Astro&109&“Lecture”&12:& Review&for&midterm&exam&ajbaker/ph109/midterm/Lec12-Oct10-F14.pdf · Astro&109&“Lecture”&12:& Review&for&midterm&exam& ... etc.&atdiﬀerentBmes.&

Astro 109 “Lecture” 12: Review for midterm exam

October 10, 2014

ConstellaBon of the day: Aries

•  Zodiac constellaBon visible in the sky at night in northern fall; all stars are relaBvely dim.

•  In Greek mythology, represented the winged ram whose fleece became the Golden Fleece pursued by the Argonauts.

•  Different cultures have idenBfied Aries as a porpoise, a farmhand, etc. at different Bmes.

Image credit: IAU/Sky & Telescope

Why astronomers like Aries •  The dwarf spheroidal galaxy Segue 2 lies 114,000 light-‐years from Earth, is about 220 light-‐years across, and contains only about one thousand stars.

•  Fewer stars than a globular cluster! – but it has its own dark ma\er, so it’s a real galaxy (the least massive known).

Image credit: Kirby et al. (2013)

Preparing for the midterm (i) •  Midterm will be in class on Wednesday, October 15th.

–  If you miss the midterm, you will need a dean’s note before I will let you take a makeup midterm.

–  PracBce explaining ideas, answers to example quesBons, etc. to a classmate, or a friend, or a stranger, or a brick wall. Don’t just listen; if you can explain something, you’ve understood it!

–  PracBce drawing pictures, not just visualizing in your head. •  Four more sets of my in-‐person office hours before the

midterm: today (3:00-‐4:30pm in Sco\ 102), tomorrow (3:00-‐4:30pm in BSC food court), Sunday (3:00-‐4:30 in LSC food court), and next Tuesday (3:00-‐4:30pm in Serin 401). Online availability in Sakai will conBnue.

•  All seven study groups next week available for signup (extra credit even if you a\end Wednesday/Thursday).

Preparing for the midterm (ii) •  You will be asked to occupy every other seat, and to

leave every third row empty (so that I can answer quesBons more easily). If you’re late, come sit at the front of the classroom (we’ll keep seats open).

•  You must bring: –  Photo ID. –  Your student ID (RUID) number. –  A number two pencil.

•  You may bring (opBonally): –  A calculator.

•  You cannot consult: –  Textbook, notes, copies of slides, smart phones, dumb phones, computers, each other, or any other resource.

Preparing for the midterm (iii) •  All material in assigned textbook readings, and all

material in lecture slides that do not have blue backgrounds.

•  Homework quesBons, clicker quesBons, and pracBce midterm quesBons are representaBve.

•  Exam has 30 quesBons, some of which will draw from more than one lecture.

•  Plan for today: –  Discuss ten hardest quesBons from last year’s midterm (from most to tenth most difficult).

–  Answer any quesBons that you have. –  Allow you to work through the remaining 20 quesBons from last year’s midterm, while I float from group to group.

Homework # 5 – QuesBon 1

Which statement about the energy levels of an atom is NOT correct?

A.  Atoms have electronic, vibraBonal, and rotaBonal energy levels. B.  An atom can emit a photon when an electron drops from a

higher energy level to a lower energy level. C.  An atom can absorb a photon when an electron jumps from a

lower energy level to a higher energy level. D.  Every atom of a parBcular element will have the same set of

energy levels.

Atoms don’t vibrate or rotate – only molecules do. It is correct that the energy levels for a given atom are always the same.

Homework # 5 – QuesBon 8 A friend shows you a black-‐and-‐white picture of the night sky taken with a camera that operates at a single wavelength. Two stars in the same constellaBon appear equally bright in this picture. What conclusion can you draw about the properBes of the stars? A.  The two stars are equally bright at all wavelengths. B.  The two stars are equally distant from us. C.  The two stars have idenBcal surface temperature. D.  It is not possible to reach any of the other conclusions

without more informaBon. We cannot tell what color a star is from a black-‐and-‐white picture, so C may not be true – and if the stars do have different temperatures, then A is not true. Stars in the same constellaBon can sBll have different distances, so B may not be true. Bo\om line: D is true by process of eliminaBon.

