Let's determine the ratio of the volumes $\frac{V_2}{V_1}$ given that the radius of sphere 2 ($r_2$) is 9 times the radius of sphere 1 ($r_1$).

Volume of a Sphere

The volume $V$ of a sphere is given by:

$V = \frac{4}{3}\pi r^3$

Where $r$ is the radius of the sphere.

Ratio of the Volumes

For sphere 1, the volume is:

$V_1 = \frac{4}{3}\pi r_1^3$

For sphere 2, the volume is:

$V_2 = \frac{4}{3}\pi r_2^3$

Given that $r_2 = 9r_1$, substitute $r_2$ into the equation for $V_2$:

$V_2 = \frac{4}{3}\pi (9r_1)^3 = \frac{4}{3}\pi \times 729r_1^3$

Now, the ratio of the volumes $\frac{V_2}{V_1}$ is:

$\frac{V_2}{V_1} = \frac{\frac{4}{3}\pi \times 729r_1^3}{\frac{4}{3}\pi r_1^3} = 729$

So, the ratio of the volumes $\frac{V_2}{V_1}$ is $\boxed{729}$.

Do you have any other questions or would you like more details?

Here are five related questions:

1. What is the relationship between the surface area and volume ratios for two spheres with different radii?
2. How does increasing the radius of a sphere by a factor of $k$ affect its volume?
3. What is the ratio of the surface area to volume for a single sphere?
4. If the radius of a sphere triples, what happens to its volume?
5. How does the volume of a sphere change if its diameter is doubled?

Tip: The volume of a sphere increases with the cube of the radius, meaning even a small change in radius leads to a significant change in volume.