Planet Earth

What is the weight of the planet Earth, if its average density is ρ = 2.5 g/cm ^ 3?

Correct result:

m =  2.70801729212E+21 t

Solution:

ρ=2.5 g/cm3 ρ1=ρkg/m3=ρ 1000 kg/m3=2.5 1000 kg/m3=2500 kg/m3  R=6371 kmm=6371 1000 m=6371000 m   V=43 π R3=43 3.1416 63710003=1.083206916851021 m3  m1=ρ1 V=2500 1.083206916851021=2.708017292121024 kg  m=m1t=m1/1000 t=2.708017292121024/1000 t=2.708017292121021 t=2.708017292121021 tρ=2.5 \ \text{g/cm}^3 \ \\ ρ_{1}=ρ \rightarrow kg/m^3=ρ \cdot \ 1000 \ kg/m^3=2.5 \cdot \ 1000 \ kg/m^3=2500 \ kg/m^3 \ \\ \ \\ R=6371 \ km \rightarrow m=6371 \cdot \ 1000 \ m=6371000 \ m \ \\ \ \\ \ \\ V=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ R^3=\dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 6371000^3=1.08320691685\cdot 10^{ 21 } \ \text{m}^3 \ \\ \ \\ m_{1}=ρ_{1} \cdot \ V=2500 \cdot \ 1.08320691685 \cdot 10^{21}=2.70801729212\cdot 10^{ 24 } \ \text{kg} \ \\ \ \\ m=m_{1} \rightarrow t=m_{1} / 1000 \ t=2.70801729212 \cdot 10^{24} / 1000 \ t=2.70801729212 \cdot 10^{21} \ t=2.70801729212\cdot 10^{ 21 } \ \text{t}



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