Homework # 5 – QuesBon 10 Why do the spectra of nearby stars show no emission at certain discrete wavelengths? A.  Atoms in the stars’ cool, outer layers absorb the light from the

stars’ ho\er interiors at wavelengths corresponding to specific energy transiBons.

B.  The stars are missing certain elements, which means there is no light at wavelengths corresponding to those elements’ specific energy transiBons.

C.  These wavelengths correspond to energy transiBons of helium, an element that is only found in stars.

D.  The features are produced by a lack of material in the stars moving at certain velociBes (i.e., they are a consequence of the Doppler effect).

Lack of emission at discrete wavelengths is due to absorpBon by atoms of many elements (not just helium) in the stars’ cooler outer layers – this is how we determine what stars (and the Sun) are made of.

2013 midterm exam – top ten quesBons

Ten most difficult quesBons from 2013 midterm exam were correctly answered by 38% or less of the class… we will go through them from most to tenth most difficult. Common theme: drawing pictures is useful!

2013 midterm exam – QuesBon 22

In any Earth year, there will be a unique night when Mars exhibits the greatest apparent retrograde moBon on the sky. On that night which of the following moBons will contribute to a shir in the wavelengths of spectral lines from the MarBan atmosphere that we can detect from the Earth?

A.  Both Mars’s orbital moBon and its rotaBon about its own axis.

B.  Both Mars’s orbital moBon and Earth’s orbital moBon. C.  Mars’s rotaBon about its own axis. D.  Mars’s orbital moBon. E.  Earth’s orbital moBon.

a bit longer than twice an XXX



Key insight # 1: apparent retrograde moBon will be greatest when Mars is at opposiBon (on opposite side of Earth from Sun), and Earth is at the exact moment of “passing” Mars.




Key insight # 2: shir in wavelengths of spectral lines is due to the Doppler effect, which requires relaBve moBon towards us or away from us along the line of sight.




Key insight # 3: at the moment when Earth is “passing” Mars, their orbital moBons are parallel. No line-‐of-‐sight component of their relaBve moBon means no contribuBon to Doppler effect.

















2013 midterm exam – QuesBon 3 Imagine that you are a Rutgers astronomer who studies supernovae (exploding stars). A new supernova has appeared in a part of the sky that – currently – is above the horizon in New Jersey only during the dayBme. As a result, you can’t study the supernova from New Jersey at visible wavelengths. What can you do to get an observaBon of the supernova at visible wavelengths as quickly as possible? A.  Based on what month it is, call a colleague in either Spain or Japan and

ask him/her to observe the supernova for you. B.  Call a colleague in California (i.e., at the same laBtude but farther west)

and ask him/her to observe the supernova for you. C.  You can’t do anything except wait for the supernova to become

observable at night. D.  Call a colleague in Spain (i.e., at the same laBtude but a quarter of the

way around the world) and ask him/her to observe the supernova for you. E.  Call a colleague in Japan (i.e., at the same laBtude but on the other side of

the world) and ask him/her to observe the supernova for you.

2013 midterm exam – QuesBon 3 Imagine that you are a Rutgers astronomer who studies supernovae (exploding stars). A new supernova has appeared in a part of the sky that – currently – is above the horizon in New Jersey only during the dayBme. As a result, you can’t study the supernova from New Jersey at visible wavelengths. What can you do to get an observaBon of the supernova at visible wavelengths as quickly as possible?

If only above horizon during the dayBme – where does the supernova have to be?


Insight # 1: supernova has to be on the far side of the Sun from the Earth.


Insight # 2: by the Bme supernova is above horizon in {California, Spain, Japan} it will be dayBme there, too.



















