Very Interesting by Discover

2 of 6 Very Interesting If we consider that the Earth is the third planet from the Sun, and 3 is a good approximation to pi, and as well that 0.3 is the earth albedo saying that the earth reflects 30% of incident light back into space, then letâ&#x20AC;&#x2122;s run my program for these values and we get near fahrenheit-centigrade equivalence: jharvard@appliance (~): cd Dropbox/descubrir jharvard@appliance (~/Dropbox/descubrir): ./stelr We determine the surface temperature of a planet. What is the luminosity of the star in solar luminosities? 3 What is the albedo of the planet (0-1)?.3 What is the distance from the star in AU? 3 The surface temperature of the planet is: 230.783005 K That is -42.216995 C, or -43.990593 F The luminosity of the star in joules per second is: 117.00E25 The habitable zone of the star in AU is: 1.732051 Flux at planet is 0.34 times that at earth. That is 459.87 Watts per square meter jharvard@appliance (~/Dropbox/descubrir): Interesting because I published in my book titled Volume 03 by Ian Beardsley: We run the program for the golden ratio and its conjugate and find we get near equivalence for the fahrenheit-centigrade scales, as well as near to the golden ratio conjugate of the flux at earth for the Sun. This is strange because that means our scales for temperature, which are fahrenheit and centigrade, as well as our scales for energy, distance, and time (joules, meters, seconds) have been defined such that they are in jive with stellar dynamics and climate dynamics, even though they evolved through a random and complex history. That is a mystery: Case 1 jharvard@appliance (~/Dropbox/descubrir): ./stelr We determine the surface temperature of a planet. What is the luminosity of the star in solar luminosities? 1.618 What is the albedo of the planet (0-1)?0.618 What is the distance from the star in AU? 1.618 The surface temperature of the planet is: 231.462616 K That is -41.537384 C, or -42.767292 F The luminosity of the star in joules per second is: 63.10E25

3 of 6 The habitable zone of the star in AU is: 1.272006 Flux at planet is 0.62 times that at earth. That is 852.66 Watts per square meter jharvard@appliance (~/Dropbox/descubrir): The foot-pound system was not derived from any relationship to nature that we know of. The Metric system was: one gram is the mass of a cube of water at STP one centimeter on each side. A centimeter is a hundredth of a meter, and a meter is a thousandth of a kilometer. One kilometer is one ten thousandth of the distance from the pole to the equator. Centigrade is derived such that there are 100 units between freezing and boiling points of water, freezing being zero and boiling, 100. Fahrenheit is not derived from anything natural: It stars at freezing of water at 32 degrees and who knows why? We do know the foot-pound system has earth gravity at 32 feet per second per second. Let us find the temperature where centigrade and Fahrenheit are the same: F=(9/5)C + 32 F=C C=(9/5)C+32 (25/25)C-(45/25)C=32 (-20/25)C=32 C=-32(25)/20=-800/20=-40 Degrees C -40 degrees C = -40 degrees F Where my program is: The Program In C #include<stdio.h> #include<math.h> int main(void) { float s, a, l, b, r, AU, N, root, number, answer, C, F; printf("We determine the surface temperature of a planet.\n"); printf("What is the luminosity of the star in solar luminosities? "); scanf("%f", &s); printf("What is the albedo of the planet (0-1)?" ); scanf("%f", &a); printf("What is the distance from the star in AU? "); scanf("%f", &AU);

4 of 6 r=1.5E11*AU; l=3.9E26*s; b=l/(4*3.141*r*r); N=(1-a)*b/(4*(5.67E-8)); root=sqrt(N); number=sqrt(root); answer=1.189*(number); printf("The surface temperature of the planet is: %f K\n", answer); C=answer-273; F=(C*1.8)+32; printf("That is %f C, or %f F", C, F); printf("\n"); float joules; joules=(3.9E26*s); printf("The luminosity of the star in joules per second is: %.2fE25\n", joules/1E25); float HZ; HZ=sqrt(joules/3.9E26); printf("The habitable zone of the star in AU is: %f\n", HZ); printf("Flux at planet is %.2f times that at earth.\n", b/1370); printf("That is %.2f Watts per square meter\n", b); } Which is based on: The luminosity of the sun is: L_0=3.9E26 J/s The separation between the earth and the sun is: 1.5E11 m The solar luminosity at the earth is reduced by the inverse square law, so the solar constant is: S_0=3.9E26/4(pi)(1.5E11)^2 = 1,370 watts/square meter That is the effective energy hitting the earth per second per square meter. This radiation is equal to the temperature, T_e, to the fourth power by the steffan-bolzmann constant, sigma. T_e can be called the effective temperature, the temperature entering the earth.

5 of 6 S_0 intercepts the earth disc, (pi)r^2, and distributes itself over the entire earth surface, 4(pi)r^2, while 30% is reflected back into space due to the earth’s albedo, a, which is equal to 0.3, so (sigma)(T_e)^4 = (S_0/4)(1-a) from (1-a)(S_0)((pi)r^2)/4(pi)(r^2) But, just as the same amount of radiation that enters the system, leaves it, to have radiative equilibrium, the atmosphere radiates back to the surface so that the radiation from the atmosphere, (sigma)(T_a)^4 plus the radiation entering the earth, (sigma) (T_e)^4 is the radiation at the surface of the earth, (sigma)(T_s)^4. However, (sigma)(T_a)^4=(sigma)(T_e)^4  and we have:  (sigma)(T_s)^4=(sigma)(T_a)^4 + (sigma)(T_e)^4 = 2(sigma)(T_e)^4 T_s=(2^(1/4)) (T_e) (sigma)(T_e)^4=(S_0/4)(1-a) sigma = 5.67E-8  S_0=1,370 (1,370/4)(1-0.3)=(1,370/4)(0.7) S_0=239.75  (sigma)(T_e)^4=239.75  (T_e)^4 = (238.75)/(5.67E-8) = 4.228E9 T_e=255 degrees kelvin So, for the temperature at the surface of the Earth: (sigma)(T_s) = 2(sigma)(T_e)^4  T_s=(2^(1/4))T_e  or T_s = 1.189(255) = 303 degrees Kelvin Let’s convert that to degrees centigrade: Degrees Centigrade = 303 - 273 = 30 degrees centigrade And, let’s convert that to Fahrenheit:  Degrees Fahrenheight = 30(9/5)+32=86 Degrees Fahrenheit In reality this is warmer than the average annual temperature at the surface of the earth, but, in this model, we only considered radiative heat transfer and not convective heat transfer. In other words, there is cooling due to vaporization of water (the formation of clouds) and due to the condensation of water vapor into rain droplets (precipitation or the formation of rain). I don’t think people realize how strange this is.

6 of 6 The Author