Challenge Math Challenge - August 2021

[Not anonymous]

The mass flow is correct. But we also need to velocities and a reason why the beam widens at the nozzle.

Sorry, I had pressed the post-reply button instead of the save button, so ended up posting an answer that I was still typing. I have updated the original post now. As for why the water bean widens on existing the nozzle, my best guesses are:

Spoiler: Question 12

1. The water stream flowing through the hose is under higher surrounding pressure - there is the atmospheric pressure exerted on the hose plus the pressure of water itself. And water being a more dense fluid would have a higher pressure than the same volume of air. So the stream of water exiting the smaller (than the hose) nozzle has not just the forward force given by the flow from the faucet but also the internal pressure of water. On existing the nozzle, the surrounding pressure is relatively lower than inside the hose, hence the water jet expands to adjust to the lower pressure.

2. Pressure is inversely proportional to the velocity. A jet of water moving at a high velocity experiences lower air pressure than water that is static or moving at a slower speed. Since water exits the nozzle with a higher speed that the speed at which it flows in the hose, the pressure experienced by water when inside moving inside the hose is higher than the pressure experienced when it exits the nozzle. The movement from a region of higher pressure to lower pressure caused the water beam to widen.

 

[fresh_42]

Sorry, I had pressed the post-reply button instead of the save button, so ended up posting an answer that I was still typing. I have updated the original post now. As for why the water bean widens on existing the nozzle, my best guesses are:

Spoiler: Question 12

1. The water stream flowing through the hose is under higher surrounding pressure - there is the atmospheric pressure exerted on the hose plus the pressure of water itself. And water being a more dense fluid would have a higher pressure than the same volume of air. So the stream of water exiting the smaller (than the hose) nozzle has not just the forward force given by the flow from the faucet but also the internal pressure of water. On existing the nozzle, the surrounding pressure is relatively lower than inside the hose, hence the water jet expands to adjust to the lower pressure.

2. Pressure is inversely proportional to the velocity. A jet of water moving at a high velocity experiences lower air pressure than water that is static or moving at a slower speed. Since water exits the nozzle with a higher speed that the speed at which it flows in the hose, the pressure experienced by water when inside moving inside the hose is higher than the pressure experienced when it exits the nozzle. The movement from a region of higher pressure to lower pressure caused the water beam to widen.

Your numbers are correct. The reason why the beam widens is easier than your explanation by pressure. I'm not sure whether your explanation can be transformed into the easier one, but here it goes:

The velocity of the water (an incompressible fluid) is slowed down due to friction and air resistance. However, mass and volume stay the same, such that velocity times cross-section is constant. That is why the beam widens.

 

[mathwonk]

hint for #7, building on the statement of #6:

Let's try it for just one polynomial. from #6, in a finite field k of q = p^r elements, a^(q-1) = 1 for every non zero element a of k, and of course a^(q-1) equals 0, if a=0. Now suppose f is a polynomial on affine n space over k. Then for every element x of the affine n space, f(x)^(q-1) = 0 or 1, depending on whether or not x is a zero of f.

Thus 1-f(x)^(q-1) = 1 or 0, depending on whether or not x is a zero of f. Thus if we add up all q^n such polynomials, namely [1-f(x)^(q-1)] summed over every point x of affine n space, we get the sum of as many 1's as there are zeroes of f.

This sum is in k, so p such terms add to 0, so this sum will equal the number of zeroes of f, mod p. Now we are asked in problem #7 to show this number is zero, and this expression for it is supposed to help. Of course we will need the hypothesis, that degree f < n.

Start by showing that any monomial of degree < n, summed over all points of k^n, is zero. First show that the sum, over all elements x of k, of any power x^j, is zero, if 1 ≤ j < q-1.

Hint: Look at the polynomial X^(q-1)-1, and recall what you know about its coefficients and roots.

 

[Not anonymous]

13. As a result of the refraction, the light bundle emanating from the slit $S$ produces two bundles that overlap in the screen area of width $D$ and appear to arise from two virtual slit images $A$ and $B$. Since the two virtual slit images originate from the same slit, the light emanating from them is coherent and can interfere in the area of overlap.

Calculate the wavelength if monochromatic light is used from the quantities given in the sketch and the distance between two adjacent interference strips $Δy$? Assume that the dimensions parallel to the optical axis can be viewed as large compared to those perpendicular to the optical axis.

Spoiler: Question 13

Diagram 1

Diagram 2Please refer diagram 1 above. Consider 2 points A and B from the 2 virtual images (named A and B in the question's diagram) such that they lie on the same vertical line. Let A' and B' be the points on the screen where the (virtual) light rays from points A and B respectively hit the screen perpendicular to the surface of the screen. A' and B' will also lie on the same vertical axis/line and distance between them will be the same as the distance between A and B, i.e. $|AB| = d = |A'B'|$ (using the notation $|AB|$ to denote length of line segment $AB$). Let P' be the midpoint of line segment A'B'. Since P' is equidistant from A and B, the light rays from the 2 points hitting P' at any time instant $t$ would be in exactly the same phase (since the 2 virtual images are coherent) with respect to one another and therefore result in a bright fringe caused by constructive interference, i.e. P' lies on the central bright fringe. Let Q' be the point on the screen where the middle line of the next bright fringe (the first bright fringe after and above the central bright fringe passing through P') intersects A'B'. Since this fringe is also caused by constructive interference but distance of Q' and its fringe from B is larger than the distance from A, it must be the case that the light rays from A and B hitting Q' at any instant $t$ differ in the distance traveled by exactly 1 wavelength (λ), i.e. $|BQ'| - |AQ'| = λ$. Hence, the wavelength can be calculated based on the distance between the slit and the screen ($a$), the distance between the 2 virtual slit image points ($d$) and the distance between the fringes ($\Delta y$), as follows.
$|BQ'| = \sqrt{|BB'|^2 + |B'Q'|^2}= \sqrt{|BB'|^2 + (|B'P'| + |P'Q'|)^2} = \sqrt{a^2 + (\dfrac{d}{2} + \Delta y)^2}$

