The mass of the earth is 5.98x10^24 kg and its radius is 6.38x10^6 m. Compute the density of the earth, using the power of 10 notations and

The mass of the earth is 5.98x10^24 kg and its radius is 6.38x10^6 m. Compute the density of the earth, using the power of 10 notations and the correct number of significant figures.

 

$$\\V=\frac{4}{3}\pi*r^3\\\\
V=\frac{4}{3}\pi*(6.38*10^6)^3\\\\
V=\frac{4}{3}\pi*6.38^3*10^{18}\\\\
Density=\frac{Mass}{Volume}\\\\
Density=\frac{5.98*10^{24}}{\frac{4}{3}\pi*6.38^3*10^{18}}\;\;Kg/m^3\\\\
Density=\frac{3*5.98*10^{6}}{4\pi*6.38^3}\;\;Kg/m^3\\\\$$

 

$${\frac{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{5.98}}\right)}{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{6.38}}}^{{\mathtt{3}}}\right)}} = {\mathtt{0.005\: \!497\: \!313\: \!929\: \!962\: \!6}}$$

 

$$Density=\frac{3*5.98*10^{6}}{4\pi*6.38^3}\;\;Kg/m^3\\\\
Density=0.0054973139299626*10^{6}\;\;Kg/m^3\\\\
Density=5.4973139299626*10^{3}\;\;Kg/m^3\\\\$$

 

Now volume and mass both have 3 significant figures so I think density has 3 significant figures too.

 

$$\\Density=5.50*10^{3}\;\;Kg/m^3\\\\$$

 

Here are some clips to help you understand

 

https://www.youtube.com/watch?v=VRtR-rv96rg

https://www.youtube.com/watch?v=__csP0NtlGI