Purpose:

Plants are living materials and are completely dependent upon solar

radiation for their existence. The amount of solar radiation the earth receives is

finite. There is an upper limit to the amount of plant material that can be raised

and hence an upper limit to our human population. The purpose of this activity

was to generate one approach to approximate a limit to population growth.

Summary:

How can so many people live on the Earth at one time? Is there a limit to

the carrying capacity of the Earth – a limit to growth? Since the size of the Earth

is finite, the answer is affirmative, but then how does one measure or

approximate this limit to growth? There are a variety of approaches. Certainly

there is a limit to the number of people that could physically fit on Earth; however

before one considers a situation where the Earth is “covered with people” – a

condition where life would be impossible – realize that the availability of food is a

limiting factor. Hence if the limit to the Earth’s food production could be

established, the limit to the Earth’s carrying capacity could be approximated.

The amount of food produced on Earth depends directly on the sun’s

energy. The value for the sun’s power radiated per unit area is well established

and known as the solar input or the solar constant. The precise numerical value

varies with the seasons, time of day, weather, and latitude, and is expressed in

units of power / area such as watts/m2 or joules/s/m2.

Data and Results:

Table 1Elapsed Time

(min)Temperature

(0C)0 101 10.12 10.33 114 11.25 11.76 127 12.28 12.69 12.8

10 1311 13.512 13.813 14.814 14.115 14.416 14.817 1518 15.219 15.320 15.6

Solve for the Value of the Solar Input

The solar input is the amount of solar power a given cross sectional area

receives from the sun when the sun is directly over that cross section. To

determine this value it is necessary to establish the rate of heat absorption of the

water in the tray. We assume that the water in the styrofoam tray absorbs all the

solar radiation which strikes it and that the tray is a perfect insulator. Convection

losses are minimized owing to the glass cover plate and the black paint reduces

radiation losses. The rate of heat absorption by the tray should equal the rate of

heat generated by the sun on that specific area. We have established the rate of

energy generated on an area at right angles to the sun.

Relationship between normal and horizontal dimension.

The cross sectional area of the styrofoam tray perpendicular to the sun’s

rays, the effective area of tray is the product of the area of the tray and the sine

of the sun’s elevation angle. (Fig 3 below.)

Determination of Sun’s Elevation Angle:

Tan = Length of stick / Length of shadow =

= 27 0

Sin = 0.454

Figure 3

In this experiment, we have assumed that the styrofoam tray is a perfect

insulator, i.e., it gains no heat from its surroundings nor loses any heat to them.

This assumption is quite reasonable. Obviously, any heat added to or taken

away from the water by its surroundings will give errors. These errors were

A

x

Length of

Stick

Length of Shadow

minimized by determining the rate of temperature rising when the water and the

surrounding air are at the same temperature.

Computation for the solar input consists of solving the equation which

states that the heat gained by the system is equal to that received from the sun.

Data Record

Date 11/05/2007 Time 6:35 p.m.

Air temperature = 13 oC

Volume of water = 600 mL

Weight of water = 600 grams

Area of tray = length x width

Area of tray = 21.5 cm x 21.5 cm = 462.25 cm2

Length of meter shadow = 196 cm

tan = length of stick 100 cm / length of shadow 196 cm

tan = 100 cm / length of shadow 196 cm

tan = 0.51

= 27 o

Sin = 0.454

( is the Sun’s elevation angle)

Calculating the Solar Input

The quantity of heat absorbed by the water per second is easily calculated

using ; that is the quantity of heat per second is given by the product

of the mass of water times its specific heat times its temperature change per

second. The result was in calories per second.

A. Rate of heat absorbed by H20 Sp. Heat of H20 rate of temperature change

Rate of temperature change per second is obtained graphically and is the

slope of the graph in the region about ambient temperature.

Now recall that 4.18 Joules = 1 calorie, so

So the heat gained by the water = 1.153x10 1 watts

The heat gained by the water per second is equal to the heat received from

the sun per second.

B. The area on which it falls is the effective area of the tray perpendicular to the sun’s rays.

Effective Area = tray area sin

= 4.623x102 cm2 4.5x10 -1 = 2.08x102 cm2

C. Heat RECEIVED from sun/sec per unit area:

The solar input S is the solar energy received per second per unit area. This

quantity can be expressed in watts / second.

S= (Heat GAINED by water/sec)/ Effective area

S = 1.153x10 1 watts /2.08x102 cm2

Now 100cm = 1 m, 1 cm = 1/100mTherefore 1 cm2 = (1/10000) m2 and

S = 5.5x10 -2 watts / (1/10000) m2

S = 5.5x10 -2 10000 watts/ m2

S = 5.522x10 2 watts / m2 (Solar Constant)

The accepted value of the solar input is about 1000 watt/ m2 on a clear day

at sea level. Above the earth’s atmosphere the solar input is higher since there

are no reflecting particles; at this point in space the solar input is approximately

1350 watt/ m2 (To put this number in perspective, an area one square mile would

receive 2.6 billion watts of power-enough energy to power a small city providing

efficient energy conversion occurred.)

Part 2: Establishing a limit to Growth for the Earth’s Population

For the following determinations I have used the sea level value of S =

1000 watt/ m2 as the solar input. Most of the world’s food growing regions are

quite close to sea level.

Our next problem was to estimate the total carrying capacity of the earth.