$|AQ'| = \sqrt{|AA'|^2 + |A'Q'|^2}= \sqrt{|AA'|^2 + (|A'P'| - |P'Q'|)^2} = \sqrt{a^2 + (\dfrac{d}{2} - \Delta y)^2}$

∴ $\lambda = \sqrt{a^2 + (\dfrac{d}{2} + \Delta y)^2} - \sqrt{a^2 + (\dfrac{d}{2} - \Delta y)^2}$ (Eq. 1)

The question does not specify the value of $d$, but it can be calculated using trigonometry based on the distances given in the question. According to the diagram given in the question, the bottom most light ray from A and the top most light ray from B cross each other at a horizontal distance of $x$ from the slit, and these 2 light rays hit the screen at points that are at a vertical distance of $D$ from each other. Refer to diagram 2 above which shows the triangles involving the 2 virtual image points (A, B), the points on the screen where the topmost and bottom-most light rays marking the boundary of the region of overlap hit the screen (F, H), and the point of intersection of the 2 rays. Since the 2 triangles are similar, $\Delta ACB \sim \Delta FCH$, the ratios of height to base of the two triangles must be the same (here, I consider the line segments $AB$ and $FH$ to be the bases of the 2 triangles and the perpendicular distance between $C$ and these segments to be the heights), i.e.
$\dfrac{x}{d} = \dfrac{a-x}{D} \Rightarrow d = \dfrac{x}{a-x}D$.

Substituting for $d$ in (Eq. 1), we get the wavelength of the light source in terms of the distances specified in the question:
$\lambda = \sqrt{a^2 + \left(\dfrac{xD}{2(a-x)} + \Delta y \right)^2} - \sqrt{a^2 + \left(\dfrac{xD}{2(a-x)} - \Delta y \right)^2}$

 

[fresh_42]

Spoiler: Question 13

View attachment 287673

Diagram 1View attachment 287674Diagram 2Please refer diagram 1 above. Consider 2 points A and B from the 2 virtual images (named A and B in the question's diagram) such that they lie on the same vertical line. Let A' and B' be the points on the screen where the (virtual) light rays from points A and B respectively hit the screen perpendicular to the surface of the screen. A' and B' will also lie on the same vertical axis/line and distance between them will be the same as the distance between A and B, i.e. $|AB| = d = |A'B'|$ (using the notation $|AB|$ to denote length of line segment $AB$). Let P' be the midpoint of line segment A'B'. Since P' is equidistant from A and B, the light rays from the 2 points hitting P' at any time instant $t$ would be in exactly the same phase (since the 2 virtual images are coherent) with respect to one another and therefore result in a bright fringe caused by constructive interference, i.e. P' lies on the central bright fringe. Let Q' be the point on the screen where the middle line of the next bright fringe (the first bright fringe after and above the central bright fringe passing through P') intersects A'B'. Since this fringe is also caused by constructive interference but distance of Q' and its fringe from B is larger than the distance from A, it must be the case that the light rays from A and B hitting Q' at any instant $t$ differ in the distance traveled by exactly 1 wavelength (λ), i.e. $|BQ'| - |AQ'| = λ$. Hence, the wavelength can be calculated based on the distance between the slit and the screen ($a$), the distance between the 2 virtual slit image points ($d$) and the distance between the fringes ($\Delta y$), as follows.
$|BQ'| = \sqrt{|BB'|^2 + |B'Q'|^2}= \sqrt{|BB'|^2 + (|B'P'| + |P'Q'|)^2} = \sqrt{a^2 + (\dfrac{d}{2} + \Delta y)^2}$

$|AQ'| = \sqrt{|AA'|^2 + |A'Q'|^2}= \sqrt{|AA'|^2 + (|A'P'| - |P'Q'|)^2} = \sqrt{a^2 + (\dfrac{d}{2} - \Delta y)^2}$

∴ $\lambda = \sqrt{a^2 + (\dfrac{d}{2} + \Delta y)^2} - \sqrt{a^2 + (\dfrac{d}{2} - \Delta y)^2}$ (Eq. 1)

The question does not specify the value of $d$, but it can be calculated using trigonometry based on the distances given in the question. According to the diagram given in the question, the bottom most light ray from A and the top most light ray from B cross each other at a horizontal distance of $x$ from the slit, and these 2 light rays hit the screen at points that are at a vertical distance of $D$ from each other. Refer to diagram 2 above which shows the triangles involving the 2 virtual image points (A, B), the points on the screen where the topmost and bottom-most light rays marking the boundary of the region of overlap hit the screen (F, H), and the point of intersection of the 2 rays. Since the 2 triangles are similar, $\Delta ACB \sim \Delta FCH$, the ratios of height to base of the two triangles must be the same (here, I consider the line segments $AB$ and $FH$ to be the bases of the 2 triangles and the perpendicular distance between $C$ and these segments to be the heights), i.e.
$\dfrac{x}{d} = \dfrac{a-x}{D} \Rightarrow d = \dfrac{x}{a-x}D$.

