Finding weight in Megagrams (Mg) given circumference and density

Question

Glaciers often deposit large rocks called erratics. The granite rock has a circumference of 9.5 m. Assuming it conforms to the shape of a sphere, what would be its weight in Megagrams (Mg), where 1 Mg = 1,000 Kg ≈ 1 US ton. The average density of granite is $2.70\ \mathrm{g} \cdot \mathrm{cm}^{-3}$.

Answer 1

HINT

1. Given a circumference of the sphere, find its volume $V$.
2. With volume and density, find the mass.
3. Convert the mass to the desired units.

UPDATE

Since the circumference is $$C=2\pi r \iff r = \frac{C}{2\pi}$$ and $V = 4\pi r^3/2$, you have $$ \begin{split} V &= \frac43\pi r^3 = \frac43 \pi \left(\frac{C}{2\pi}\right)^3 = \frac{4}{3 \cdot 8} \frac{\pi}{\pi^3} C^3 = \frac{C^3}{6\pi^2} \\ &= \frac{(9.5 \mathrm{m})^3}{6 \pi^2} = \frac{9.5^3}{6 \pi^2} \textrm{m}^3 \\ &\approx 14.478 \mathrm{m}^3. \end{split} $$

Your density $\rho$ is in the wrong units, so to find the mass $M$ you have to do the following: $$ \begin{split} M &= \rho V = 2.7 \frac{\mathrm{g}}{\mathrm{cm}^3} \times 14.478 \mathrm{m}^3 \\ &= 2.7 \frac{\mathrm{g}} {\mathrm{\left(cm \times \frac{1 \mathrm{m}} {100 \mathrm{cm}}\right)}^3} \times 14.478 \mathrm{m}^3 \\ &= \frac{2.7 \mathrm{g}}{\left(\frac{1}{100} \mathrm{m}\right)^3} \times 14.478 \mathrm{m}^3 \\ \\ &= 2.7 \times 100^3 \times 14.478 \frac{\mathrm{g} \cdot \mathrm{m}^3}{\mathrm{m}^3} \\ &= 39090.6 \mathrm{g}. \end{split} $$ Can you take it from here?