆周运动的人造卫星线速度都要比第一宇宙速度要小．
当卫星在地球表面附近做圆周运动时线速度是第一宇宙速度，利用万有引力提供向心力求解第一宇宙速度．
设：卫星质量m，地球质量M，由万有引力提供向心力得：
，解得
将黄金代换：GM=gR2代入得：
，在高空绕地球做匀速圆周运动的人造卫星线速度都要比第一宇宙速度要小，所以只有②③项满足
故选： B
点评：本题主要考察第一宇宙速度的求解方法和第一宇宙速度是卫星的最大环绕速度这个知识点．
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