A rotation rate, or frequency, of 500 nHz corresponds to a rotation period of 23 days—it takes 23 days for material to follow one complete circle around the Sun. Furthermore, the frequency and period are inversely proportional to one another: if one is doubled, the other is halved. Given these facts, what is the approximate rotation period (in days) for equatorial material at the Sun's surface? (Use your answer to the previous question as the rotation rate.)

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A rotation rate, or frequency, of 500 nHz corresponds to a rotation period of 23 days—it takes 23 days for material to follow one complete circle around the Sun. Furthermore, the frequency and period are inversely proportional to one another: if one is doubled, the other is halved. Given these facts, what is the approximate rotation period (in days) for equatorial material at the Sun's surface? (Use your answer to the previous question as the rotation rate.)

ASI Course Mc X
* VAL Helios X
https://asu X
Chapter 7 X
Study Guic x
Chapter 5 x
Study Guic x
A American ×
Соса-Cola X
+
webassign.net/web/Student/Assignment-Responses/submit?dep=25854829&tags=autosave#question4313115_1
Update
Consider the previous graph, duplicated below, as well as the figure below.
500
Radiative
Convective
Zone i Zone
450
30°
30°
400
60°
<+ 60°
350
0.4
0.6
0.8
1.0
R/R,
sun
Solar Interior Rotation Rates at Three Latitudes
w
60°
30°
0°
-30°
· -60°
2:48 PM
2/18/2021
Surface
Doppler
Solar Latitude
Rotation Rate (nHz)
Transcribed Image Text:ASI Course Mc X * VAL Helios X https://asu X Chapter 7 X Study Guic x Chapter 5 x Study Guic x A American × Соса-Cola X + webassign.net/web/Student/Assignment-Responses/submit?dep=25854829&tags=autosave#question4313115_1 Update Consider the previous graph, duplicated below, as well as the figure below. 500 Radiative Convective Zone i Zone 450 30° 30° 400 60° <+ 60° 350 0.4 0.6 0.8 1.0 R/R, sun Solar Interior Rotation Rates at Three Latitudes w 60° 30° 0° -30° · -60° 2:48 PM 2/18/2021 Surface Doppler Solar Latitude Rotation Rate (nHz)
ASI Course Mc X
* VAL Helios X
https://asu X
Chapter 7 X
Study Guic x
Chapter 5 x
Study Guic x
A American
Coca-Cola X
+
webassign.net/web/Student/Assignment-Responses/submit?dep=25854829&tags=autosave#question4313115_1
Update
rotation periods.
• Just below the photosphere, the indirect rotation measurements agree reasonably well with the rotation rates that Doppler shifts reveal for the very surface.
• The measurements do not reach all the way to the center of the Sun. Rather, they probe two regions outside of the energy-producing core: the radiative zone, in which
photons percolate outward, and the convective zone, in which hot gas rises, gives off energy, and sinks, over and over in a repeating cycle.
Now take on the role of a solar scientist and interpret the graph!
(e) In which zone does the Sun rotate more or less as a solid body, with all latitudes taking a similar amount of time to circle the Sun's axis?
radiative zone
convective zone
(f) Throughout most of the Sun's outer zones, equatorial material (0°) has a greater rotation rate, or frequency, than material that is closer to the poles (30° or 60°). This
means that the material near the Sun's equator takes fewer
Earth days to circle the Sun's axis once.
(g) At one of the three plotted latitudes, solar plasma takes roughly the same amount of time to circle the Sun's axis-no matter how far it lies below the surface. Which
latitude is this?
w
0°
30°
60°
(h) What is the approximate rotation rate (in nanohertz, or nHz) for equatorial material at the Sun's surface
R
= 1.0 )?
R.
sun
350 nHz
400 nHz
450 nHz
く
500 nHz
A rotation rate, or frequency, of 500 nHz corresponds to a rotation period of 23 days–it takes 23 days for material to follow one complete circle around the Sun.
Furthermore, the frequency and period are inversely proportional to one another: if one is doubled, the other is halved. Given these facts, what is the approximate
rotation period (in days) for equatorial material at the Sun's surface? (Use your answer to the previous question as the rotation rate.)
(i)
2:49 PM
2/18/2021
The photosphere's equator has a lower rotation rate than 500 nHz. Should this imply that it will take more than 23 days, or fewer, to complete one circular motion? What
is the ratio of 500 nHz to the equator's rotation rate? Should you divide or multiply 23 by that ratio? days
Transcribed Image Text:ASI Course Mc X * VAL Helios X https://asu X Chapter 7 X Study Guic x Chapter 5 x Study Guic x A American Coca-Cola X + webassign.net/web/Student/Assignment-Responses/submit?dep=25854829&tags=autosave#question4313115_1 Update rotation periods. • Just below the photosphere, the indirect rotation measurements agree reasonably well with the rotation rates that Doppler shifts reveal for the very surface. • The measurements do not reach all the way to the center of the Sun. Rather, they probe two regions outside of the energy-producing core: the radiative zone, in which photons percolate outward, and the convective zone, in which hot gas rises, gives off energy, and sinks, over and over in a repeating cycle. Now take on the role of a solar scientist and interpret the graph! (e) In which zone does the Sun rotate more or less as a solid body, with all latitudes taking a similar amount of time to circle the Sun's axis? radiative zone convective zone (f) Throughout most of the Sun's outer zones, equatorial material (0°) has a greater rotation rate, or frequency, than material that is closer to the poles (30° or 60°). This means that the material near the Sun's equator takes fewer Earth days to circle the Sun's axis once. (g) At one of the three plotted latitudes, solar plasma takes roughly the same amount of time to circle the Sun's axis-no matter how far it lies below the surface. Which latitude is this? w 0° 30° 60° (h) What is the approximate rotation rate (in nanohertz, or nHz) for equatorial material at the Sun's surface R = 1.0 )? R. sun 350 nHz 400 nHz 450 nHz く 500 nHz A rotation rate, or frequency, of 500 nHz corresponds to a rotation period of 23 days–it takes 23 days for material to follow one complete circle around the Sun. Furthermore, the frequency and period are inversely proportional to one another: if one is doubled, the other is halved. Given these facts, what is the approximate rotation period (in days) for equatorial material at the Sun's surface? (Use your answer to the previous question as the rotation rate.) (i) 2:49 PM 2/18/2021 The photosphere's equator has a lower rotation rate than 500 nHz. Should this imply that it will take more than 23 days, or fewer, to complete one circular motion? What is the ratio of 500 nHz to the equator's rotation rate? Should you divide or multiply 23 by that ratio? days
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