We've all seen majestic pictures of big rockets blasting off from the earch, burning millions of dollars a second, and carrying interesting payloads into space, such as experiments on how spiders build webs in space. Fascinating.
But there are some general things about this picture that might strike you as odd. One is that payloads are pretty small. Also the size of these big rockets is absolutely enormous. Is there a reason for this? Yes I know what you're thinking, it's boys playing with their toys. But no, that's not the only reason. They're big because they have to be. We're now in a poistion to understand rockets, and you'll see why this happens.
What is a rocket? It's this thing that spews out lots of fuel at high
velocity. The expulsion of fuel speeds up the rocket. The more fuel
expelled, the faster the thing goes. We want to know how fast the rocket
will be going after total mass of the rocket has been reduced from it's
initial mass 
to its final mass 
. This will cleary have to
depend on the velocity that fuel is expelled at. Let's call that U.
Let's try to understand this with a simple low tech example. A small child called Cindy sits on top of her little red go-cart loaded up with a huge pile of bricks. She's a strong girl and can hurl a large number of these out in one go.
Suppose she starts at rest with 128 bricks each weighing 1kg. We'll
ignore Cindy and the go cart's mass. She hurls out 64 of these
bricks, with velocity of 
. What will be the final velocity of the
cart?
Well the initial momentum was zero, so assuming a frictionless go-cart, the final momentum must be the same. So if half the brickes are travelling to the left at 1m/s the other half must be travelling to the right at the same speed, 1m/s.
OK, now they're 64 bricks left in the cart. Cindy hurls out half of these, again with the same velocity relative to her. Now what will the final velocity of the cart be? To figure this out, go to the reference frame of the cart, which was initially travelling at 1m/s. In this frame, it looks like the initial problem, where the cart was stationary, but with 64 bricks inside it. After 32 bricks are hurled, by the same reasoning as before, the cart will be going an additional 1m/s, so that relative to the ground the cart is going 2m/s.
Now let's ask what happens when Cindy again hurls out half of these 32 bricks, at the same velocity she did before. Going to to a reference frame of 2m/s. we see as we did before, that the cart will be going 1m/s faster than before.
You get the idea. Every time half the mass is thrown out, the velocity
increases by 1m/s. So if the mass decreases by 
, the velocity
will increase by 
. In this case there is a logarimthic
dependence of mass on velocity. You can check that the final velocity
Here U is 1m/s, and 
is the ratio of initial
to final masses. What does this say? It's pretty hard
to speed up a rocket. If you want the speed to go up
by U, you have to expell half half your mass.
We can rewrite the above equation as
So if you have a payload of mass , the amount of mass you have
to start with, depends exponentially on the final velocity.
Exponentials are very rapidly increasing functions. So you're
initial mass normally has to be very large.
Now that we understand this simplified problem, we can ask what happens in a real rocket. Now we don't have a little girl hurling bricks, but a rocket called Cindy. Cindy expells fuel at a velocity -U (to the left), but does so continuously, not in the punctuated way we just analyzed. So if initially, we choose a reference frame where the rocket is at rest. What happens after it expells a little fuel of mass dM? The initial mass of the rocket is M. After expulsion, the mass will have gone down to M-dM. Let's call the increase in velocity dv. So conservation of momentum says that
This says that the velocty after this is
Since dM << M (formally its an infinitessimal, we can safely ignore it. So we get the equation
This is the increase in velocity after a very short time. If we now add up all these velocity increases, we'd formally be integrating the left hand side:
which says that
This is almost the same form as our previous analysis except now we get a logarithm base e instead of base 2. Qualitatively it acts the same way.
