Download Homework Solutions: Chapter 7, The Sun

Homework Solutions: Chapter 7, The Sun - Our Star
[ RQ: 3 , 5 , 13
Problems: 1 , 3 , 5 ]
RQ: 7-3
Q: How are granules and supergranules related? How do they differ?
A: Granules and supergranules appear to be the result of convection within the sun.
Rather small convection cells just below the photosphere produce the granules. The
supergranules appear to be caused by convection cells deeper in the interior of the sun.
The granules are associated with the patterns we see in the photosphere, while the
supergranules are associated with the pattern we see in the spicules observed in the solar
chromosphere.
RQ: 7-5
Q: What evidence do we have that corona has a very high temperature?
A: The spectral emission lines of the corona are strongly broadened in wavelength,
indicating extremely high temperature. Furthermore in the corona’s spectrum we find
emission lines of highly ionized gases which also give us more evidence that corona has a
very high temperature.
RQ: 7-13
Q: Why does nuclear fusion in the sun occur only near the center?
A: Nuclear fusion does not occur in regions far from the core, because only the core is
hot enough to sustain nuclear fusion. The temperature of the interior of the sun decreases
from the center to the photosphere. At that point the temperature will be lower than is
required to cause the nuclei to approach each other with sufficient energy to overcome
the coulomb repulsion. Therefore, only near the center of the sun, where the temperature
is above 10 million Kelvin does nuclear fusion occur.
PR. 7-1
Q: The radius of the sun is 0.7 million km. What percentage of the radius is taken up to
the chromosphere?
A: We know that the chromosphere is approximately 3000 km thick. So the percentage x
of the radius of the sun taken up to the chromosphere is equal to,
x=
3000 km
⋅ 100 (% ) ⇒ x = 0.43%
0.7 ⋅ 10 6 km
PR. 7-3
Q: What is the angular diameter of a star like the sun located 5 ly from Earth? Is the
Hubble Space Telescope able to resolve detail on the surface of such a star?
A: According to the small-angle formula,
angular diameter linear diameter
linear diameter
=
⇒ angular diameter = 206,265 ⋅
206,265
dis tan ce
dias tan ce
But,
dis tan ce = 5 ly = 5 ⋅ 9.46 ⋅ 1012 = 4.73 ⋅ 1013 km
linear dis tan ce = 1.392 ⋅ 10 6 km
Hence,
1.392 ⋅ 10 6
⇒ ang. diam. = 0.006 sec onds of arc.
1 ⇒ ang . diam. = 206,265 ⋅
4.73 ⋅ 1013
We know that the resolving power of the Hubble Space Telescope is 0.048 sec of arc
and therefore the H.S.T. is not able to resolve detail on the surface of such a star.
PR. 7-5
Q: A solar flare can release 10 25 J. How many megatons of TNT would be equivalent?
(Hint: A 1- megaton bomb produces about 4 × 1015 J.)
A: Suppose that x is the number of megatons bombs.
Then,
1 megaton
=
4 ⋅ 1015 J
x megaton
=
10 25 J
Hence,
x = 1⋅
10 25 J
⇒ x = 2.5 ⋅ 10 9 megatons of TNT.
15
4 ⋅ 10 J
1