2013 midterm exam – QuesBon 7 The planet Endor is famous for its beauBful eclipsing moons – an inner red moon and an outer green moon – that have the same angular size when viewed from the planet’s surface. The two moons are known to have the same mass; one has an orbital period of three Earth days, while the other has an orbital period of 24 Earth days. What conclusion can you draw about the moons’ mean densiBes?

A.  The mean density of the red moon is 512 Bmes higher than that of the green moon.

B.  The mean density of the red moon is 64 Bmes higher than that of the green moon.

C.  The mean density of the red moon is 8 Bmes higher than that of the green moon.

D.  It is not possible to reach a conclusion without knowing whether the red moon has the 3-‐day orbital period or the 24-‐day orbital period.

E.  It is not possible to reach a conclusion without knowing the mass of Endor.

2013 midterm exam – QuesBon 7 The planet Endor is famous for its beauBful eclipsing moons – an inner red moon and an outer green moon – that have the same angular size when viewed from the planet’s surface. The two moons are known to have the same mass; one has an orbital period of 3 Earth days, while the other has an orbital period of 24 Earth days. What conclusion can you draw about the moons’ mean densiBes?






2013 midterm exam – QuesBon 7 The planet Endor is famous for its beauBful eclipsing moons – an inner red moon and an outer green moon – that have the same angular size when viewed from the planet’s surface. The two moons are known to have the same mass; one has an orbital period of 3 Earth days, while the other has an orbital period of 24 Earth days. What conclusion can you draw about the moons’ mean densiBes? [D is wrong since we know Kepler’s third law.]







Red moon’s density ρred = Mred/(4πrred3) Green moon’s density ρgreen = Mgreen/(4πrgreen3) RaBo ρred/ρgreen = (Mred/Mgreen) x (4πrgreen3/4πrred3)


Red moon’s density ρred = Mred/(4πrred3) Green moon’s density ρgreen = Mgreen/(4πrgreen3) RaBo ρred/ρgreen = (Mred/Mgreen) x (4πrgreen3/4πrred3)

= (rgreen3/rred3) = (rgreen/rred)3


Density raBo ρred/ρgreen = (rgreen/rred)3 but what is the raBo of the moons’ radii?


Density raBo ρred/ρgreen = (rgreen/rred)3 but what is the raBo of the moons’ radii?


Density raBo ρred/ρgreen = (rgreen/rred)3 but what is the raBo of the moons’ radii? rgreen/rred = agreen/ared = raBo of their orbital semi-‐major axes.


Density raBo ρred/ρgreen = (agreen/ared)3 but what is the raBo of the moons’ semi-‐major axes? Kepler’s third law:

Pred2 = ared3/MEndor

Pgreen2 = agreen3/MEndor


Density radio ρred/ρgreen = (agreen/ared)3 = (Pgreen/Pred)2 = (24/3)2 = 82 = 64 (doesn’t depend on MEndor!)

Kepler’s third law: Pred2 = ared3/MEndor

Pgreen2 = agreen3/MEndor

2013 midterm exam – QuesBon 7 The planet Endor is famous for its beauBful eclipsing moons – an inner red moon and an outer green moon – that have the same angular size when viewed from the planet’s surface. The two moons are known to have the same mass; one has an orbital period of three Earth days, while the other has an orbital period of 24 Earth days. What conclusion can you draw about the moons’ mean densiBes?






2013 midterm exam – QuesBon 24 A rogue planet is moving at a constant speed into our solar system. Its path bends a li\le bit as it passes by the Sun, but it is travelling so fast that it conBnues to move at a nearly constant speed unBl it exits the solar system. What is the best descripBon of the force of the Sun on the rogue planet?

