Calculate the average density of the Earth in g/cm^3 given that Earth's mass is 5.97*10^24 kg, Earth's radius is 6.38*10^6 m, and Earth is a perfect sphere.

Question

lgbasallote · Answer

$$\rho = \frac MV$$
$$V = \frac 43 \pi r^3$$
$$\implies \rho = \frac{M}{\frac 43 \pi r^3}$$
$$\implies \rho = \frac{3M}{4\pi r^3}$$
does that help?

anonymous · Answer

You see, I did that. I suppose I am just having issues with converting from $$kg/m ^{3} \to g/cm ^{3}$$I try and keep coming up with about 55,000; any help? By the way, before the conversion I head roughly 5,500. So, it multiplied by ten. Am I just making a clumsy mistake? Thanks for the help. It is much appreciated.

lgbasallote · Answer

you were specifically told to convert to g/cm^3?

lgbasallote · Answer

oh yeah i see

lgbasallote · Answer

im going to use dimension analysis to convert...
$$\frac{kg}{m^3} \times \frac{1000g}{1 kg} \times \frac{ 1 m}{100 cm} \times \frac{1 m}{100 cm} \times \frac{1m}{100 cm}$$
is that what you did?

anonymous · Answer

Oh goodness, I completely ignored the cubed meters, so I only did it once. It was just a clumsy mistake then. Thank you so much. :D

lgbasallote · Answer

haha welcome ^_^