Question

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c. How does the chemical (i.e. elemental) composition of the centre of the Sun today compare to the composition at its surface? Why?

d.Describe the rings of Saturn?

e. How do astronomers determine the radius of extrasolar planets?

Answer 1

c) Spectroscopy shows that hydrogen makes up about 94% of the solar material, helium makes up about 6% of the Sun, and all the other elements make up just 0.13% (with oxygen, carbon, and nitrogen the three most abundant ``metals''---they make up 0.11%). In astronomy, any atom heavier than helium is called a ``metal'' atom. The Sun also has traces of neon, sodium, magnesium, aluminum, silicon, phosphorus, sulfur, potassium, and iron. The percentages quoted here are by the relative number of atoms. If you use the percentage by mass, you find that hydrogen makes up 78.5% of the Sun's mass, helium 19.7%, oxygen 0.86%, carbon 0.4%, iron 0.14%, and the other elements are 0.54%

d)

They consist of countless small particles, ranging in size from micrometres to metres, that form clumps that in turn orbit about Saturn. The ring particles are made almost entirely of water ice, with some contamination from dust and other chemicals. There are several gaps within the rings: two opened by known moons embedded within them. It seems likely that the ring system is composed of debris from the disruption of a moon about 300 km in diameter

e)

The first calculation comes from Kepler's Third Law (shown below), where 'G' is Newton's Gravitational Constant.The period, 'P', is the orbital period of the exoplanet, and comes directly from the measured period using, for example, the transit or radial velocity detection methods (Detection Methods page). The mass of the star, 'M', was calculated above using the mass-luminosity relationship of stars. Finally, the mass of the exoplanet, 'm', in the equation can be ignored, since it is much smaller than the mass of the parent star. As an example, since the Sun is about three hundred thousand times heavier than the Earth, ignoring the mass of the Earth in this calculation woud introduce an error of less than 0.001%. The equation can be solved for the only remaining variable which is the orbital radius, 'a'