Problems

#01

We are told that there are $10^3$ H atoms/m$^3$ in the Local Bubble, typically. The mass of one H atom is $1.7\times 10^{-27}$ kg. We want to know the mass of a typical volume in the Local Bubble equal to the volume taken up by the Earth. So find the volume of the Earth. Multiple by the number density of the Local Bubble to get the number of H atoms that would normally occupy that volume. Then multiple by the mass of 1 H atom to get the mass in that volume.

The first step is that the volume of the Earth (being a sphere) is $4\pi R_E^3/3$. The Earth radius is $R_E = 6.4\times 10^6$ m, so the volume is $1.1\times 10^{21}$ m$^3$. The number of H atoms in such a volume for the Local Bubble is $1.1\times 10^{24}$. The mass then is a mere $0.0019$ kg, or about 2 grams. Pretty small!

#02

Same approach as last problem, just a different volume to consider. From Table 5 in the appendix, Alpha Cen is at a distance of 1.35 parsecs. The volume of a cylinder is its length times the cross-sectional area. The distance is the length, and the cross-section is given at 1 m$^2$. Let's convert the distance to meters and find the volume.

The distance is $1.35 \times 3.09\times 10^{16} = 4.17\times 10^{16}$ m, so the volume of the cylinder is $4.17\times 10^{16}$ m$^3$.

The number of atoms is thus $4.17\times 10^{19}$, and the mass is only $7.1\times 10^{-8}$ kg.

#03

From Section 18.1, the average number density of the ISM is $10^6$ H atoms/m$^3$. The average density is then $1.7\times 10^{-21}$ kg/m$^3$. The density of air at Earth is 1.2 kg/m$^3$. The question is, how much ISM volume would have to be compressed to get the same mass density as air at Earth.

Let's rephrase. If 1 m$^3$ of air at Earth has 1.2 kg of mass, when volume in the ISM is required to get a total of 1.2 kg of mass? That is easy - we take 1.2 kg and divide by the density of the ISM gas, which gives $1.2 / 1.7\times 10^{-21} = 7 \times 10^{20}$ m$^3$. This is equivalent to a cube that is about 9,000 km on each side!

#13

This makes use of the Doppler formula. For wavelengths we use $\Delta \lambda/\lambda_0 = v/c$, where $\lambda_0$ is 21.1 cm, and for frequencies we use $\Delta \nu / \nu_0 = v/c$, where $\nu_0$ is 1420 MHz. The range of velocities is 75 km/s for receding and 50 km/s for approaching.

Let's do wavelengths first. The biggest redshift from 21.1 cm will be $21.1\times (75/300,000) = 5.3\times 10^{-3}$ cm, and the biggest blueshift will be $3.5\times 10^{-3}$ cm, so the range will be from 21.1 cm minus 0.0035 cm to 21.1 cm plus 0.0053 cm.

Similarly the frequency shifts are 0.35 MHz (redshift) and 0.24 MHz (blueshift), so the frequency range is from 1420 minus 0.35 MHz to 1420 plus 0.24 MHz.

The trick is to get the adding and subtracting sense correct. Blueshifts make wavelength shorter but frequencies larger.

#14

Rather similar to problem #3. We need the mass of the Sun and the mass density of the ISM, but we are given the number density. Look up the mass of the Sun to get $M_\odot = 2.0 \times 10^{30}$ kg. The number density of H atoms is given as $10^{12}$ atoms/m$^3$. The mass density is then $1.7\times 10^{-15}$ kg/m$^3$. Dividing the mass density into the Sun's mass gives the volume $V$ of a molecular cloud of one solar mass, or $V = 2.0 \times 10^{30} / 1.7\times 10^{-15} \approx 10^{45}$ m$^3$. We are asked to get the cloud's radius. The volume for a sphere is $V=4\pi r^3/3$. Solving for the radius gives $r = \sqrt[3]{3V/4\pi} \approx 2 \times 10^{14}$ m. This is only about 0.01 parsecs.