Substituting for $d$ in (Eq. 1), we get the wavelength of the light source in terms of the distances specified in the question:
$\lambda = \sqrt{a^2 + \left(\dfrac{xD}{2(a-x)} + \Delta y \right)^2} - \sqrt{a^2 + \left(\dfrac{xD}{2(a-x)} - \Delta y \right)^2}$

Can you simplify this to $\lambda =\lambda (a,d,\Delta y)$ under the assumption $a \gg d$ and that the beams that run from $A$ and $B$ in direction $P$ on the screen are approximately parallel? And where did you use the interference pattern?

 

[Not anonymous]

Can you simplify this to $\lambda =\lambda (a,d,\Delta y)$ under the assumption $a \gg d$ and that the beams that run from $A$ and $B$ in direction $P$ on the screen are approximately parallel? And where did you use the interference pattern?

Thanks for reviewing the answer. (Eq. 1) in the solution I gave is already an expression involving only $a, d, \Delta y$, so I am not sure I understand what is meant by simplifying to $\lambda (a,d,\Delta y)$. Do you mean a simpler expression that does not involve difference of square roots? I couldn't get any satisfactory simplification by extracting $a$ out of the square root expressions, but I will try again tomorrow.

I am using the interference pattern to derive the difference between the lengths traveled by the 2 rays to reach Q'. The central bright fringe is equidistant from A and B and so the 2 rays interfering there will be in exactly the same phase, i.e. the 2 waves reaching and positively interfering at Q' at any instant $t_1$ would have originated from the 2 images at the same instant $t_0$, and hence the same number of waves would have been emitted along the 2 paths. In case of the next bright fringe (in the solution, I considered the one immediately above the central fringe, but the same applies to the first bright fringe below too), the 2 constructively interfering waves would differ in phase by 1 wavelength, i.e. they would have originated from the sources at time instants that are $\dfrac{1}{f}$ seconds apart, where $f = \dfrac{c}{\lambda}$ is the frequency of the emitted light (assuming that it is okay to ignore the minor variations in speed of light as it passes through prism and non-vacuum air). In general, if we consider the bright fringe that is $n$ fringes above or below the central bright fringe, the waves interfering there would different in phase by $n$ times the wavelength, i.e. they would have been emitted from the source $\dfrac{n}{f}$ seconds apart. Am I missing something?

 

[kshitij]

Do you mean a simpler expression that does not involve difference of square roots?

You can get a simpler expression as follows,

$$\begin{align} |BQ'|^2-|AQ'|^2&= (a^2 + (\dfrac{d}{2} + \Delta y)^2)-(a^2 + (\dfrac{d}{2} - \Delta y)^2)\nonumber\\ (|BQ'|-|AQ'|)(|BQ'|+|AQ'|)&=2 \Delta y \cdot d \nonumber \end{align}$$
Now, $|BQ'|-|AQ'|$ is the path difference which for a maxima (bright fringe) is $n \lambda$ (where $n$ is an integer)
and for $a \gg d$ $|BQ'| \approx |AQ'| \approx a$ hence we get, $$n \lambda= \frac{ \Delta y \cdot d}{a}$$ Which is pretty much the case for every YDSE, and if the question gives the value of angle of prism $A$ and its refractive index $\mu$ we will get $$n \lambda= \frac{2 \Delta y \cdot (2x(\mu-1)A)}{a}$$

 

[kshitij]

You can get a simpler expression as follows,

$$\begin{align} |BQ'|^2-|AQ'|^2&= (a^2 + (\dfrac{d}{2} + \Delta y)^2)-(a^2 + (\dfrac{d}{2} - \Delta y)^2)\nonumber\\ (|BQ'|-|AQ'|)(|BQ'|+|AQ'|)&=2 \Delta y \cdot d \nonumber \end{align}$$
Now, $|BQ'|-|AQ'|$ is the path difference which for a maxima (bright fringe) is $n \lambda$ (where $n$ is an integer)
and for $a \gg d$ $|BQ'| \approx |AQ'| \approx a$ hence we get, $$n \lambda= \frac{ \Delta y \cdot d}{a}$$ Which is pretty much the case for every YDSE, and if the question gives the value of angle of prism $A$ and its refractive index $\mu$ we will get $$n \lambda= \frac{2 \Delta y \cdot (2x(\mu-1)A)}{a}$$

Note that $d$ is equal to the distance between A and B in the figure

which is equal to $\dfrac{x}{a-x}D$ according to the question

 

[Not anonymous]

14. A galaxy is away and oriented in space, such that its rotation axis is perpendicular to the line of sight. The $\alpha$ line of hydrogen is measured to occur at $\lambda_1 = 658.003 \,\text{nm}$ instead of $\lambda_0 = 656.28 \, \text{nm}$ widened to $b = 0.438 \, \text{nm}$.

Assume that the main cause of the widening is the rotation of the stars around the center of the galaxy. Assume further that the different wavelength is only due to the radial motion of the galaxy compared to our solar system.

