Even now, there are those who claim that the long-jump record of 8.9 meters that Bob Beamon set in 1968 was so crazy awesome because he accomplished it in Mexico City, which is almost 8,000 feet above sea level. The argument is that the air is thinner, and so there is less air resistance, and Mexico City is further from the center of the earth, and so the gravitational forces are smaller. Does any of this have any impact? And if so, does it really matter?
Gravity
First, let’s look at gravity. On the surface of the Earth, the usual model for gravitational force is the object’s mass times the gravitational field (represented by g) where g is about 9.8 Newtons per kilogram. So, a 1 kg object would have a gravitational force of 9.8 Newtons (directed down).
However, this model doesn’t work if you get too far from the surface. Really, the gravitational force is an interaction between two objects with mass, and the magnitude of this force decreases as the two objects get farther away. For an object interacting with Earth, the magnitude could be written as:
In this expression, G is the gravitational constant (not to be confused with “g”). ME an RE are the mass and radius of the Earth and h is the height above the surface. If you put in a height of zero meters as well as the mass and radius of the Earth, you would find:
Which gets you back to the gravitational force being “mg.” Also, since the radius of the Earth is around 6,000 km, a height of 100 meters above the surface doesn’t change the force too much. But what about a place like Mexico City with an elevation of 2,240 meters above sea level? With that value for h, an object would have weight that is 99.93% the weight of the object at sea level. Not a big difference, no. But is it a big enough difference to mean a new world-record long jump?
More Than Gravity
The above comparison of weights at sea level and at elevation would be valid if that were all that mattered. In terms of the apparent gravitational force, there are two other issues. First, the Earth is not a uniform sphere with uniform density. If you are near a mountain, the mass of that mountain can affect the gravitational field in the area - even if you are at sea level.
The second consideration is the rotation of the Earth. The closer a location is to the equator, the faster that location has to move in a circle as the Earth rotates each day. Mexico City is about 19.5 degrees above the equator, so it must move fairly fast. Of course, if you move in a circle you aren’t exactly in a non-accelerating reference frame. In order to treat it like a stationary frame (which is what it seems like), you would have to add a fake force called a centrifugal force pointing away from the axis of rotation. The combination of this fake force and the actual gravitational force would be the apparent weight.
If Mexico City were at sea level, this rotational motion would cause the apparent weight to be 99.69% the value if the Earth were not rotating (like at the North Pole). Putting both the gravitational and rotational effects together, the apparent weight at the elevation of Mexico City would be 99.62% the expected value. So, not much. In fact, if you compare the apparent weight at the same location on the Earth but at sea level, Mexico City has a gravitational field value of just 99.92% smaller.
In other words, there’s no discernible difference in the gravitational pull.
OK, Fine. What About the Lower-Density Air?
First, let’s think about a person moving through the air during the long jump. If we are going to consider small variations in the gravitational force during the jump, we also should consider other small forces. One such small force (small for this speed) would be air resistance. Typically, the magnitude of the air resistance can be modeled as:
In this model, the A and C parameters are the shape and size of the object. The important variable for this discussion is ρ, the density of the air. As you move higher in elevation, air density decreases. Air density isn’t the simplest thing to model. It depends on the pressure and the temperature (both of which change with weather). However, this is an expression for the density of air that will be close enough.
With this density model, I find that at sea level the density of air is about 1.22 kg/m3 compared to 0.98 kg/m3 at an elevation of 2240 meters. Would this decrease in density have as much of an impact as the decrease in the gravitational force?
Numerical Modeling
The motion of an object moving through the air with air resistance isn’t really a simple problem. Why? Without the air resistance, the acceleration of the object would be constant. With constant acceleration, the following kinematic equations are valid:
But with air resistance, there is now a force that depends upon the velocity of the object. Of course, the velocity depends upon the acceleration so perhaps you can see how this could cause some problems.
There is a solution. The answer is to create a numerical calculation of the motion. An analytic solution (like the case with no air resistance) is solvable with some algebraic manipulations - or sometimes with calculus. The analytic solution is what you would typically see in an introductory physics textbook. For the numerical calculation, you need to break the problem into a bunch of small steps in time. For each step, you can assume that the forces (and thus the acceleration) are constant. This means the typical constant acceleration solutions will work.
The smaller the time steps the problem is broken into, the better the solution. Of course, if you break a long jump into time steps 1 nanosecond in length, you are going to have to do 109 calculations for a 1 second jump. Even a time step of 0.01 seconds would require 100 steps. Even this is too many for a person to reasonably do. The best bet is to use a computer. They rarely complain.
Modeling a Long Jump
In order to see how much changes in gravity and the density of air affect a jumper, we need to start with a basic model. If we look at Beamon’s record-setting jump, we can get some information about the initial velocity assuming there was no air resistance. From the video (and by counting frames), Beamon was aloft 0.93 seconds. Since he traveled 8.39 meters horizontally, this would put his horizontal velocity at 10.1 m/s (22.6 mph).
It will also be useful to know the initial vertical velocity (y-velocity). I can use the trick that the initial vertical velocity has the same magnitude (but opposite direction) as the final velocity. Now, I can use the time he was in the air and the following kinematic equation:
This gives an initial y-velocity of about 4.5 m/s. Now that I have both the starting x- and y-velocities, I can use these as my initial values in my numerical model.
Here is a plot showing three different cases of this model. The first case is at sea level (so the acceleration is 9.8 m/s2) with a typical density of air. The second case shows a trajectory at sea level with no air resistance at all. The third case is for a jump in Mexico City with a lower apparent weight and a lower density of air.
There’s not much of a difference, but there is a difference. The model with air resistance and at sea level gives a jump distance of 8.89 meters compared to Mexico City (with air) at 8.96 meters. That is just 7 cm further - but every little bit counts. But in Beamon’s case, it wouldn’t have mattered if he made the jump at sea level or at 5,000 feet. He beat the previous record by an astounding 55 centimeters. That’s truly an incredible feat.
__Update (11:34 AM 8/4/12) __The original graph showing the three cases for a long jump (No Air at Sea level, Air at Sea Level, and Mexico City) had the wrong labels on the axes. I have replaced the graph with the correct axes labels.
