Q1. Describe the three main layers of the Sun’s interior.
The three main layers of the Sun’s interior are the core, the radiative envelope and the connective envelope.
The core is the center of the Sun, with a density of 150 g/cc and a temperature of 15 million K. This is the region where high energy fusion reactions take place and energy is produced. The energy produced is mainly in the form neutrino and gamma rays, which are actually high energy protons which travel several light years and supply energy to the Earth and other planets. The energy produced in the core then radiates outwards through the radiative zone. The temperature here is quite lower, about 4 million K, and so is the density. This zone is responsible for maintaining the core’s temperature. Next comes the convection zone where the energy travels swifter than the radiative envelope. In this layer energy is transmitted by convection as with boiling water. These convective motions are responsible for the granular surface of the Sun .
Q2. What do astronomers mean by a “model of the Sun”?
A model of the Sun is nothing but a miniature version of the Sun build by scientists. It can be made out of any material such as paper or iron. It is build to serve a practical purpose. Such a standard model makes it easier for the scientists to understand the properties and characteristics of the Sun. The models have resulted in scientific discoveries such as the protoplasmic composition of the Sun. The nuclear fusion reactions of the Sun can also be studied more thoroughly with the help of these models. The properties researched include the temperature profile of the Sun, its density, pressure, etc. In the standard model the Sun is treated as a spherical form of gas. The standard solar model has been able to account for the Helium abundance. The model also makes it easier for the scientists to analyze the physical properties of the Sun and build up more complex models viz. magnetic fields, diffusion, rotation, convection, and so on .
Q3. What is a neutrino, and why are astronomers so interested in detecting neutrinos from the Sun?
A neutrino is a fundamental particle similar to the electron, with the main difference being neutrinos do not carry charge. Being electrically neutral, the neutrinos are devoid of electromagnetic interactions. Hence they can pass through electromagnetic fields without much disturbance. This is one of the main reasons why astronomers are so interested in detecting neutrinos from the Sun . Neutrinos are very hard to detect since they are not much affected by any matter through which they pass. So it would be quite easy to detect them from the Sun because our normal techniques of observation are affected by the magnetic field of the Sun, making it difficult for us to look at the core. However a neutrino would not be easily affected by it, and we would be able to look at the Sun’s core. Many arrangements have been made recently under the Earth to collect neutrinos. This is due to the fact that detection of neutrinos would prove the existence of a thermonuclear source of power which is responsible for the Sun’s running .
Q4. Stellar parallax measurements are used in astronomy to determine which of the following properties of stars? a. speeds, b. rotation rates, c. distances, d. Colors, e. Temperatures.
Stellar parallax measurements are used in astronomy mainly to determine the distance of the stars. There are two types of stellar parallax, geocentric and heliocentric, used to measure the distance of the star from the Earth and the Sun respectively. The distance is obtained from a well-framed distance equation. The main theory behind is, a triangle is formed by two positions of the observer and the object. To calculate the distance of the star, the base distance of the two observation points and the direction of the object in each case have to be known. Thus we can find out the apex angle and hence the distance of the star easily. Various stars can be measured by varying the length of the base line. The result of the distances found from the stellar parallax calculations are used for the calculation of the absolute magnitude of the nearby stars as well. Thus these measurements serve a much wider purpose .
Q5. Briefly describe how you would determine the absolute magnitude of a nearby star.
The Fraunhofer lines generated from the star and also its color of light gives more information about the star than we actually think. Actual magnitude of a star is actually its amount of brightness. It is essential to specify the type of electromagnetic radiation while measuring absolute magnitude of the star. Stellar parallax measurements are also once again essential in the measurements of this absolute magnitude. To measure absolute magnitude one needs to measure the luminosity distance of the star. For nearby stars, the luminosity distance is nearly equal to the actual distance of the star. This distance can be measured using stellar parallax measurements.
The absolute magnitude M of a nearby star can be measured using Euclidean adjustments by using the equation: M= m + 5(1+logp), where m is the apparent magnitude of the star and p is the parallax measurement. The parallax p in seconds is measured in arcs using the equation p = plx/1000 .
Q6. What is the mass-luminosity relation? To what kind of stars does it apply?
The mass-luminosity relation is an equation which gives us an idea of the relationship between the mass and luminosity of a star. To achieve this, simple mathematical calculations and graph-plotting are required. When the logarithmic value of luminosity is plotted against the logarithmic value of a mass of the star, both in solar units, we get almost a straight line. This theory is however applicable to only the main-sequence stars.
The relationship is given by log (L/Ls) = a log (M/Ms) where Ls and Ms are the luminosity and mass of the Sun respectively. For main-sequence stars the value of a is normally taken as 3.5. Thus doubling the mass of such a star would mean its luminosity would increase by a multiple of 11.3. Thus it can be seen that such a straight line is only possible for main sequence stars and there is a strong dependant relationship between the luminosity and the mass of such stars .
Q7. What is the lowest mass that a star can have on the main sequence? a. There is no lower limit, b. 0.003 Mo, c. 0.08 Mo, d. 0.4 Mo, e. 2.0 Mo.
There is an upper bound and a lower limit to the mass that a star in the main sequence can have. The upper bound is called the Eddington limit and is about 80 Mo. According to scientific experiments, the lowest mass that a main sequence star can have is 0.08 Mo. This is due to the fact that the conversion of hydrogen to helium in the thermonuclear process becomes an almost impossible task thanks to the extremely low internal pressure and temperature profiles of stars which have mass below this limit. Thus in the absence of a source of the thermonuclear energy, the stars loses its power of self-illumination, and thus it is termed as a ‘failed star’. However these bodies do exist and radiate wavelengths of infrared levels. This is possible due to the heat energy that is generated when these bodies undergo contraction under gravity. These bodies are called brown dwarfs and some bodies of less mass become planetary bodies, for example Jupiter .
