If the full moon was covered with snow, how bright would it be compared to the sun?
In one case light reaches the earth directly and in the other case it first reaches the moon and then a part of the light energy gets reflected on the surface of moon and reaches the earth If I is the intensity of sunlight on earth and if the distance to be covered by it in the second case x times the distance to be covered in directly reaching the earth and if y is a fraction of light that is reflected, then the intensity of light reflected from the surface of the moon will be I * (y/x²) which would be obviously much less than the intensity of sunlight directly reaching the earth.
1/120th Both the sun and the moon occupy appx 1/120th of the sky. Since the moon is in roughly the same orbit as the earth, the sun would also occupy 1/120th of its sky. This that, per unit surface area, it would receive and reflect energy at an intensity that was 1/120th of the sun. Since, as mentioned above, from the earth’s perspective, both the sun and the moon occupy the same percentage of the sky, the brightness level of the moon would be 1/120th that of the sun. As a note, the moon is one of the darkest objects in the solar system. It has an albedo roughly that of an asphault parking lot. (~0.07). Fresh snow is roughly ~0.90 albedo. This would make the moon 10-12x brighter. *********** Stupid mistake. 1/120th is the arc distance or degrees of the moon in the sky. The answer is 1/120^2 or 1/14400. *********** Second Stupid mistake. John C wrote me re: I had the moon’s arc length wrong. He is right. My memory has befuddled me. The portion the moon occupies of the sky is (1/221)^2.
The scale of apparent magnitude is base on a difference of 1 magnitude is a difference of luminosity of: 100^(1/5) Apparent Magnitude of Sun = -26.73 Apparent Magnitude of Full Moon = -12.6 Albedo of Full Moon = 0.12 Difference in magnitude = -26.73 – 12.6 = -14.13 ________ If moon was covered with snow: New Albedo = 1.00 (given in problem) d = Difference in Brightness (Ratio Sun to Moon) d = (0.12/1.00) [(100^(1/5)]^14.13 = 53,849 The sun is about 54,000 times as bright as the snow covered full moon. _______ Edit: Looks like you beat me to the methodology.
The area of sphere surrounding the Sun at the distance of the Earth & Moon is 4(pi)(Rs)^2. Rs is distance from earth to sun. Area of section (through center) of the moon pi(Rm)^2. Rm is the radius of the moon. The moon will be reflecting this fraction of the light of the sun: pi(Rm)^2/4(pi)(Rs)^2 =(Rm)^2/4(Rs)^2. This is the ratio of the light energy leaving the moon vs the energy leaving the sun. Intensity varies inversely as the square of the distance. The relative energies disperse by this ratio:(Rs)^2/(Rem)^2. Rem is rapid eye movement; NO! it’s distance from the Eath to the Moon. Multiply the two ratios (Rm)^2/4(Rs)^2 x (Rs)^2/(Rem)^2 = (Rm)^2/4(Rem)^2 = (Rm/2Rem)^2 = (1737.4/2*384,403)^2=5.1×10^-6 The reciprocal 1/5.1×10^-6 = 196078 196078/56000 = 3.5 This is off by a factor of ~3.5 from the imperical calculation. But so far this model has treated the reflection from the moon as coming equally from the whole surface. But, of course, half of the surface is not illuminated at all.
