Concept explainers
(a)
The average speed of a Hydrogen atom in the Sun’s atmosphere.
Answer to Problem 47Q
Explanation of Solution
Given:
Temperature of atmosphere of present day Sun is
Mass of hydrogen atom is
Boltzmann constant
Formula used:
Average speed of an atom is given by the formula,
Where
Calculation:
Substitute the values
Conclusion:
The average speed of a Hydrogen atom in the Sun’s atmosphere where temperature is
(b)
The average speed of a Hydrogen atom in the
Answer to Problem 47Q
Explanation of Solution
Given:
Temperature of atmosphere of
Mass of hydrogen atom is
Boltzmann constant
Formula used:
Average speed of an atom is given by the formula,
Where
Calculation:
Conclusion:
The average speed of a Hydrogen atom in the
(c)
Comparison of the two average values of a Hydrogen atom in the atmospheres of present-day Sun and the
Explanation of Solution
Given:
Escape speed from a red giant =
Escape speed from the present-day sun =
According to the calculations, the escape speed of an atom in the atmosphere of a red giant is much lower than the escape speed from the atmosphere of the present-day Sun. However, there is not much difference in the average speeds in the respective atmospheres as they are
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Chapter 19 Solutions
Universe: Stars And Galaxies
- A main sequence star of mass 25 M⊙has a luminosity of approximately 80,000 L⊙. a. At what rate DOES MASS VANISH as H is fused to He in the star’s core? Note: When we say “mass vanish '' what we really mean is “gets converted into energy and leaves the star as light”. Note: approximate answer: 3.55 E14 kg/s b. At what rate is H converted into He? To do this you need to take into account that for every kg of hydrogen burned, only 0.7% gets converted into energy while the rest turns into helium. Approximate answer = 5E16 kg/s c. Assuming that only the 10% of the star’s mass in the central regions will get hot enough for fusion, calculate the main sequence lifetime of the star. Put your answer in years, and compare it to the lifetime of the Sun. It should be much, much shorter. Approximate answer: 30 million years.arrow_forwardWe learned in class that, when stars collapse under their own gravity, they conserve angular momentum, which is proportional to mass times radius times rotational speed. Suppose the entire sun (radius 695,700 km) were to collapse to a neutron star with a radius of only 10 km. Before the collapse, the rotational speed at the equator = 2.0 km/s, and the rotational period is 25 days. Using the same steps that you used for the white dwarf calculations, calculate the final rotation period if the entire sun were to collapse to a 10 km radius neutron star. Give your answer in units of seconds. Answer: Checkarrow_forwardA star has initially a radius of 680000000 m and a period of rotation about its axis of 26 days. Eventually it changes into a neutron star with a radius of only 40000 m and a period of 0.2 s. Assuming that the mass has not changed, find Assume a star has the shape of a sphere. (Suggestion: do it with formula first, then put the numbers in) [Recommended time : 5-8 minutes] (a) the ratio of initial to final angular momentum (Li/Lf) Oa. 3.25E+15 Ob. 25.7 Oc. 0.0389 Od. 3.08E-16 (b) the ratio of initial to final kinetic energy Oa. 2.74E-23 Ob. 437000 Cc. 2.29E-6 FUJITSUarrow_forward
- Consider a star with more brightness at 280 pc from the Sun. Suppose this star gets exploded as a supernova at a temperature of 18000 K. The absolute bolometric magnitude of this supernova is-12.24. Calculate its diameter by assuming a sphere at maximum light. (Assume the luminosity of Sun as 3.8×1026 W, the mass of thesun as 1.9 ×1030 kg, and surface temperature of Sun as 5778 K).