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Mobile App Chapter 14: Problem 24
(For all of these problems, where necessary assume the normal ratio of total- to-selective extinction.) Calculate the mass of an HI cloud with a circular appearance, with a radius of \(10 \mathrm{pc}\) and an average H column density of \(10^{21} \mathrm{cm}^{-2}\).
Short Answer
Expert verified
The mass of the HI cloud is approximately 2500 solar masses.
Step by step solution
01
- Understand the Given Values
Identify the given values in the problem. The radius of the HI cloud is given as 10 parsecs (pc), and the average hydrogen column density is given as \( 10^{21} \; \text{cm}^{-2} \).
02
- Calculate the Area of the Cloud
Since the cloud is circular, calculate its area using the formula for the area of a circle. The area (A) is given by \[ A = \text{πr}^2 \]where \( r = 10 \; \text{pc} \). First, convert the radius to centimeters. Knowing that 1 parsec is approximately \( 3.086 \times 10^{18} \; \text{cm} \), you can find \[ r = 10 \; \text{pc} \times 3.086 \times 10^{18} \; \text{cm} = 3.086 \times 10^{19} \; \text{cm} \]. Then find \[ A = \text{π} (3.086 \times 10^{19} \; \text{cm})^2 \approx 2.992 \times 10^{39} \; \text{cm}^2 \].
03
- Calculate the Volume of the Cloud
Assume the cloud is a sphere for simplicity. The volume (V) is given by \[ V = \frac{4}{3} \text{πr}^3 \].Using \( r = 10 \; \text{pc} \) from Step 2, convert to centimeters and find \[ V = \frac{4}{3} \text{π} (3.086 \times 10^{19} \; \text{cm})^3 \approx 1.232 \times 10^{59} \; \text{cm}^3 \].
04
- Find the Column Density to Mass Conversion
Use the column density to find the total number of hydrogen atoms in the area. Column density \( N_H \) is \[ 10^{21} \; \text{cm}^{-2} \]. The number of hydrogen atoms (N) is given by \[ N = N_H \times \text{Area} \]. So, \[ N = 10^{21} \; \text{cm}^{-2} \times 2.992 \times 10^{39} \; \text{cm}^2 \approx 2.992 \times 10^{60} \; \text{atoms} \].
05
- Calculate the Mass
The mass of the cloud is found by converting the number of hydrogen atoms to mass. Knowing the mass of a hydrogen atom is approximately \( 1.67 \times 10^{-24} \; \text{g} \), find \[ \text{Mass} = 2.992 \times 10^{60} \; \text{atoms} \times 1.67 \times 10^{-24} \; \text{g/atom} \]. Thus, the mass is \[ \text{Mass} \approx 5.00 \times 10^{36} \; \text{g} \].
06
- Convert Mass to Solar Masses
Convert the mass from grams to solar masses for a more convenient astronomical unit. Knowing that one solar mass \( M_\odot \) is approximately \( 2 \times 10^{33} \; \text{g} \), find \[ \text{Mass} = \frac{5.00 \times 10^{36} \; \text{g}}{2 \times 10^{33} \; \text{g/M}_\odot} \approx 2500 \, M_\odot \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
hydrogen column density
Hydrogen column density is a measure used in astronomy to quantify the number of hydrogen atoms along a line of sight through a cloud or other structure. It is typically expressed in units of \( \text{cm}^{-2} \) and is a crucial parameter for understanding the mass and composition of interstellar clouds.
To calculate the column density, astronomers often measure the amount of absorption or emission lines from hydrogen atoms in the cloud. The column density (\( N_H \)) represents the cumulative number of hydrogen atoms per unit area along the line of sight.
To calculate the column density, astronomers often measure the amount of absorption or emission lines from hydrogen atoms in the cloud. The column density (\( N_H \)) represents the cumulative number of hydrogen atoms per unit area along the line of sight.
parsecs to centimeters conversion
In astronomy, distances are often measured in parsecs (pc). One parsec is approximately equal to \( 3.086 \times 10^{18} \) centimeters. Converting parsecs to centimeters is essential for various calculations, like determining the size, volume, or area of astronomical objects.
For example, if you have a cloud radius of 10 parsecs, you convert it to centimeters by multiplying: \[ 10 \, \text{pc} \times 3.086 \times 10^{18} \, \text{cm/pc} = 3.086 \times 10^{19} \, \text{cm} \].
For example, if you have a cloud radius of 10 parsecs, you convert it to centimeters by multiplying: \[ 10 \, \text{pc} \times 3.086 \times 10^{18} \, \text{cm/pc} = 3.086 \times 10^{19} \, \text{cm} \].
cloud mass in solar masses
The mass of astronomy objects like HI clouds is often expressed in solar masses (\( M_\odot \)). One solar mass is equivalent to approximately \( 2 \times 10^{33} \) grams. Converting the mass from grams to solar masses provides a more intuitive sense of scale for astronomers.
After calculating the mass of the HI cloud in grams, divide by the mass of the sun. For instance, if the mass of the cloud is \( 5 \times 10^{36} \) grams, the mass in solar masses is: \[ \text{Mass} = \frac{5 \times 10^{36} \, \text{g}}{2 \times 10^{33} \, \text{g/M}_\odot} \approx 2500 \, M_\odot \].
After calculating the mass of the HI cloud in grams, divide by the mass of the sun. For instance, if the mass of the cloud is \( 5 \times 10^{36} \) grams, the mass in solar masses is: \[ \text{Mass} = \frac{5 \times 10^{36} \, \text{g}}{2 \times 10^{33} \, \text{g/M}_\odot} \approx 2500 \, M_\odot \].
astronomy problem solving
Solving astronomy problems often involves multiple steps and various conversions. Key skills include identifying the given values, understanding necessary formulas, and converting units.
Typically, you start by recognizing provided data and required units, like parsecs or hydrogen column density. Next, you convert units where needed and apply appropriate mathematical formulas. These steps include calculating areas, volumes, and masses, combined with unit conversions for an accurate understanding of the celestial object in question.
Breaking down complex problems into manageable steps makes it easier to come to a clear, correct solution.
Typically, you start by recognizing provided data and required units, like parsecs or hydrogen column density. Next, you convert units where needed and apply appropriate mathematical formulas. These steps include calculating areas, volumes, and masses, combined with unit conversions for an accurate understanding of the celestial object in question.
Breaking down complex problems into manageable steps makes it easier to come to a clear, correct solution.