What is the maximal rotational velocity of the observed stars, and what is the maximal velocity the galaxy is moving, and in which direction as seen from our solar system?

I have not solved before or been taught problems of this type (or anything related to stellar astronomy for that matter, except for a brief introduction to types of galaxies by shape), so I might be totally wrong in this attempted solution. I do not know the real meaning of $\alpha$ line of hydrogen though I surmise it is related to the spectrum of light emitted by that element at certain temperatures and is probably a signature of that element. And I hope that I am correct at least in assuming that the change in wavelength is due to the Doppler effect, which I have studied in the context of sound waves.

Spoiler: Question 14

Let $v_x$ be the velocity at which the galaxy is moving along the line of sight, with a negative value meaning it is moving away from the solar system and a positive value indicating it is moving towards us. Assume that $v_r$ is the maximum rotational velocity of the stars in that galaxy. Since the rotational axis is perpendicular to the line of sight, there would be some (many, in fact) stars rotating towards us (solar system) and others rotating away from us. The stars rotating fastest in a direction away from us would have an effective velocity of $v_A = v_x - v_r$ along the line of sight, while the stars rotating fastest in a direction towards us have an effective velocity of $v_B = v_x + v_r$ along the line of sight. Among all stars in the galaxy, these two groups are stars are the ones that would appear to be moving the fastest and the slowest (or vice-versa, depending on whether the galaxy is moving away or towards us) along the line of sight and hence they should correspond to the 2 extremes of the $\alpha$ line shifted due to Doppler effect. I assume that the widening happens due to the $\alpha$ line of different stars getting shifted by different amounts due to their differing relative velocities w.r.t. solar system. Let $\lambda_{1A}, \lambda_{1B}$ be the shifted wavelengths of the stars moving with effective velocities $v_A, v_B$ respectively. Using Doppler effect formula, we get:

$\lambda_{1A} = \lambda_{0} \dfrac{c - (v_x - v_r)}{c}$
$\lambda_{1B} = \lambda_{0} \dfrac{c - (v_x + v_r)}{c}$
$\lambda_1 = \dfrac{\lambda_{1A} + \lambda_{1B}}{2} = \lambda_{0} \dfrac{c - v_x}{c}$
$\Rightarrow v_x = c - \dfrac{c \lambda_1}{\lambda_0} = c (1 - \frac{658.003}{656.28}) ≈ -787.621 \,\text{km/s}$

Thus the galaxy is moving away from the solar system with a velocity of 787.621km/s along the line of sight.

To derive $v_r$, we use the observation that the width of the Doppler-effected $\alpha$ line would be $b = \lambda_{1A} - \lambda_{1B} = \lambda_0 \dfrac{2 v_r}{c} \Rightarrow v_r = c \dfrac{b} {2 \lambda_0} = c \frac{0.438} {656.28} ≈ 200.219 \,\text{km/s}$.

 

[fresh_42]

I have not solved before or been taught problems of this type (or anything related to stellar astronomy for that matter, except for a brief introduction to types of galaxies by shape), so I might be totally wrong in this attempted solution. I do not know the real meaning of $\alpha$ line of hydrogen though I surmise it is related to the spectrum of light emitted by that element at certain temperatures and is probably a signature of that element. And I hope that I am correct at least in assuming that the change in wavelength is due to the Doppler effect, which I have studied in the context of sound waves.

Spoiler: Question 14

Let $v_x$ be the velocity at which the galaxy is moving along the line of sight, with a negative value meaning it is moving away from the solar system and a positive value indicating it is moving towards us. Assume that $v_r$ is the maximum rotational velocity of the stars in that galaxy. Since the rotational axis is perpendicular to the line of sight, there would be some (many, in fact) stars rotating towards us (solar system) and others rotating away from us. The stars rotating fastest in a direction away from us would have an effective velocity of $v_A = v_x - v_r$ along the line of sight, while the stars rotating fastest in a direction towards us have an effective velocity of $v_B = v_x + v_r$ along the line of sight. Among all stars in the galaxy, these two groups are stars are the ones that would appear to be moving the fastest and the slowest (or vice-versa, depending on whether the galaxy is moving away or towards us) along the line of sight and hence they should correspond to the 2 extremes of the $\alpha$ line shifted due to Doppler effect. I assume that the widening happens due to the $\alpha$ line of different stars getting shifted by different amounts due to their differing relative velocities w.r.t. solar system. Let $\lambda_{1A}, \lambda_{1B}$ be the shifted wavelengths of the stars moving with effective velocities $v_A, v_B$ respectively. Using Doppler effect formula, we get:

$\lambda_{1A} = \lambda_{0} \dfrac{c - (v_x - v_r)}{c}$
$\lambda_{1B} = \lambda_{0} \dfrac{c - (v_x + v_r)}{c}$
$\lambda_1 = \dfrac{\lambda_{1A} + \lambda_{1B}}{2} = \lambda_{0} \dfrac{c - v_x}{c}$
$\Rightarrow v_x = c - \dfrac{c \lambda_1}{\lambda_0} = c (1 - \frac{658.003}{656.28}) ≈ -787.621 \,\text{km/s}$

Thus the galaxy is moving away from the solar system with a velocity of 787.621km/s along the line of sight.

