Week 9, Post 5

Worksheet 7, Problem 1

a)

The question asks us to set up the problem, which describes a small, cylindrical parcel of gas, oriented vertically in the Earth's atmosphere. The drawing is shown below.

$P_{up} = P(r)$
$P_{down} = P(r + \Delta r)$

b)

The other force acting on the parcel of gas is gravity. The Earth has a mass of $M_{\oplus}$, while the parcel has a mass of $m = \rho V = \rho(r) A \Delta r$.

$$F_G = \frac{GMm}{r^2}$$
$$F_G = \frac{G M_{\oplus} \rho(r) A \Delta r}{r^2}$$

c)

We can draw a free-body diagram to show all the forces acting on the parcel of air.

Note that pressure is force per unit area, so the forces exerted by $P_{up}$ and $P_{down}$ are therefore:

$F_{up} = A \cdot P_{up}$

$F_{down} = A \cdot P_{down}$

Since it is not moving, we know that the sum of the forces is 0. 

$$F_{net} = F_{up} - F_{down} - F_G = 0$$

$$A(P_{up}-P_{down}) = F_G$$

$$A(P(r) - P(r + \Delta r)) = \frac{G M_{\oplus} \rho(r) A \Delta r}{r^2}$$

d)

$$g = \frac{F_G}{m} = \frac{GM_{\oplus}}{r^2}$$

e)

First, we start with the result from part c)

$$A(P(r) - P(r + \Delta r)) = \frac{G M_{\oplus} \rho(r) A \Delta r}{r^2}$$

Substitute the result from d)

$$A(P(r) - P(r + \Delta r)) = \left(\frac{GM_{\oplus}}{r^2}\right) \rho(r) A \Delta r = g \rho(r) A \Delta r$$

Solve:

$$P(r + \Delta r) - P(r) = -g \rho(r) \Delta r$$

$$\lim_{x \rightarrow \infty} \frac{P(r + \Delta r) - P(r)}{\Delta r} = -g \rho(r)$$

$$\frac{dP(r)}{dr} = -g\rho(r)$$

f)

We refer to the ideal gas law given at the beginning of the worksheet: $P = nk_B T$.

We note that $n = \frac{\rho(r)}{m}$, where $n$ is the number density of particles (with units $cm^{-3}$) and $m$ is the mass of a particle of gas.

This means we can rewrite the ideal gas law as: 

$$P(r) = \frac{\rho(r) k_B T}{m}$$

$$\frac{dP(r)}{dr} = \frac{k_B T}{m} \frac{d\rho(r)}{dr} = -g\rho(r)$$

$$\frac{d\rho(r)}{dr} = \frac{-gm\rho(r)}{k_B T}$$

Integrate:

$$\int \frac{d\rho(r)}{\rho(r)} = \int\frac{-gm}{k_B T}dr$$

$$\ln(\rho(r)) = \frac{-gmr}{k_B T} + C$$

$$\rho(r) = \rho_0 e^{\frac{-gmr}{k_B T}}$$

g)

The scale height is given by: $H = \frac{kT}{\bar m g}$.

Let's check the units on this:

$$\frac{[erg][K]^{-1}[K]}{[g][cm][s]^{-2}} = \frac{[erg][s]^2}{[g][cm]}$$

$$= \frac{[g][cm]^2}{[s]^2}\frac{[s]^2}{[g][cm]} = [cm]$$

The scale height is the height $H$ added to $r$ that would make the density fall off by a factor of $1/e$. We can express this mathematically as:

$$\frac{\rho(r + H)}{\rho(r)} = \frac{1}{e}$$

$${\displaystyle \frac{\rho_0 e^{\frac{-gm(r+H)}{k_B T}}}{\rho_0 e^{\frac{-gmr}{k_B T}}} = \frac{1}{e}}$$

$${\displaystyle \exp(\frac{-gm(r+H)}{k_B T} + \frac{gmr}{k_B T}) = \exp(-1)}$$

$$1 = \frac{gmH}{k_B T}$$

$$H = \frac{k_B T}{mg}$$

h)

We are told that the Earth's atmosphere is mostly molecular nitrogen, $\text{N}_2$. Each molecule contains a total of 14 protons and 14 neutrons, for a total mass of 28 protons.

$$H_{\oplus} = \frac{kT}{\bar m g}$$

$$ = \frac{1.4 \times 10^{-16} \text{ erg K}^{-1} \cdot 300 \text{ K}}{28 \cdot 1.7 \times 10^{-24} \text{ g} \cdot 980 \text{ cm/s}^{2}} = 900360 \text{ cm} = 9 \text{ km}$$

Worksheet 8, Problem 2

a)

To solve this problem, we start with a diagram of the situation:

Depicted is a star, with a radius of $r$ and a mass shell of width $\Delta r$. The mass shell has a density of $u + \Delta u$, while the next mass shell out, at radius $r + \Delta r$, has an energy density of $u$. Notably, as radius increases, energy density decreases.

