Question

The radius of the earth is $6.37\times {10}^{6}m$ and its mass is $5.975\times {10}^{24}kg$. Find the earth's average density to appropriate significant figures.

• A. $5\times {10}^{3}kg{m}^{-3}$
• B. $5.52\times {10}^{3}kg{m}^{-3}$
• C. $2\times {10}^{3}kg{m}^{-3}$
• D. $5.52\times {10}^{3}kg{m}^{-4}$

Answer 1

The correct option is B $5.52\times {10}^{3}kg{m}^{-3}$

$\begin{array}{c}\text{radius}=6.37\times {10}^{6}m\\ \text{Volume}=\frac{4}{3}\pi r^3\\ V=\frac{4}{3}\times \pi \times {(6.37\times {10}^6)}^3\\ \text{Mass}=5.975\times {10}^{24}kg\\ \text{density}\left(\rho \right)=\frac{\mathrm{mass}\left(m\right)}{V}\end{array}$

$\begin{array}{cc}\rho & =\frac{5.975\times {10}^{24}}{4\times \pi \times {(6.37\times {10}^6)}^3}\\ & =5.52\times {10}^3{kgm}^{-3}\end{array}$