To derive $v_r$, we use the observation that the width of the Doppler-effected $\alpha$ line would be $b = \lambda_{1A} - \lambda_{1B} = \lambda_0 \dfrac{2 v_r}{c} \Rightarrow v_r = c \dfrac{b} {2 \lambda_0} = c \frac{0.438} {656.28} ≈ 200.219 \,\text{km/s}$.

You must use $\Delta \lambda =0.5 b$ so the rotational velocity is only $\approx 100\text{ km/s}.$

 

[Not anonymous]

You must use $\Delta \lambda =0.5 b$ so the rotational velocity is only $\approx 100\text{ km/s}.$

Sorry, I forgot to include 2 in the product expression of the denominator in the final step of the calculation and hence the mistake.

 

[Not anonymous]

15. The spiral galaxy $\text{M81}$ near Ursa Major can already be viewed by a small telescope. It has an apparent magnitude of $\text{M} = 6.9$. The angle to the celestial pole is about Is 21°. it possible to observe $\text{M81}$ the entire year, if you live in Toronto?

Calculate our distance from $\text{M81}$ in lightyears. (Use an average value of magnitude $22.3$ at a pulsation rate of 30 per day and the relation $M = -1.67 - 2.54 \, . \log_{10} p$.)

Spoiler: Question 15 - Part 1

The celestial north pole is visible from the whole of northern hemisphere throughout the year and for people living at a latitude of θ° north, the celestial north pole will appear at an elevation angle of θ°. Stars located at an angle of not more than θ° on any side of the celestial north pole must be visible above the horizon always from a place located at a latitude of θ° north on Earth's surface (though it must be noted that Arctic regions have 24-hours sunshine during summer making other stars invisible even when above horizon). Since Toronto is located at a latitude of approximately 43.5° N, stars that located less than 43.5° from the celestial north pole should be visible in the night sky throughout the year, and hence $\text{M81}$ that is located at a smaller deviation of 21° should also be visible throughout the year.

To calculate the distance of a cepheid variable, we can use the formula $d = 10^\frac{m - M + 5}{5}$ parsecs where $m$ is the apparent magnitude and $M$ is the absolute magnitude. The question tells us to use 22.3 as the average apparent magnitude, so $m = 22.3$. To compute the absolute magnitude, we substitute $p=1/30$ (since question says that the pulsation rate is 30 per day, the pulsation period is $\frac{1}{30}$ days) in the periodicity-luminosity relation formula $M = -1.67 - 2.54 \, . \log_{10} p$, giving $M \approx 2.0819$. And from this we derive the distance to the cepheid C27 (and therefore also an estimate of the distance between Earth and M81): $d \approx 10^\frac{22.3-2.0819+5}{5} \approx 110566.2 \text{parsecs}$, i.e. approximately 360618.73 light years.

Thanks to the websites of some astronomical observatories for making the cepheid luminosity-distance formula easily available on search and for explaining the formula too.

Verify that if a celestial body orbits a center of great mass, then we can calculate the central mass approximately by $M = \dfrac{v^2 \, . r}{G}$. Show by choosing two data-points that the rotation curve of $\text{M81}$ is approximately $v \sim \dfrac{1}{\sqrt{r}}$ for $r = 10 \, \text{kpc}$. What does that mean for the mass distribution in $\text{M81}$? Estimate the mass of M81 within the optical spectrum in units of sun masses.

Spoiler: Question 15 - Part 2

From Newton's laws of motion, we know that the centrifugal force of an object in circular motion is given by $F_c = \dfrac{mv^2}{r}$, where $m$ is the mass of the object, $v$ its orbital velocity and $r$ is the radius of the circular motion. For a celestial object orbiting around a center of great mass $M$ (such as a star orbiting around the center of gravity of a galaxy), when its circular motion is stable (i.e. its orbit is of fixed radius and its orbital velocity is also constant), the outward centrifugal force must be balanced by the gravitational force pulling the body towards the center of great mass. Newton's law of gravity gives the gravitational force acting on the celestial body to be $F_g = \dfrac{GMm}{r^2}$. And since the forces balance out, we get $F_g = F_c$ and this gives a way to estimate the great mass $M$ which the celestial body (of mass $m$) is orbiting. $F_g = F_c \Rightarrow \dfrac{GMm}{r^2} = \dfrac{mv^2}{r} \Rightarrow M = \dfrac{v^2 \, . r}{G}$.

I am not too familiar with the $\sim$ notation in the context of $v \sim \dfrac{1}{\sqrt{r}}$ and I assume it means "proportional too". Since the question asks to validate this relation for $r = 10 \, \text{kpc}$, we choose 2 points from the rotation curve that are fairly close to $r = 10$ and also fall in a region where the curve follows a similar trend. Since the curve for $\text{M81}$ shows a different relationship ($v$ increasing with increasing $r$) between $v$ and $r$ for $r$ up to around 8 kpc compared to the trend for $r \geq 8 \, \text{kpc}$, we choose the points corresponding to $r = 8\, \text{kpc}$ and $r = 12.5\, \text{kpc}$ and calculate $v\sqrt{r}$ at each.

For $r=8$, $v$ is approximately 270 and we get $v\sqrt{r} \approx 763.7$. For $r=12.5$, $v$ is approximately 235 (or is it 240? I find it difficult to find the better approximation from the plot as the axes ticks do not make it easy to interpolate) and we get $v\sqrt{r} \approx 830.4$. Since 763.7 and 830.4 are of similar magnitude and may be treated as approximately equal considering the scale and range of values that $v, r$ take, it seems acceptable to use the approximate relation $v \sim \dfrac{1}{\sqrt{r}}$ for $r = 10 \, \text{kpc}$.

