Concept explainers
(a)
The average speed of a hydrogen atom having mass =
(a)
Answer to Problem 47Q
Solution:
Explanation of Solution
Given data:
For present-day Sun,
Mass of hydrogen atom is
Formula used:
Write the expression for the average speed for a gas atom or molecule.
Here,
Explanation:
Recall the equation for the average speed.
Substitute
Conclusion:
Therefore, the average speed of the hydrogen atom is
(b)
The average speed of a hydrogen atom having mass =
(b)
Answer to Problem 47Q
Solution:
Explanation of Solution
Given data:
For
Mass of hydrogen atom is
Formula used:
Write the expression for the average speed for a gas atom or molecule.
Here,
Explanation:
Recall the equation for the average speed.
Substitute
Conclusion:
Therefore, the average speed of the hydrogen atom in the atmosphere of the red giant is
(c)
The extent to which the present day Sun and a
(c)
Answer to Problem 47Q
Solution: In both cases, the value of the escape speed is higher than the average speed of the atoms but is close in the case of a red giant. Therefore, one can say that for speed above the average speed, hydrogen will leave the red giant.
Explanation of Solution
Introduction:
The escape velocity is the minimum velocity required to escape the gravitational attraction of a planet or star.
Explanation:
Calculate the escape velocity for both cases.
For present-day Sun,
Write the expression for the escape speed.
Here,
Recall the equation for the escape speed.
Substitute
Calculate the escape speed for the red giant;
Recall the equation for the escape speed.
Substitute
The escape velocity for present-day Sun is
Conclusion:
Hence, both the Sun and the red giant will retain hydrogen in their cores but the red giant is more likely to lose that hydrogen as compared to the Sun.
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Chapter 19 Solutions
Universe
- 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_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_forwardA 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_forward
- A 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 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_forwardIf a neutron Star has a radius of 12 km and a temperature of 8.0 x 10^6 K, how luminous is it? Express your answer in watts and also in solar luminosity units. (Hint: Use the relation L/L= (R/R)^2(T/T)^4 . Use 5,800 K for the surface temperature of the Sun. The luminosity of the sun is 3.83 x 10^26W) luminosity in watts ________ W luminosity in solar luminosity units ______ Larrow_forward
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