In astronomy and planetary science we often need to consider areas and volumes. We might be interested in the light hitting the surface of a planet or the distribution of craters over the surface - how many craters per unit area, for example. Then, there are processes that depend on the volume - the amount of heat generated by radioactivity inside, for example.

First, let's start with something easy - squares and cubes:

Consider a square that is some distance "D" on a side - the area
is D x D= D^{2} (with the appropriate units - meters^{2} or
km^{2} or feet^{2} (if you cannot handle metric))

Now, double the size of the square so that each side is 2D - the area is then
2D x 2D = (2D)^{2} = 4 x D^{2} (with same appropriate units).

The** ratio of sizes** of the squares is (Size Big)/(Size Small) = (2D)/(D)
= 2.

The **ratio of the areas** of the squares is (Area Big)/(Area Small) = (2D)^{2}
/ D^{2} = 2^{2} = 4.

Next , let's look at **volume.** Consider a volume of a cube that is D on
a side. The volume of the cube is D x D x D = D^{3} (with the appropriate
units - meters^{3} or km^{3} or feet^{3} (if you cannot
handle metric))

Again, consider a cube that is twice the size - 2D on a side. The volume of
the cube is then (2D)^{3} = 2^{3} D^{3} = 8 D^{3}.

Taking** ratios **we can summarize as follow (labelling the square or cube
that is of size of D as "small" and that the one that is twice the
size, 2D, as "big"):

(Size Big)/(Size Small) = (2D)/ D = 2

(Area Big)/(Area Small) = (2D)^{2}/ D^{2} = 2^{2} =
2 x 2 = 4 = [(Size Big)/(Size Small)]^{2}

(Volume Big)/ (Volume Small) = (2D)^{3}/ D^{3} = 2^{3}
= 2 x 2 x 2 = 8 = [(Size Big)/(Size Small)]^{3}

OK - but planets and stars are spheres not cubes. But let's begin with a circle
of radius R. The area =p R^{2}.

If we double the size of the circle - double the radius (and/or double the
diameter) - Area =p (2R)^{2} = =p
2^{2} R^{2} = 4 R^{2}

Moving from 2 dimensions to 3 dimensions, let's thing about spheres. Remember
the formula for the area of a sphere of radius R? It is 4p R^{2}

Let's go for a sphere twice the size - radius = 2D. The area will then be 4p
(2R)^{2} = 4p 2^{2}R^{2}
= 16 pR^{2}

Volume? The volume of the smaller sphere is 4/3 pR^{3}

Double the radius of the sphere and we have a volume of 4/3 p(2R)^{3}=
4/3 p2^{3}R^{3}=
4/3 x 8 pR^{3}=
32/3 pR^{3}

Taking** ratios** again, we get

(Area Big Circle)/ (Area Small Circle) = (p (2R)^{2})
/ ( p R^{2}) = 2^{2}= 4

(Area Big Sphere)/ (Area Small Sphere) = (4p (2R)^{2})
/ ( 4p R^{2}) = 2^{2}= 4

(Volume Big Sphere)/ (Volume Small Sphere) = (4/3 p
(2R)^{3}) / ( 4/3 p R^{3}) = 2^{3}=
8

OK - so, you have mastered squares, cubes, circles and spheres - let's move
on to **frogs**. What's the formula for the area of a frog?

Area of a frog = "something" x H^{2}

What if the frog is twice as big? Area of big frog = "something" x (2H)^{2}=
"something" x 2^{2}H^{2} = "something" x 4 H^{2}

And, the volume of an frog?

Volume of a frog = "something else" x H^{3}

But, while we might have no idea about the "something" or the "something else", look what happens when we take ratios:

(Area Big Frog)/ (Area Small Frog) = (something x (2H)^{2}) / (something
x H^{2}) = 2^{2}= 4

(Volume Big Frog)/ (Volume Small Frog) = (something else (2H)^{3})
/ (something else H^{3}) = 2^{3}= 8

So, What do we conclude from all this? That if you are not interested in the actual formula for an area of volume of an object, whatever its shape, but interested in what happens when you consider an object of the same shape but some factor bigger or smaller, then all you need to think about is the ratio.

Furthermore, if you look at the above, you will see a pattern. You will see
that we can generalize. There are trends. You will notice that the ratio of
areas is always (Area Big)/ (Area Small) = [(Size Big)]/(Size Small)]^{2}.
And that

(Volume Big)/(Volume Small) = [(Size Big)]/(Size Small)]^{3}.

In the above cases we always doubled the size of the object (to keep it simple)
- but the math would be the same if we multiplied by 3 or 8 or 42.456788. The
area would then increase by a factor of 3^{2} or 8^{2} or 42.456788^{2} and the volume by
a factor of 3^{3} or 8^{3} or 42.456788^{3}.

We can say

**Area** is** proportional** to **Size ^{2}**

**Volume** is **proportional** to **Size ^{3}**

This works for squares, this works spheres, this works for frogs, elephants, you name it. Furthermore - it means that if you are just looking at the same shape and scaling bigger or smller, you never need to plug p into your calculator. In fact, you can probably do all the math in your head. It also means that you are less likely to hit the wrong key on your calculator.

For example, **the marbles problem**. Jupiter is 10 times the size of Earth.
So, if we model Jupiter with a volley ball and Earth with a marble, how many
marbles fit into the Jupiter-volley ball? The ratio of volumes is just

(Volume of Jupiter)/(Volume of Earth) = [(radius of Jupiter)/(radius of earth)] ^{3}
= (10)^{3} = 1000

BUT.... IN REALITY..... there will be space between the marbles so that we will probably not be able to get as much as 1000 marbles into the Jupiter ball - there will be gaps. How much? You guess! 10%? 20%? 30%? 50%....? what do you think?