Being able to apply the relation $v \sim \dfrac{1}{\sqrt{r}}$ for some value of $r$ means that the orbital velocity is largely obeying the gravitational force-determined circular motion (the centrifugal force-gravitational force balance described above). This implies that most of the mass of the galaxy is concentrated well within a radius of $10 , \text{kpc}$ from the center of (mass of) the galaxy, so the celestial body can be viewed as orbiting one center of great mass (instead of orbiting around or between 2 or more centers of great masses far apart from each other).

To find the mass of M81 within he optical spectrum, we use the formula $M = \dfrac{v^2 \, . r}{G}$ and substitute $r = 15 \, \text{kpc}$ (since question hints that $r \geq 16 \, text{kpc}$ is outside optical spectrum) and the corresponding value of $v$ taken from the rotation curve for $\text{M81}$ (I take $v \approx 220 \, \text{km/s} =220000 \, \text{m/s}$ for $r = 15 \, \text{kpc} \approx 462851600000000000 \, \text{m}$). This gives $M \approx \dfrac{220000^2 \timex 462851600000000000} {6.674 \times 10^{-11}} \approx 3.356610345 \times 10^{38} \, \text{kg}$, which is approximately equal to 168803670.3908 solar masses.

The rotation curves of $\text{M81}$ and the Milky Way differ a lot for great distances from the center. What does that mean for the mass distribution in our Milky Way?

Spoiler: Question 15 - Part 3

We notice that while the rotation curve of $text{M81}$ follows $v \sim \dfrac{1}{\sqrt{r}}$ from around $r = 8 \, \text{kpc}$ well into large distances (radius distance beyond optical spectrum), the rotation curve for Milky way does not show a similar strong inverse proportionality relation between $v$ and $\sqrt{r}$ even for large values of $r$. Rather, $v$ appears almost constant for $8 < r < 15 \, \text{kpc}$ and decreasing very gradually as $r$ increases beyond $15 \, \text{kpc}$. This suggests that the orbiting around once center of great mass relation described above cannot be applied the same way for Milky way as it could be done for $\text{M81}$. This is most likely because the mass of Milky way is distributed (spread out) over a much larger distance and is not as highly concentrated in a relatively small region (within $r = 10 \, \text{kpc}$ (or even within $r = 20 \, \text{kpc}$ from its center) unlike in case of $\text{M81}$. So the orbiting of many stars cannot be modeled by gravity-controlled circular motion around one center of great mass located well within their center of orbit.

 

[fresh_42]

Spoiler: Question 15 - Part 1

The celestial north pole is visible from the whole of northern hemisphere throughout the year and for people living at a latitude of θ° north, the celestial north pole will appear at an elevation angle of θ°. Stars located at an angle of not more than θ° on any side of the celestial north pole must be visible above the horizon always from a place located at a latitude of θ° north on Earth's surface (though it must be noted that Arctic regions have 24-hours sunshine during summer making other stars invisible even when above horizon). Since Toronto is located at a latitude of approximately 43.5° N, stars that located less than 43.5° from the celestial north pole should be visible in the night sky throughout the year, and hence $\text{M81}$ that is located at a smaller deviation of 21° should also be visible throughout the year.

To calculate the distance of a cepheid variable, we can use the formula $d = 10^\frac{m - M + 5}{5}$ parsecs where $m$ is the apparent magnitude and $M$ is the absolute magnitude. The question tells us to use 22.3 as the average apparent magnitude, so $m = 22.3$. To compute the absolute magnitude, we substitute $p=1/30$ (since question says that the pulsation rate is 30 per day, the pulsation period is $\frac{1}{30}$ days) in the periodicity-luminosity relation formula $M = -1.67 - 2.54 \, . \log_{10} p$, giving $M \approx 2.0819$. And from this we derive the distance to the cepheid C27 (and therefore also an estimate of the distance between Earth and M81): $d \approx 10^\frac{22.3-2.0819+5}{5} \approx 110566.2 \text{parsecs}$, i.e. approximately 360618.73 light years.

Thanks to the websites of some astronomical observatories for making the cepheid luminosity-distance formula easily available on search and for explaining the formula too.

You have to use $p=30$ not $1/30$ but this might be due to a misleading translation on my side. The answer is, therefore, $11.4\, Mly.$

Spoiler: Question 15 - Part 2

From Newton's laws of motion, we know that the centrifugal force of an object in circular motion is given by $F_c = \dfrac{mv^2}{r}$, where $m$ is the mass of the object, $v$ its orbital velocity and $r$ is the radius of the circular motion. For a celestial object orbiting around a center of great mass $M$ (such as a star orbiting around the center of gravity of a galaxy), when its circular motion is stable (i.e. its orbit is of fixed radius and its orbital velocity is also constant), the outward centrifugal force must be balanced by the gravitational force pulling the body towards the center of great mass. Newton's law of gravity gives the gravitational force acting on the celestial body to be $F_g = \dfrac{GMm}{r^2}$. And since the forces balance out, we get $F_g = F_c$ and this gives a way to estimate the great mass $M$ which the celestial body (of mass $m$) is orbiting. $F_g = F_c \Rightarrow \dfrac{GMm}{r^2} = \dfrac{mv^2}{r} \Rightarrow M = \dfrac{v^2 \, . r}{G}$.

