Why are so many rocket launches done from Cape Canaveral in Florida? There are two main reasons. The first is energy.
Suppose I want to put an object in low-Earth orbit (about 300 km above the surface). There are two things to consider. How fast does it have to be in orbit so that it doesn't crash? How much energy would it take to get it there?
Here is the force diagram for an object in orbit around the Earth (not to scale):
The only force on the space object (rocket or whatever) is the gravitational force from the Earth. It can be expressed as a force with a magnitude:
For this case, the force is perpendicular to the velocity and only changes the direction. For an object moving in a circle, it has an acceleration towards the center of the circle and a magnitude of:
Putting these two together with Newton's Second law (everything is in the same direction - the r-hat direction and I will use positive to mean away from the center of the circle):
Note that r in this case is the distance from the center of the Earth to the object in orbit.
First, let me assume the Earth is not rotating (or that we are launching rockets from the North pole). This means that the rocket (or object or whatever) starts from rest. If I apply the work-energy principle and include the rocket plus the Earth as my system, then I get:
Remember that the gravitational potential energy has a negative sign in it. Also, (maybe it is obvious) RE is the radius of the Earth (where the object starts). So that work is the amount of energy you would need to put something in orbit.
But wait. There's more. What if the object is already moving? In this case the initial kinetic energy is not zero. How much kinetic energy would it have? Well, it depends on where it is on the Earth. You know the Earth spins, right? It's rotational rate is 1rpd (revolution per day). Actually, that isn't true. 24 hours is the time for the Sun to get back in the same position. It actually takes the Earth a little bit less time to rotate (Sidereal vs. solar day). Let me say the Earth has an angular velocity of ω. And suppose you are at a location on the Earth like this:
If you are at a latitude of θ degrees above the equator, then you are moving around the Earth in a circle of radius REcosθ. This means that the object at that location would have a speed of:
This changes the amount of work needed to get the thing into orbit since the object will not be starting with zero kinetic energy. The new work-energy equation will be:
Ok. How about some graphs. Here is a plot of the energy needed to put something in low-Earth orbit (altitude of about 300 km) as a function of latitude. Oh, actually I will plot the energy per kg of mass to put in orbit.
Just a couple of values. The energy per kg at the equator is 3.25 x 107 J/kg compared to 3.26 x 107 J/kg at the North Pole. What about Cape Canaveral at a latitude of 28.5 degrees? The energy per kg for a launch from there would be 3.252 x 107 J/kg. This is 0.3% less than the energy per kg from the North Pole. Big deal, right?
What about cost? How much does it cost to get something in orbit? According to Wikipedia, the Space Shuttle has a payload price of $260/kg (in 1972 dollars). Odd that I don't find that value in today's dollars. Oh well. So if you have a payload of 20,000 kg that would be a payload cost of 5.2 million 1972 dollars. Using this conversion that would be 27.3 million 2010 dollars (and $1,366 per kg). Still seems low. But if you moved that launch to the North pole, it would cost about $800,000 more.