Q8. On what grounds are astronomers able to say that the Sun has about 5 billion years remaining in its main sequence stage?
It is possible to calculate the energy output of the Sun accurately along with its mass. The stars use energy created through nuclear fusion reactions to maintain hydrostatic equilibrium. The amount of energy released due to nuclear fusion reactions can be known easily. Thus several calculations done on that basis show that the Sun is only halfway through its stage in the main sequence, and it was formed about 4.6 billion years ago. This means that the Sun has about 5 billion years remaining in that stage. The nuclear reactions which take place involve the conversion of hydrogen to helium. The energy formed is essential for the support of the star under gravitational collapse. Actually the conversion of hydrogen to helium is essential for the existence of the Sun. But it can be estimated that there is only enough hydrogen left in its core to serve for about another 5 billion years. Thus the period of existence of the Sun can be ascertained .
Q9. How is the evolution of a main-sequence star with less than 0.4 Mo fundamentally different from that of a main-sequence star with more than 0.4 Mo?
It is known that the stars exist due to their nuclear energy reactions which involve conversion of hydrogen into helium. This leads to a substantial amount of energy release which supports the star. Stars of a minimum of 0.4 Mo on finishing their hydrogen quantity cool off soon; especially the outer layers and they form a body known as the red giant. These stars have a convective interior. On finishing their main-sequence lifetime, they are rich in their quantity of helium. The main-sequence stars having less than 0.4 solar mass go through the process of convection all through their masses. It is now well-observed that more massive the star is, the shorter is its life span. Lighter stars become white dwarfs relatively, slightly heavier ones become red giants, while heavier stars form supernova or can even result in a black hole formation. Thus main-sequence stars differ in evolution according to their masses .
Q10. A white dwarf is composed of primarily a. Neutrons, b. hydrogen and helium, c. iron, d. cosmic rays, e. carbon and oxygen.
A white dwarf is nothing but the remains of a normal star. Stars require hydrogen to be converted to helium so that they can produce the energy required for their support against collapse under gravity. In the ends of their main sequence stage, the stars use up all their hydrogen or the fuel necessary for the nuclear fusion reactions. These lead to them they lose their shining and begin to cool off, being known as a white dwarf. Thus they are primarily composed of the waste products remaining from the nuclear fusion reactions of the normal stars, which consist mainly of carbon and oxygen. They also contain other elements in trace amounts. Hydrogen and helium form the outer part of the white dwarf. The high amount of gravitational force which acts in these stars is responsible for this differential of the elements according to masses. The heaviest elements are situated most deep down into the white dwarf .
Q11. What is the Chandrasekhar limit?
The Chandrasekhar limit has been named after Subrahmanyan Chandrasekhar, a renowned Indian astrophysicist. This limit is actually the maximum mass of a body created from electron-degenerate matter. This matter is nothing but it actually consists of nuclei which are immersed in an electron gas. This is also the maximum mass which is non-rotating, that can be supported against collapsing under gravity by the pressure formed due to electron degeneracy. The limit is commonly set at 1.4 Mo. White dwarfs are formed by degeneration of electrons and hence this is also the upper limit of the mass of a non-rotating white dwarf. In the main sequence the stars produce energy by nuclear fusion reactions. This energy prevents the star from collapsing under gravity. The stars which are below 8 Mo when finish their main-sequence stage they form bodies of mass less than Chandrasekhar limit and continue to lose mass until they become a white dwarf .
Q12. Compare a white dwarf and a neutron star. Which of these stellar corpses is more common? Why?
The major difference between a white dwarf and a neutron star lies in the process of their formation. White dwarfs are formed out of stars having very low mass (less than ten times the mass of the Sun). On the other hand neutron stars are formed from the core of giant stars. A white dwarf is prevented from collapsing under gravity by electron degeneracy pressure, while a neutron star is supported by neutron degeneracy pressure. The radius of a white dwarf is larger; about six hundred times more than a neutron star, even though a neutron star consists of a stronger gravitational field. The neutron stars also boast of higher temperature, stronger magnetic fields and more spin.
Among these two bodies, white dwarfs are more common. This is due to the fact that white dwarfs require lesser mass to form compared to the neutron stars. The gigantic stars which are required to form neutron stars are lesser in number than the ones which form white dwarfs .
Q13. Super massive black holes are found in which of the following locations? a. in the centers of galaxies, b. in globular clusters, c. in open (or galactic) clusters' d. between galaxies, e. in orbit with a single star.
The super massive black holes are found in the centers of the galaxies. These black holes are the energy sources of high amount of strength that these galaxies boast of. These structures have mass of the order of a billion solar masses. Photographic evidences have proved that the galaxy M87 has a black hole of a mass of 3 billion Mo concentrated at its center. This hole is responsible for the ejection of tremendously high frequency electrons which lead to enormous amount of energy. In the M84 galaxy too, photographic evidences have shown a black hole at its centre. The deviation in the velocity distribution along with taking into account the Doppler shift of the light, have made the existence of the black hole amply clear .
Q14. If the Sun suddenly became a black hole, how would Earth's orbit be affected?
Black holes boast of extremely strong gravitational forces which do not let even light leave them, thus being invisible to us. It does not live up to its name of being a hole in any sense and fully consists of all sorts of matters. The black hole actually contains of matter squeezed into a small object. The effect of a black hole depends mostly on the distance of the body and the mass, having nothing to do with its size. Thus if the Sun suddenly became a black hole, it would still have the same mass and remain at the same distance from the Earth as it is now. Hence the Sun would not exert any more gravitational force than what it is exerting at present and it would be safe to say that the Earth’s orbit would remain unaffected.
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