Physics of Stars Essay examples

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Stars are phenomenal glowing spheres that everyone has noticed in the night sky. Long ago they were poorly understood. Today, with the help from astronomers, physicists, and other space scientists, we have discovered a large amount of information about stars.

These huge balls of flaming gas have many different ranges of characteristics. We can observe the many fascinating colors that may be displayed from stars. Some of them are not stars themselves, but the trillions of fragments left behind after they explode into supernova (Moreau, 2000).

There is a huge variation in sizes of stars as well. They range from super giants to small dwarfs. Most often their sizes correlate to their age or the particular cycle they are beginning to
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These two diagrams show the action of subatomic particles smashing into each other creating other elements and energy as a byproduct. Both of these pictures demonstrate the occurrence between the most common elements, hydrogen and helium, undergoing fusion.

Sometimes when dealing with mathematical descriptions of specific phenomenon concerning stars the complexity may become intense. However, we will explore common occurrences and changes of stars that can be described by simple mathematical relationships.

One primary occurrence between stars that are approaching their later stages is a reduction in their radius. This occurs when elements such as hydrogen and helium in the center of the star change into gaseous iron from electron exchanges and other processes. Eventually the star becomes very dense and the radius of the star is reduced. By the conservation of angular momentum in this case, the velocity must increase due to the decreased radius (Kippenhahn, 194-95). Lets do a calculation to show this.

By the conservation of angular momentum we get this equation: Ii wi = If wf, where "Ii" is the initial moment of inertia, "wi" is the initial angular speed, and "If" and "wf" are the final products of what I just listed. The moment of inertia of a sphere is (2/5)mR^2, where "m" is the mass and "R" is the radius. Plugging these into the equation Ii wi = If wf, we get, (2/5)mRi^2 wi = (2/5)mRf^2 wf.
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