I am not too familiar with the $\sim$ notation in the context of $v \sim \dfrac{1}{\sqrt{r}}$ and I assume it means "proportional too". Since the question asks to validate this relation for $r = 10 \, \text{kpc}$, we choose 2 points from the rotation curve that are fairly close to $r = 10$ and also fall in a region where the curve follows a similar trend. Since the curve for $\text{M81}$ shows a different relationship ($v$ increasing with increasing $r$) between $v$ and $r$ for $r$ up to around 8 kpc compared to the trend for $r \geq 8 \, \text{kpc}$, we choose the points corresponding to $r = 8\, \text{kpc}$ and $r = 12.5\, \text{kpc}$ and calculate $v\sqrt{r}$ at each.

For $r=8$, $v$ is approximately 270 and we get $v\sqrt{r} \approx 763.7$. For $r=12.5$, $v$ is approximately 235 (or is it 240? I find it difficult to find the better approximation from the plot as the axes ticks do not make it easy to interpolate) and we get $v\sqrt{r} \approx 830.4$. Since 763.7 and 830.4 are of similar magnitude and may be treated as approximately equal considering the scale and range of values that $v, r$ take, it seems acceptable to use the approximate relation $v \sim \dfrac{1}{\sqrt{r}}$ for $r = 10 \, \text{kpc}$.

Being able to apply the relation $v \sim \dfrac{1}{\sqrt{r}}$ for some value of $r$ means that the orbital velocity is largely obeying the gravitational force-determined circular motion (the centrifugal force-gravitational force balance described above). This implies that most of the mass of the galaxy is concentrated well within a radius of $10 , \text{kpc}$ from the center of (mass of) the galaxy, so the celestial body can be viewed as orbiting one center of great mass (instead of orbiting around or between 2 or more centers of great masses far apart from each other).

To find the mass of M81 within he optical spectrum, we use the formula $M = \dfrac{v^2 \, . r}{G}$ and substitute $r = 15 \, \text{kpc}$ (since question hints that $r \geq 16 \, text{kpc}$ is outside optical spectrum) and the corresponding value of $v$ taken from the rotation curve for $\text{M81}$ (I take $v \approx 220 \, \text{km/s} =220000 \, \text{m/s}$ for $r = 15 \, \text{kpc} \approx 462851600000000000 \, \text{m}$). This gives $M \approx \dfrac{220000^2 \timex 462851600000000000} {6.674 \times 10^{-11}} \approx 3.356610345 \times 10^{38} \, \text{kg}$, which is approximately equal to 168803670.3908 solar masses.

I took $16\,kpc$ and $210\,kms^{-1}$ but that shouldn't lead to a difference of magnitude $10^3$. I assume that you have forgotten the kilo in $kpc$.

My number is $3.2628 \cdot 10^{41}\,\rm kg \approx 164\cdot 10^{9} \,\rm m_{sun}\,.$

Spoiler: Question 15 - Part 3

We notice that while the rotation curve of $text{M81}$ follows $v \sim \dfrac{1}{\sqrt{r}}$ from around $r = 8 \, \text{kpc}$ well into large distances (radius distance beyond optical spectrum), the rotation curve for Milky way does not show a similar strong inverse proportionality relation between $v$ and $\sqrt{r}$ even for large values of $r$. Rather, $v$ appears almost constant for $8 < r < 15 \, \text{kpc}$ and decreasing very gradually as $r$ increases beyond $15 \, \text{kpc}$. This suggests that the orbiting around once center of great mass relation described above cannot be applied the same way for Milky way as it could be done for $\text{M81}$. This is most likely because the mass of Milky way is distributed (spread out) over a much larger distance and is not as highly concentrated in a relatively small region (within $r = 10 \, \text{kpc}$ (or even within $r = 20 \, \text{kpc}$ from its center) unlike in case of $\text{M81}$. So the orbiting of many stars cannot be modeled by gravity-controlled circular motion around one center of great mass located well within their center of orbit.

It can. It only means that we did not see all the mass (dark matter):

The orbital velocity of the Milky Way is almost constant for large distances from its center. So there must be considerable (non-luminous) masses at these distances.

The last answer that you have forgotten is:

Since the $\alpha$-line of hydrogen is blue-shifted, $\lambda_0>\lambda_1,$ Hubble's law does not apply. $\rm M81$ is approaching the Milky Way.

 

[Not anonymous]

I assume that you have forgotten the kilo in $kpc$.

Sorry. Yes, you are right - I forgot to multiply by 1000. Also forgot to answer to the last part of the question relating to Hubble's law - the question was quite huge and I missed including the last part of the answer before posting the reply. That said, I thought that the 2 wavelength values indicate a redshift rather than a blueshift.

Since the $\alpha$-line of hydrogen is blue-shifted, $\lambda_1 > \lambda_0$, Hubble's law does not apply

Assuming that $\lambda_0$ is the original wavelength of the $\alpha$-line of hydrogen (i.e. the wavelength that would be seen by an observer if the light source is relatively stationary w.r.t. the observer), and $\lambda_1$ is the actually observed wavelength of light (or more precisely, that of the $\alpha$-line in that light) from the center of $\text{M81}$ (which is in all likelihood moving, not stationary w.r.t. earth) as observed from earth, doesn't $\lambda_1 > \lambda_0$ indicate a shift towards the red end of the spectrum rather than towards the blue end? My understanding has always been that an increase in wavelength (or in other words, a decrease in frequency) is indicative of a redshift rather than a blueshift. Did I swap the meanings of $\lambda_0$ and $\lambda_1$ leading to this confusion?

