Meteorites, Meteors and Meteoroids

My roommate and I were talking about the weather earlier tonight because of how bizarre the weather has been in Riverside lately.  This conversation led us to wonder about the study of meteorology and then we both asked ourselves why in the world it is actually called 'meteorology'.  How important could meteors actually be in studying the weather?

It goes to show how little I know about my Greek roots. I soon discovered the word meteorology comes from the Greek word 'meteoros' meaning high in the air, meaning many things up in the high atmosphere could be described with that root word.

So after learning about all this, I took the next logical step and decided to once and for all get it all sorted out what is the difference between meteors, meteorites and meteoroids.

Meteoroids are small pieces of space debris floating around the Solar System minding their own business. Sometimes when meteoroids are just cruising along a massive object will swoop by them and gravity will yank it out of its own orbit. When Earth is that massive object that swoops by, often times, the meteoroids will have the misfortune of being pulled right into Earth's mesosphere and massively slowed down until they burn up and fall to Earth's surface. When burning up through the atmosphere a large path will generally show up in the sky behind the burning meteoroid. This large ribbon of color in the night sky is called the meteor. If the meteoroid isn't completely burnt up when it touches down on Earth, the surviving solid is called a meteorite.

Most meteoroids are only the size of your pinky toe and burn up completely in the atmosphere. However if they're large, they have a higher chance of creating meteorites impacting the ground.

One great way to remember the difference between meteoroids, meteors and meteorites is to do what one of my favorite musicians, Joanna Newsom, did in a section of her song, "Emily".

    That the meteorite is a source of the light
    and the meteor's just what we see
    and the meteoroid is a stone that's devoid of the fire that propelled it to thee.


    And the meteorite's just what causes the light
    and the meteor's how it's perceived
    and the meteoroid's a bone thrown from the void that lies quiet in offering to thee.

Look Up for a Change

Lucianne Walkowicz talks about Looking Up for a Change and how amateur astronomy is helping the Kepler Mission find exoplanets

Kepler's Second Law of Planetary Motion

Diagram of Kepler's laws of planetary motion

 I wanted to play around with Celestia some more so I decided to see if I could practice understanding Kepler's Second Law of Planetary Motion that says

"A line joining a planet and the Sun sweeps out equal areas during equal intervals of time."

Diagram of Kepler's second law of planetary motion

To do this I decided to measure the area swept out of two sections of equal time of a planet's orbit around the Sun. To do this, I wanted to find the angle across the solar sky that the planet swept across in a certain amount of days and compare it to a different area of the orbit swept across in the same amount of days.


I also wanted to pick a planet with a high eccentricity and the planet with the most eccentric orbit is Mercury at .206.

According to Celestia, Mercury just passed aphelion on January 18 and will reach perihelion on March 2. So I arbitrarily chose two dates both 4.5 Earth days on either side of these dates and measured the distance from the Sun to Mercury and the angle on the Ecliptic grid Mercury was located at on these days. Because these dates are equally spaced from perihelion and aphelion, the distance from Mercury to the Sun will be equal in each pair.

ecliptic longitude = λ, distance = d

• t1 = January 13: λ1 = 245˚, d1 = 0.45954 AU
• t2 = January 22: λ2 = 269˚, d2 = 0.45954 AU
• t3 = February 26: λ3 = 49˚, d3 = 0.30962 AU
• t4 = March 2: λ4 = 105˚, d4 = 0.30962 

∆t12 = 9 days, ∆λ12 = 24˚ = 2π/15, ∆d12 = 0 AU
∆t34 = 9 days, ∆λ34 = 56˚ = 14π/45, ∆d34 = 0 AU


The equation to find these swept out areas is given by


A = (1/2)*(r^2)*(θ)


A12 = (1/2)*((0.45954 AU)^2)*(2π/15) = 0.0468483 (AU)^2
A34 = (1/2)*((0.30962 AU)^2)*(14π/45) = 0.0442288 (AU)^2


So indeed, it does seem that with equal time an equal elliptical area will be swept out by the orbiting planet in different sections of the orbit.


There was definitely some rounding and estimation with ecliptic longitude measurements done here to lose information which would account for my error, but I am satisfied with my results. Also I do realize it's pretty silly to get my "observational readings" from a computer simulation based on the actual laws I'm trying to prove, but this is pretty much the only feasible means I can make these measurements right now. I hope that someday I can use much more advanced equipment. :D

The Amazing Binary Systems of Castor

Looking at Gemini from Earth with the naked eye, one might conclude that Alpha Geminorum, aka Castor is one very bright star. However, astronomers have known since the 17th century that Castor is at least a binary system, meaning two stars gravitationally bound to each other. A little later, it was discovered that each of the two stars of Castor were themselves binary systems, making the whole thing at least a quadruple system. Today we now recognize a faint third binary system that is also gravitationally bound with the quadruple system. So, in all, we can call Castor a sextuple system.

Using my favorite 3D astronomy program Celestia with an add-on created by "Chuft-Captain" from the Celestia Motherlode, I've recorded how the system looks when sped up to 10 billion times the normal rate.