How large a population can the earth sustain based on the total amount of

calories per day available as food? To do this we have computed the total

calories per day available to people or the total world food production in terms of

available calories per day. We have divided this by the number of calories per

day that a person needs, it has given us the number of people the earth can

sustain each day.

A. Determination of the Total World Food Production in calories/day.

First let’s compute the total amount of power received from the sun each

day in watts. Given radius of the Earth = 6.38x106 meters, and the solar input

S = 1000 watts m2. Then since the cross sectional area of the earth is A = ;

Total Power = Solar INPUT Cross Sectional Area of the Earth

Total Power = 1000 w/m2 m2 = 1.27876644x10 19 watts. (Received by the Earth)

The photosynthesis process is not 100% efficient. Actually only

approximately 0.0757% of the solar energy that reaches the Earth is captured by

the chlorophyll of plants and used in the photosynthesis process.

Total Potential Photosynthesis Power = 1.27876644x10 19 watts

TPPP = 9.680261951x10 15 watts

Only 10% of the total photosynthesis power is ultimately converted to food for

humans.

Solar Power Ultimately Converted to Food = .10 TPPP

= .10 9.680261951x10 15 = 9.680261951x1014 watts

To obtain the caloric equivalent of the food we used the fact that 1 watt = 0.24

cal/sec. Amount of food/day = Solar Power Ultimately Converted to Food 0.24

cal/sec.

9.680261951x10 14 watts (0.24 cal/sec watt)= 2.323262868x10 14 cal/day.

Recall that there are 86,400 seconds in one day (i.e. 1 day = 60 s/min x60 min/hr

x 24 hr = 86,400 s)

So, Solar Power Ultimately Converted to Food/day = 2.323262868x10 14 cal/

(1/86400) day

= 2.323262868x10 14 86400 cal/day

World Food Production / Day (WFP) = 2.007299118x10 19

(Note: the dietitian’s Calorie has a capital C and is equal to 1000 normal calories)

B. Determination of average number of calories required for one person:

An average American with a healthy appetite has a daily food intake of

approximately 3,240 dieting Calories (plant and animal food stuff). However, this

figure rises to approximately 11,000 primary Calories/day when the conversions

of plant to animal food is considered since five or more plant calories are

required for every calorie generated of animal food.

The process of nutrition however is much more complicated than mere

calories since the right kind of calories and their distribution in terms of

carbohydrates, protein, and fat are important. With this restriction in mind, and

an awareness of the nature of this approximation, we can estimate the Earth’s

carrying capacity.

Food requirement for one well fed person.= 11,000 primary Calories/day= 11,000,000 calories/day= 11x106 cal/day for one person = 1.1x107

C. Earth’s Carrying Capacity:

Carrying Capacity = (World food production/day)/ (food requirement of well

fed person/day)

Carrying Capacity of the Earth =

Carrying Capacity of the earth = 1.822x1010 people. 18,220,000,000 people.

For Consideration and Discussion / Extra 10 points :

The equation below

T2 = 70/R,

Allows one to compute either the doubling time, T2, (the time it takes for

something to double in size) or R (the percentage rate of growth of a particular

process) when T2 or R is supplied

The Earth’s population growth is approximately 2% per year. Calculate the

doubling time for the Earth’s population growth using the equation above.

Answer:

Using the equation above of T2 = 70/R and the R of 2% per year I have found

that the earths doubling time will be every 35 years. Or .

Look up the Earth’s population:

Answer

The earth’s current population according to Live Science’s web site

livescience.com is about 6.5 billion humans.

1. Based on your calculation of the Earth’s carrying capacity, the populations doubling time, and the Earths current population, estimate how many doubling periods it will take to reach or exceed the Earth’s carrying capacity. When will this occur?

Answer:

The earth’s carrying capacity is 1.822x1010 or 18,220,000,000 people. The

doubling time is 35 years and the current population is 6.500x109 or

6,500,000,000. It will take 98.5 years approximately to reach and exceed the

earth’s carrying capacity. I have figured this by first dividing the current

population by the earth’s doubling time to get 1.857142857x108 people per

year over the 35 year span. I then realized that the current population

multiplied twice gives you 1.3x1010 or 13,000,000,000 million people. I began

to add the individual years on to this number until I had received a number

just greater then the earth’s carrying capacity.

Step 1)

Step 2) 6.500x109 x 2 = 1.3x1010

Step 3.) I gradually multiplied the individual years to get 28.5 years which gave me 1.857142857x108 x 28.5 = 5.292857142x109

Step 4.) I then added 1.3x1010 and 5.292857142x109 to get =1.829285714x1010.

Step 5.) I then figured out the years 35+35+28.5 = 9.85x101 or 98.5 years.

This will occur mid year 2105 or June of 2105.

2. What errors are associated with this approximation of the Earth’s carrying capacity?

Answer:

The errors that have to be taken into consideration are weight of structures

and other substances like cars that are on the earth which add to the weight, they

also take up space. We also need to factor in the amount of population growth

percentage per year. The 2% point is not an exact figure one feels and every

year the population may either exceed what is calculated or not meet the 2 %

point. These are just some of the errors that have to be determined.

3. How can the carrying capacity of the Earth be altered and to what degree?

Answer:

The carrying capacity of the Earth can be altered by the amount of food

intake that we each have. If we were to portion the amount of calories per day

each person received then we would be lighter people and in better shape. Most

of the earth’s population is overweight; we could lessen this amount and provide

extra room for more people, but not necessarily change the earth’s carrying

capacity.