We are told that the net outwards flow of energy, $L(r)$, must equal the total excess energy in the inner shell divided by the amount of time needed to cross the shell's width $\Delta r$.

$$L(r) = \frac{\text{excess energy}}{\text{time to cross} \Delta r}$$

Since we have the relation in which energy density decreases as radius increases, we need to be careful of the sign.

$$\frac{du}{dr} = \lim_{\Delta r, \Delta u \rightarrow 0} -\frac{\Delta u}{\Delta r}$$

First, we need to solve for change in total energy. We know that change in volume is $\Delta V = \Delta r \cdot SA = 4 \pi r^2 \Delta r$. Then we can multiply this change in volume by energy density.

$$\text{energy} = u \cdot 4\pi r^2 \Delta r$$

Next, we need to solve for time to cross $\Delta r$, which is $\Delta t$.

$$\Delta t = \frac{\Delta r}{v_{diff}}$$

In question 1, we found that $v_{diff} = \frac{cD \rho \kappa}{N}$ and $N = \left(\frac{R_{\odot}}{l}\right)^2$

In this question, we know that $D = R_{\odot} = \Delta r$, so we can rewrite $v_{diff}$:

$$v_{diff} = \frac{cD\rho\kappa}{N} = \frac{c\Delta r \rho \kappa l^2}{(\Delta r)^2} = \frac{c\rho\kappa}{\Delta r \sigma^2 n_e^2}$$

Now, we can plug in and solve:

$$L(r) = \frac{energy}{\Delta t} = \frac{u \cdot 4\pi r^2 \Delta r}{\frac{\Delta r}{v_{diff}}}$$

$$= \Delta u 4 \pi r^2 v_{diff} = \frac{4\pi r^2 c \rho \kappa}{\sigma^2 n_e^2} \frac{\Delta u}{\Delta r}$$

$$L(r) = -\frac{4\pi r^2 c \rho \kappa}{\sigma^2 n_e^2}\frac{du}{dr}$$

b)

From the diffusion equation, we use the fact that the energy density of a blackbody is $u(T(r)) = aT^4$ to derive the differential equation:

$$\frac{dT(r)}{dr} \propto -\frac{L(r)\kappa \rho(r)}{\pi r^2 acT^3}$$

where $a$ is the radiation constant.

Starting with $u(T(r)) = aT^4$, we take the derivative with respect to r:

$$\frac{du(T(r))}{dT} \cdot \frac{dT(r)}{dr} = 4aT^3 \cdot \frac{dT}{dr}$$

$$\frac{du(T(r))}{dr} = 4aT^3 \frac{dT}{dr}$$

$$\frac{dT}{dr} = \frac{1}{4aT^3}\frac{du(T(r))}{dr}$$

Using $L(r)$ from part a), we can solve for $\frac{du}{dr}$:

$$\frac{du}{dr} = \frac{\sigma^2 n_e^2}{4\pi r^2 c \rho \kappa} L(r)$$

Plug this in and solve:

$$\frac{dT}{dr} = \frac{1}{4aT^3}\frac{\sigma^2 n_e^2}{4\pi r^2 c \rho \kappa} L(r)$$

$$= \frac{\sigma^2 n_e^2}{16 \pi r^2 a T^3 c \rho \kappa} L(r)$$

Then, substitute in $\sigma = \frac{\rho \kappa}{n_e}$:

$$\frac{dT}{dr} = \frac{\rho\kappa}{16\pi r^2 aT^3c} L(r)$$

This matches the form given at the beginning of the problem.

Acknowledgement: Worked with Awnit, Elton, and Simon on these problems.

General Post

https://www.nytimes.com/2019/04/10/science/black-hole-picture.html

Today’s exciting news item comes from one of our very own! A team of astronomers led by Shep Doeleman, an astronomer at the Harvard-Smithsonian Center for Astrophysics, announced that they had finally captured an image of a black hole, once thought to be unobservable because it is so dense that light cannot escape from it.

The image captured was of Messier 87 (M87), a galaxy in the constellation Virgo. In this galaxy is a black hole billions of times more massive than the sun, which unleashes a jet of energy into space about 5000 light-years away. The image of the black hole provides confirmation of a strange conclusion from Einstein’s theory of general of relativity that even he was reluctant to accept. When too much matter is forced into a small volume, the density becomes so extreme that space-time collapses. As Einstein’s theory predicts, the shadow formed by the black hole is circular in shape.

Many black holes are the remains of stars that burned through their fuel and collapsed into themselves. However, there are also supermassive black holes millions or billions of times larger than the sun that reside in the center of almost every galaxy. Scientists don’t know the origins of these black holes. They also don’t know exactly what would happen to something to an object that falls into a black hole.

The image was created from two years of computer analysis of observations from a large telescope the size of Earth, the Event Horizon Telescope. In reality, this telescope is actually a network of radio antennas comprising eight radio observatories on six mountains across four continents. These observatories recorded data on M87 for 10 days in April 2017. In order to prove that the black hole really was a black hole, scientists needed to measure the size of its shadow. This is difficult because of their distance, which makes it very hard to resolve them, even with a very large telescope. However, by combining data from multiple radio telescopes using a technique called very long baseline interferometry, the team was able to create a very large “virtual” telescope, the Event Horizon Telescope.

The accretion disk of a black hole is where matter swirls around a black hole, just at its edge. The ring of light in the image is the innermost orbit of light in the accretion disk. This provides a way for astronomers to measure the size of a black hole. This size measurement also provides an estimate for the mass of M87, which is 6.5 billion solar masses.