gravitation – Mean Green Math

This is one of most creative diagrams that I’ve ever seen: the depth of various solar system gravity wells. A large version of this image can be accessed at http://xkcd.com/681_large/.

From the fine print:

Each well is scaled so that rising out of a physical well of that depth — in constant Earth surface gravity — would take the same energy as escaping from that planet’s gravity in reality.

Depth = $\displaystyle \frac{G M}{g r}$

It takes the same amount of energy to launch something on an escape trajectory away from Earth as it would to launch is 6,000 km upward under a constant $9.81 \hbox{m}/\hbox{s}^2$ Earth gravity. Hence, Earth’s well is 6,000 km deep.

Here’s some more details about the above formula.

Step 1. The escape velocity from the surface of a spherical planet is

$v = \displaystyle \sqrt{ \frac{2GM}{r} }$ ,

where $G$ is the universal gravitational constant, $M$ is the mass of the planet, and $r$ is the radius of the planet. Therefore, the kinetic energy needed for a rocket with mass $m$ to achieve this velocity is

$E = \displaystyle \frac{1}{2} m v^2 = \frac{GMm}{r}$

Step 2. Suppose that a rocket moves at constant velocity upward near the surface of the earth. Then the force exerted by the rocket exactly cancel the force of gravity, so that

$F = mg$ ,

where $g$ is the acceleration due to gravity near Earth’s surface. Also, work equals force times distance. Therefore, if the rocket travels a distance $d$ against this (hypothetically) constant gravity, then

$E = mgd$

The depth formula used in the comic is then found by equating these two expressions and solving for $d$ .

The principle of diminishing return states that as you continue to increase the amount of stress in your training, you get less benefit from the increase. This is why beginning runners make vast improvements in their fitness and elite runners don’t.

J. Daniels, Daniels’ Running Formula (second edition), p. 13

In February 2013, I began a serious (for me) exercise program so that I could start running 5K races. On March 19, I was able to cover 5K for the first time by alternating a minute of jogging with a minute of walking. My time was 36 minutes flat. Three days later, on March 22, my time was 34:38 by jogging a little more and walking a little less. During that March 22 run, I started thinking about how I could quantify this improvement.

On March 19, my rate of speed was

$\displaystyle \frac{5000 \hbox{~m}}{36 \hbox{~min}} = \frac{5000 \hbox{~m}}{36 \hbox{~min}} \times \frac{1 \hbox{~min}}{60 \hbox{~sec}} \approx 2.3148 \hbox{~m/s}$ .

On March 22, my rate of speed was

$\displaystyle \frac{5000 \hbox{~m}}{34 \times 60 + 38 \hbox{~sec}} \approx 2.4062\hbox{~m/s}$ .

That’s a change of $2.4062 - 2.3148 \approx 0.0913 \hbox{~m/s}$ over 3 days (accounting for roundoff error in the last decimal place), and so the average rate of change is

$\displaystyle \frac{0.0913 \hbox{~m/s}}{3 \hbox{~d}} = \frac{0.0913 \hbox{~m/s}}{3 \hbox{~d}} \times \frac{1 \hbox{~d}}{24 \times 60 \times 60 \hbox{~sec}} \approx 3.524 \times 10^{-7} \hbox{~m/s}^2$ .

By way of comparison, imagine a keg of beer floating in space. The specifications of beer kegs vary from country to country, but I’ll use the U.S. convention that the mass is 72.8 kg and its height is 23.3 inches = 59.182 cm. Also, for ease of calculation, let’s assume that the keg of beer is a uniformly dense sphere with radius 59.182/2 = 29.591 cm. Under this assumption, the acceleration due to gravity near the surface of the sphere is the same as the acceleration 29.591 cm away from a point-mass of 72.8 kg. Using Newton’s Second Law and the Law of Universal Gravitation, we can solve for the acceleration:

$ma = \displaystyle \frac{GMm}{r^2}$ ,

where $G \approx 6.67384 \times 10^{-11} \hbox{~m}^3/\hbox{kg} \cdot \hbox{s}^2$ is the gravitational constant, $M = 72.8 \hbox{~kg}$ is the mass of the beer keg, $r = 0.29591 \hbox{~m}$ is the distance of a particle from the center of the beer keg, $m$ is the mass of the particle, and $a$ is the acceleration of the particle. Solving for $a$ , we find

$a = \displaystyle \frac{6.67384 \times 10^{-11} \times 72.8}{0.29591^2} \approx 5.5487 \times 10^{-8} \hbox{~m/s}^2$ .

Since this is only an approximation based on a hypothetical spherical keg of beer, let’s round off and define 1 beerkeg of acceleration to be equal to $5.5 \times 10^{-8} \hbox{~m/s}^2$ .

With this new unit, my improvement in speed from March 19 to March 22 can be quantified as

$\displaystyle 3.524 \times 10^{-7} \hbox{~m/s}^2 \times \frac{1 \hbox{~beerkeg}}{5.5 \times 10^{-8} \hbox{~m/s}^2} \approx 6.35 \hbox{~beerkegs}$ .

I love physics: improvements in physical fitness can be measured in kegs of beer.

I chose the beerkeg as the unit of measurement mostly for comedic effect (I’m personally a teetotaler). If the reader desires to present a non-alcoholic version of this calculation to students, I’m sure that coolers of Gatorade would fit the bill quite nicely.

For what it’s worth, at the time of this writing (June 7), my personal record for a 5K is 26:58, and I’m trying hard to get down to 25 minutes. Alas, my current improvements in fitness have definitely witnessed the law of diminishing return and is probably best measured in millibeerkegs.

Tag: gravitation

Defining Gravity

Sphere

Gravity wells

Fun with Dimensional Analysis