Whenever I talk to 6th graders about astronomy, I can always elicit a groan by talking about anything that sounds like mathematics. When I talk to adults, it’s really the same reaction, just silently. But if you are interested in astronomy, some math is indeed required. But hey, it’s not too bad. A lot of the math is pretty simple. Here are a few simple “maths” that are key to understanding astronomical concepts.
The Kelvin Scale
K = C + 273.15 (usually just 273 is good enough).
This translates Celsius (or Centigrade) temperatures into kelvins. Although Kelvin is someone’s name (Lord Kelvin) we type it in lowercase if it doesn’t start a sentence as in “the temperature of the sun is about 5800 kelvins.” Note that the word degree is not use with kelvins. That’s because kelvins are on an absolute scale while Celsius and Fahrenheit are relative scales. Think of it this way, 50 degrees Fahrenheit or 10 degrees Celsius (which do represent the same temperature) is not twice as hot as 25 degrees Fahrenheit or 5 degrees Celsius (which do not represent equal temperatures). But 6000K is twice as hot as 3000K thanks to the fact that 0K is an absolute number, absolute zero. http://en.wikipedia.org/wiki/Kelvin
2898000 nm/Temp(K) = Wavelength of peak emission
Let’s see how that works. Your body temperature is 310K (273 + 37 Celsius). That means that you radiate at 2898000/310 or about 9350 nm. This puts you far into the infrared. This shouldn’t be a surprise. You don’t glow in visible light unless you think you have an aura. On the other hand, you have seen infrared cameras that make humans glow in a ghostly fashion. The sun appears to be nearly white which means it radiates across the visible spectrum somewhat equally. If we take the middle of that spectrum to be 500 nm and reorder this equation to solve for the temperature, we compute that the temperature of the sun is 5800 K. Remember that’s the temperature of the surface. The internal temperature is quite a bit different. Ref: http://feps.as.arizona.edu/outreach/bbwein.html
Kepler’s 3rd Law
P2 relates to a3
Without dissing Kepler’s first 2 laws, the third law of planetary orbits is the one you need to know. It is simply this: if you take the ratio of a planet’s orbital period squared and the radius of the planet’s orbit cubed, that ratio will be the same for all planets around the same star. Let’s see if that works. If we use 1 AU as the earth’s orbital radius around the sun and 1 year as the orbital period, we get a ratio of 1. Well, that seems too easy doesn’t it? How could all of the other planets end up with the same ratio? Well, the orbital radius of Venus expressed in AU is .72 and cubed that is .373. The orbital period is .6156 years and squared that is .378. Pretty darn close. In rough terms, the orbit of Jupiter is 5 AU which cubed is 125 and each Jovian orbit takes 12 years which squared is 144. Close but if you use more exact terms (like 11.86 years for the orbital period and 5.2 AU for the radius) you get a ratio very close to 1. The same will be true no matter how elliptical the orbit is. To be specific the orbital radius described here is actually the semimajor axis of an elliptical orbit. http://www-istp.gsfc.nasa.gov/stargaze/Skepl3rd.htm
d = 10 ^ (m-M+5)/5
The most important usage of the distance modulus is to determine the distance (d) to an object if you know it’s apparent magnitude (m) and it’s absolute magnitude (M). The light curve of variable stars is one way to determine the absolute magnitude of an object. The distance is given in parsecs (1 parsec = 3.26 light years). Let’s try one. The absolute magnitude of Deneb is -7 and the apparent magnitude is 1.25. That gives 10 to (13.25/5) power. That gives 446 parsecs. http://www.daviddarling.info/encyclopedia/D/distance_modulus.html
mag = focal length obj. / focal length eyepiece
Whenever you are at a star party and the first question someone asks is about how much magnification your scope can do, you immediately know that you are talking to an unsophisticated astronomer. But that’s because we know that magnification is just not that important for most applications and that department store ads for telescopes with 1000x magnification are probably talking about really awful telescopes. Still, you need to know how to compute the magnification of a telescope. If you know the distance of your telescope’s light path (from lens or mirror to eyepiece, in a refractor this is the focal length of the objective lens, in a Cassegrain design the light path is folded) and divide by the focal length of the eyepiece, you get the magnification. So you can improve magnification by using an eyepiece with a shorter (often cheaper) focal length. Hence the literally low value of higher magnification made more so by the fact that magnification takes a little bit of light and stretches it out so that the object appears dimmer.
Previous | Contents | Next