A.  The Sun exerts a force on the planet that is always directed along a line connecBng the Sun and the planet.

B.  The Sun exerts a force that is always anBparallel (i.e., opposite to) the direcBon of the planet’s moBon.

C.  The Sun exerts a force that is always parallel to the direcBon of the planet’s moBon.

D.  The Sun exerts zero force on the planet, since the planet does not accelerate.

E.  The Sun exerts a force that is always perpendicular to the direcBon of the planet’s moBon.


Does the rogue planet accelerate? YES: it changes direcBon.








What direcBon is the Sun’s gravitaBonal pull (i.e., force)?







2013 midterm exam – QuesBon 23 If the mass of Jupiter were doubled, which distance(s) would need to change for the orbital periods of Jupiter around the Sun and Europa around Jupiter to remain the same as they are today?

A.  The distance of Europa from Jupiter, which would need to be increased, and the distance of Jupiter from the Sun, which would need to be increased.

B.  Only the distance of Jupiter from the Sun, which would need to be increased.

C.  Only the distance of Jupiter from the Sun, which would need to be decreased.

D.  The distance of Europa from Jupiter, which would need to be increased, and the distance of Jupiter from the Sun, which would need to be decreased.

E.  Only the distance of Europa from Jupiter, which would need to be increased.


Remember: Kepler’s third law depends on the mass of the orbitee, not the orbiter. We could replace Jupiter with a potato, and the potato would take the same Bme to orbit the Sun that Jupiter does now.


















E.  Only the distance of Europa from Jupiter, which would need to be increased. [Increases because P2 is proporBonal to a3/M.]


New Brunswick is at 40.5 degrees north laBtude. What is the alBtude of Polaris, the pole star, when seen from New Brunswick?

A.  90 degrees. B.  40.5 degrees. C.  It depends on what Bme of year we are observing it. D.  49.5 degrees. E.  It depends on what Bme of day we are observing it.






A.  90 degrees. [only true if we are at the North Pole!] B.  40.5 degrees. C.  It depends on what Bme of year we are observing it. D.  49.5 degrees. E.  It depends on what Bme of day we are observing it.



What is the alBtude of Polaris if we are at 90 degrees north laBtude? 90 degrees



What is the alBtude of Polaris if we are at 90 degrees north laBtude? 90 degrees

What is the alBtude of Polaris if we are at 0 degrees laBtude (on the equator)? 0 degrees




2013 midterm exam – QuesBon 6 The planet Venus is someBmes nicknamed the “Morning Star” or the “Evening Star” because it only appears in the sky just arer sunset or just before sunrise. What is the best explanaBon for this phenomenon?

A.  When Venus is above the horizon in the middle of the night, it’s in its new phase and therefore not bright enough to see.

B.  Venus rotates on its axis in a retrograde sense. C.  At sunrise and sunset, light from the Sun reflects more easily

off Venus’s thick atmosphere. D.  Venus’s orbit is closer to the Sun than the Earth’s. E.  Venus is only bright when it is close enough to the Sun to be

heated to a high temperature.













morning star


evening star











2013 midterm exam – QuesBon 12 Professor Baker is observing a distant galaxy with the Robert C. Byrd Green Bank Telescope and detects a spectral line of the carbon monoxide molecule at an observed frequency of 30 GHz. What is the observed wavelength of this spectral line?

A.  30 cm. B.  It is not possible to answer this quesBon without more

informaBon about the galaxy being observed. C.  1 cm. D.  1 m. E.  3 cm.


λ = c/ν = (3 x 108 m/s) / (30 GHz) = (3 x 108 m/s) / (30 x 109 Hz) = (3 x 108 m/s) / (3 x 1010 s-‐1) = 10-‐2 m = 1 cm


A.  30 cm. B.  It is not possible to answer this quesBon without more

informaBon about the galaxy being observed. C.  1 cm. D.  1 m. E.  3 cm.