Thanks for posting an interesting set of physics questions this month! It was a different type of challenge to solve these and truly a delight, though I look forward to seeing math questions too.

 

[fresh_42]

Sorry. Yes, you are right - I forgot to multiply by 1000. Also forgot to answer to the last part of the question relating to Hubble's law - the question was quite huge and I missed including the last part of the answer before posting the reply. That said, I thought that the 2 wavelength values indicate a redshift rather than a blueshift.Assuming that $\lambda_0$ is the original wavelength of the $\alpha$-line of hydrogen (i.e. the wavelength that would be seen by an observer if the light source is relatively stationary w.r.t. the observer), and $\lambda_1$ is the actually observed wavelength of light (or more precisely, that of the $\alpha$-line in that light) from the center of $\text{M81}$ (which is in all likelihood moving, not stationary w.r.t. earth) as observed from earth, doesn't $\lambda_1 > \lambda_0$ indicate a shift towards the red end of the spectrum rather than towards the blue end? My understanding has always been that an increase in wavelength (or in other words, a decrease in frequency) is indicative of a redshift rather than a blueshift. Did I swap the meanings of $\lambda_0$ and $\lambda_1$ leading to this confusion?

Thanks for posting an interesting set of physics questions this month! It was a different type of challenge to solve these and truly a delight, though I look forward to seeing math questions too.

You are right. I made a mistake. The source where I had the problem from defined $\lambda_0=656.47 \,\rm nm$. I corrected it. My difficulty was, that I couldn't find a source that said the $\alpha$-line is ... If I changed the website, I found another value.

Sorry.

 

[Not anonymous]

It can. It only means that we did not see all the mass (dark matter)

Sorry, I am a bit confused by this. Even if there is a huge amount of unseen mass from dark matter, if most of that dark matter were concentrated near the center of the galaxy, would not the orbit of stars and other celestial bodies orbiting around the galactic center at a medium to large radius approximately obey the relation $v \sim \dfrac{1}{\sqrt{r}}$? Since the mass of anyone star, even if it be a very large one, would be far too small compared to the mass (of both ordinary and dark matter) closer to the center of the galaxy, I thought that the orbital motion of the relatively further out stars around the galactic center could be approximately modeled as a gravity-controlled circular motion of one small mass object around a large mass object if most of the mass exerting significant gravity on the star's orbit is concentrated at the galaxy's center. If the mass is more spread out, i.e. there is large amount of matter (of any type) both near orbit's center and outside the orbit but at a distance comparable to the radius (so not too far out that its garvitational effect on the stars is very small), then can the rotation of stars around the galactic center still be approximated well enough as 1 small mass object orbiting around another much more massive object that can be viewed as almost stationary? If yes, why do we not see the relation $v \sim \dfrac{1}{\sqrt{r}}$ in case of Milky way? I did not mean to say that we cannot explain the orbits using gravitational effects, but rather meant that gravity could still explain the motion but the model would be more complex than 1 small mass object orbiting 1 very large mass object.

I am happy to have got a chance to attempt solving a problem that is related to dark matter, even it were only a very tangential relation and even though I didn't get the answers correct for all parts of the question . I have read about dark matter and dark energy in science articles and the enigma surrounding them only gets deeper every time!

 

[fresh_42]

Sorry, I am a bit confused by this. Even if there is a huge amount of unseen mass from dark matter, if most of that dark matter were concentrated near the center of the galaxy, would not the orbit of stars and other celestial bodies orbiting around the galactic center at a medium to large radius approximately obey the relation $v \sim \dfrac{1}{\sqrt{r}}$? Since the mass of anyone star, even if it be a very large one, would be far too small compared to the mass (of both ordinary and dark matter) closer to the center of the galaxy, I thought that the orbital motion of the relatively further out stars around the galactic center could be approximately modeled as a gravity-controlled circular motion of one small mass object around a large mass object if most of the mass exerting significant gravity on the star's orbit is concentrated at the galaxy's center. If the mass is more spread out, i.e. there is large amount of matter (of any type) both near orbit's center and outside the orbit but at a distance comparable to the radius (so not too far out that its garvitational effect on the stars is very small), then can the rotation of stars around the galactic center still be approximated well enough as 1 small mass object orbiting around another much more massive object that can be viewed as almost stationary? If yes, why do we not see the relation $v \sim \dfrac{1}{\sqrt{r}}$ in case of Milky way? I did not mean to say that we cannot explain the orbits using gravitational effects, but rather meant that gravity could still explain the motion but the model would be more complex than 1 small mass object orbiting 1 very large mass object.

I am happy to have got a chance to attempt solving a problem that is related to dark matter, even it were only a very tangential relation and even though I didn't get the answers correct for all parts of the question . I have read about dark matter and dark energy in science articles and the enigma surrounding them only gets deeper every time!

I think you should ask this in a Cosmology thread. I am no physicist, so I don't know whether my guess that it could be due to DM was correct, or if was due to any other reason.