两颗人造卫星质量之m1:m2=3:1 绕地球做匀速园周运动的轨道半径之比为r1:r2=1:2 求线速度大小之比__ 角速
人气:131 ℃ 时间:2019-09-22 06:03:19
解答
与质量无关
a=v^2/r=g'
g'=GM/(r^2)
故v1/v2=(r2/r1)^(1/2)
w=v/r
w1/w2=(r2/r1)^(3/2)
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