Inch Diameter
What is the temperature of a diameter of 1 inch of space rock that orbits 0.1 AU from the Sun?
The laws that apply to planets also applies to space rocks / meteors? The temperature on Mercury is 800 degrees Fahrenheit during the day! But what about a space rock. Do you have a heat trap? What is the temperature of a space rock 1 inch in diameter orbit of 0.1 astronomical units of distance Sun?
Let's make some assumptions. For example, suppose that the rock is spherical (which are rarely, but makes calculations easier.) The amount intercepting solar energy is related to its cross section. If you look at the sunlight on the (3 dimensional) neighborhood of the rock, this is what is lacking behind the rock: a "tube" solar light with a cross section of 1 inch in diameter (the shadow of the rock). Let us assume that the rock is as dark as the moon (It reflects only 12% of the incident energy – this value is called the albedo). Therefore, the rock retains 88% of the energy it receives (1 – albedo). Suppose that the rock behaves as a black body, as it absorbs energy and is heated until it reaches equilibrium (re-emitting energy at the same fast as it is). We also assume that the rock is homogeneous, at least in terms of the energy is dissipated through the inside (for its size, is a reasonable approximation) — I use international units because they are easier to use and who unabashedly figures are rounded. 1 inch = 2.5 cm section section of a sphere = ^ a circle = pi * R ^ 2 = pi * (D / 2) 2 = Pi * (2.5 / 2) ^ 2 = 4.9 cm ^ 2 = 0.00049 m ^ 2 – I 0.0005 m ^ use 2. total exterior surface of the sphere = 4 * pi * R ^ 2 = 0.002 m ^ 2 Solar UA flow in a Sun = 1.370 W / m ^ 2 (watts per square meter). Flow varies as the inverse distance squared. Flow of 0.1 m ^ UA = 100 * 1370 W / 2 = 137 000 W / m ^ 2 of energy absorbed by the rock: the flow * * cross section (1-albedo) E_in = 137,000 W / m ^ 2 * 0.0005 m ^ 2 * 0.88 = 60 W (= 60 joules per second) Once equilibrium is achieved, the rock absorbs and emits 55 W 55 W Otherwise, or While it is warming or cooling. E_out = E_in should emit this energy in your outdoor area all: the surface flow of rock = E_out / outdoor = 60 W / m ^ 2 0002 = May be 30,000 W / m ^ 2 The surface temperature of the rock determined from the surface flow, using the Stefan-Boltzmann constant: T ^ 4 = 30,000 W / m ^ 2 / (5.67×10 ^ -8 W / m ^ 2 K ^ 4) (T is the temperature, the energy varies in temperature to 4) T = FourthRoot [30,000 W / m ^ 2 * 10 ^ 8 m ^ 2 ^ K 4 / 5.67 W) watts The square meters are canceled, leaving us ^ K 4 where K is the unit of absolute temperature (K is Kelvin, from the scientist's name) T = FourthRoot [5300 x 10 ^ 8 ^ K 4] = 85 x 10 ^ 2 = 850 K (Remember, I am shamelessly rounding) — Mercury does not behave the same way as the rock. The energy is dissipated evenly throughout the interior. Dissipation takes time. It keeps the same face toward the sun for a long time (very slow rate of rotation), etc. So, although Mercury is 4 times farther from the rock used in the exercise, it is "noon" the temperature reaches comparable values. However, this is offset by the "side Night "where the surface temperature can drop to 100 K = -300 F (the freezing point of water is 273 K)
Turning a 25 inch Diameter Cylinder
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