设地球表面的平均重力加速度为g，地球的半径为R，万有引力常数为G，则地球质量为gR2GgR2G地球的密度ρ为3g4πGR3g4πGR．

Question

设地球表面的平均重力加速度为g，地球的半径为R，万有引力常数为G，则地球质量为

(gR2)/(G)

(gR2)/(G)

地球的密度ρ为

(3g)/(4πGR)

(3g)/(4πGR)

．

(gR2)/(G)

(gR2)/(G)(gR2)/(G)gR2gR2R2R22GG

(gR2)/(G)

(gR2)/(G)(gR2)/(G)gR2gR2R2R22GG

(3g)/(4πGR)

(3g)/(4πGR)(3g)/(4πGR)3g3g4πGR4πGR

(3g)/(4πGR)

(3g)/(4πGR)(3g)/(4πGR)3g3g4πGR4πGR

Answer 1

不考虑地球的自转时，物体的重力等于地球对物体的万有引力，则得：
mg=G(Mm)/(R2)
解得地球的质量为：M=(gR2)/(G)
地球的密度为：ρ=(M)/(V)=((gR2)/(G))/((4)/(3)πR3)=(3g)/(4πGR)
故答案为：(gR2)/(G)，(3g)/(4πGR)． (Mm)/(R2)MmMmMmR2R2R22
解得地球的质量为：M=(gR2)/(G)
地球的密度为：ρ=(M)/(V)=((gR2)/(G))/((4)/(3)πR3)=(3g)/(4πGR)
故答案为：(gR2)/(G)，(3g)/(4πGR)． (gR2)/(G)gR2gR2gR22GGG
地球的密度为：ρ=(M)/(V)=((gR2)/(G))/((4)/(3)πR3)=(3g)/(4πGR)
故答案为：(gR2)/(G)，(3g)/(4πGR)． (M)/(V)MMMVVV=((gR2)/(G))/((4)/(3)πR3)=(3g)/(4πGR)
故答案为：(gR2)/(G)，(3g)/(4πGR)． ((gR2)/(G))/((4)/(3)πR3)(gR2)/(G)(gR2)/(G)(gR2)/(G)gR2gR2gR22GGG(4)/(3)πR3(4)/(3)πR3(4)/(3)444333πR33=(3g)/(4πGR)
故答案为：(gR2)/(G)，(3g)/(4πGR)． (3g)/(4πGR)3g3g3g4πGR4πGR4πGR
故答案为：(gR2)/(G)，(3g)/(4πGR)． (gR2)/(G)gR2gR2gR22GGG，(3g)/(4πGR)． (3g)/(4πGR)3g3g3g4πGR4πGR4πGR．