一道关于万有引力的高一物理题
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\u4e14H\uff1da*t^2/2 \u5f97 F\uff1d6.2*10\uff3e6\u725b\u987f
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\u7531\u4e07\u6709\u5f15\u529b\u63d0\u4f9b\u5411\u5fc3\u529b\uff0c\u77e5 G*m*M/(h+R)^2\uff1dm*(2*3.14/T)^2*(h+R)\uff0c\u4e14mg\uff1dG*m*M/R^2
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首先明确，物体在地球两极处不需要向心力，故可得第一个方程P=G*M*m/R^2
又因为物体在赤道处需要有一个向心力才能随着地球一起自转，故可得第二个方程0.9P=G*M*m/R^2-m*ω^2*R=G*M*m/R^2-m*(2π/T）^2*R,m*4π^2*R/T^2=0.1P
解得m=(P*T^2)/(40π^2*R)
将其代入第一个方程可得P=(G*M*P*T^2)/(40π^2*R^3)
化简可得G*M*T^2=40π^2*R^3，M=(40π^2*R^3)/(G*T^2)
又因为地球的体积为V=4πR^3/3
所以地球的平均密度为ρ=M/V=（120π^2*R^3)/(4πR^3*G*T^2)=30π/(G*T^2)
万有引力等于重力
P=GMm/R^2
合力做向心力
P-0.9P=mω^2R
可得M/R^3=ω^2/0.1G
ω=2π/T
G=6.67x10－11 N·m2 /kg2
ρ=M/V=M/(4πR^3/3)=ω^2/(4*0.1πG/3)
=30π/GT^2
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