(a) 1.7×108 km(6) 3.5x108 km(c) 5.2x108 km(d) 6.9 x108 kmarrow_forwardFor the PP chain 0.7% of the mass participating in nuclear fusion is liberated as energy which produces a star's luminosity. Assume that the core of a main sequence star consists of 10% of its total mass. Hence, estimate the lifetime of a star on the main sequence in terms of its luminosity L/L. Give your answer in years. You may use the observed mass-luminosity relation L x M³.5, where M is the star's total mass. Using typical values, calculate estimates for the main sequence lifetime of a KO star and a 05 star. Describe briefly why your estimate might be more accurate for K stars compared to O stars.arrow_forwardWhat is the escape velocity (in km/s) from the surface of a 1.5 M neutron star? From a 3.0 M neutron star? (Hint: Use the formula for escape velocity, Ve = 2GM r ; make sure to express quantities in units of meters, kilograms, and seconds. Assume a neutron star has a radius of 11 km and assume the mass of the Sun is 1.99 ✕ 1030 kg.) 1.5 M neutron star km/s3.0 M neutron star km/sarrow_forward
- A star has initially a radius of 780000000 m and a period of rotation about its axis of 22 days. Eventually it changes into a neutron star with a radius of only 25000 m and a period of 0.1 s. Assuming that the mass has not changed, find Assume a star has the shape of a sphere. (Suggestion: do it with formula first, then put the numbers in) [Recommended time : 5-8 minutes] (a) the ratio of initial to final angular momentum (Li/Lf) a. 1.85E+16 b. 51.2 c. 0.0195 d. 5.4E-17 (b) the ratio of initial to final kinetic energy a. 2.84E-24 b. 371000 c. 2.69E-6 d. 3.52E+23arrow_forwardFor a 1 solar mass white dwarf with a radius of 6000km, calculate the momentum and kinetic energy (in units of electronvolts) of an electron in the highest occupied electron state. Assume the star consists only of carbon and has a uniform density. Calculate the momentum and kinetic energy of a neutron in the highest occupied neutron state for a 2 solar mass neutron star with a radius of 10km. Assume that the star has a uniform density and consists only of neutrons.arrow_forwardA star has initially a radius of 640000000 m and a period of rotation about its axis of 20 days. Eventually it changes into a neutron star with a radius of only 50000 m and a period of 0.2 s. Assuming that the mass has not changed, find Assume a star has the shape of a sphere. (Suggestion: do it with formula first, then put the numbers in) [Recommended time : 5-8 minutes] (a) the ratio of initial to final angular momentum (Li/Lf) Oa. 1.42E+15 Ob. 19 Oc. 0.0527 Od. 7.06E-16 (b) the ratio of initial to final kinetic energy Oa. 8.18E-23 Ob. 456000 Oc. 2.19E-6 Od. 1.22E+22 52%arrow_forward
- A supernova’s energy is often compared to the total energy output of the Sun over its lifetime. Using the Sun’s current luminosity, calculate the total solar energy output, assuming a 1010 year main-sequence lifetime. Using Einstein’s formula E=mc2 calculate the equivalent amount of mass, expressed in Earth masses. [Hint: The total energy output of the Sun over its lifetime is given by its current luminosity times the number of seconds in a year times its ten billion-year lifetime; ; mass of earth = 6×1024kg; c = 3×108m/s. Your answer should be 200-300 Earth masses.]arrow_forwardA star has initially a radius of 660000000 m and a period of rotation about its axis of 34 days. Eventually it changes into a neutron star with a radius of only 35000 m and a period of 0.2 s. Assuming that the mass has not changed, find Assume a star has the shape of a sphere. (Suggestion: do it with formula first, then put the numbers in) [Recommended time : 5-8 minutes] (a) the ratio of initial to final angular momentum (Li/Lf) Oa. 5.22E+15 Ob. 24.2 Oc. 0.0413 Od. 1.91E-16 (b) the ratio of initial to final kinetic energy Oa. 1.3E-23 Activate V Go to Setting Ob. 607000 Oc. 1.65E-6 e here to searcharrow_forwardA supernova's energy is often compared to the total energy output of the Sun over its lifetime. Using the Sun's current luminosity, calculate the total solar energy output, assuming a 1010 year main-sequence lifetime. Using Einstein's formula E = mc? calculate the equivalent amount of mass, expressed in Earth masses. [Hint: The total energy output of the Sun over its lifetime is given by its current luminosity times the number of seconds in a year times its ten billion-year lifetime; Week 5 slide 4; mass of earth = 6x1024kg; c = 3x10®m/s. Your answer should be 200-300 Earth masses.]arrow_forward
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