Finding weight in Megagrams (Mg) given circumference and density 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 $2.70\ \mathrm{g} \cdot \mathrm{cm}^{-3}$. 
 A: HINT


*

*Given a circumference of the sphere, find its volume $V$.

*With volume and density, find the mass.

*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?