2013 midterm exam – QuesBon 30 Pluto completes two orbits around the Sun in the same Bme that Neptune completes three orbits around the Sun. Is a collision between Pluto and Neptune a possibility? A.  No, since Pluto’s orbit is completely outside Neptune’s. B.  Yes, since the orbital resonance ensures a close encounter

every other orbit by Pluto. C.  No, since Pluto’s orbit is Bpped at 17 degrees to the plane of

Neptune’s orbit. D.  Yes, since their orbits cross, and calculaBons show that they

will collide in about 1 million years. E.  No, because Neptune will never be in Pluto’s path when the

orbits cross.

2013 midterm exam – QuesBon 30 Pluto completes two orbits around the Sun in the same Bme that Neptune completes three orbits around the Sun. Is a collision between Pluto and Neptune a possibility?

From lecture 9: the 3:2 resonance keeps Pluto out of Neptune’s way.

2013 midterm exam – QuesBon 30 Pluto completes two orbits around the Sun in the same Bme that Neptune completes three orbits around the Sun. Is a collision between Pluto and Neptune a possibility? A.  No, since Pluto’s orbit is completely outside Neptune’s. B.  Yes, since the orbital resonance ensures a close encounter

every other orbit by Pluto. C.  No, since Pluto’s orbit is Bpped at 17 degrees to the plane of

Neptune’s orbit. D.  Yes, since their orbits cross, and calculaBons show that they

will collide in about 1 million years. E.  No, because Neptune will never be in Pluto’s path when the

orbits cross.


Why does the Earth have a Bdal bulge on the side opposite the Moon?

A.  The Sun is pulling on the Earth in that direcBon. B.  That is where the Moon’s repulsive force is strongest. C.  The Earth rotates fastest on that side. D.  Lunar gravity is weaker there than at the center of the Earth. E.  Lunar gravity is stronger there than at the center of the Earth.



From lecture 8: the Moon pulls on the far side of the Earth less strongly than on the center of the Earth. (This is true for both rock and water!)



A.  The Sun is pulling on the Earth in that direcBon. B.  That is where the Moon’s repulsive force is strongest. C.  The Earth rotates fastest on that side. D.  Lunar gravity is weaker there than at the center of the Earth. E.  Lunar gravity is stronger there than at the center of the Earth.

Other quesBons?

Example quesBon: lunar phases

Assume that on a certain day, the Moon rises at sunrise and sets at sunset. Seven days later, what would the phase of the Moon be?

A.  full B.  waning crescent C.  new D.  first quarter E.  last quarter

Example soluBon: lunar phases


Top view – not to scale!



Top view – not to scale!

W E



Top view – not to scale! StarBng day:

W E



Top view – not to scale! StarBng day:

Seven days later:



A.  full B.  waning crescent C.  new D.  first quarter E.  last quarter

Example quesBon: gravity Suppose that on the day arer your nephew’s birth, Jupiter (whose mass is about 2 x 1027 kg) has a distance of 1012 m from the Earth. How strong is the gravitaBonal force of Jupiter on your nephew compared to the force he feels from his parents’ 2000 kg car, which is 10 cm away from him in the hospital parking lot?

A.  The gravitaBonal force exerted by Jupiter is 100 Bmes larger than that of the car.

B.  The gravitaBonal force exerted by Jupiter is 10 Bmes larger than that of the car.

C.  The gravitaBonal force exerted by Jupiter is the same as that of the car.

D.  The gravitaBonal force exerted by Jupiter is 100 Bmes smaller than that of the car.

E.  The gravitaBonal force exerted by Jupiter is 10 Bmes smaller than that of the car.

Example soluBon: gravity Suppose that on the day arer your nephew’s birth, Jupiter (whose mass is about 2 x 1027 kg) has a distance of 1012 m from the Earth. How strong is the gravitaBonal force of Jupiter on your nephew compared to the force he feels from his parents’ 2000 kg car, which is 10 cm away from him in the hospital parking lot?

Force due to gravity is F = G m1 m2 / r2 (from cover sheet). But: G is constant, and m1 (nephew’s mass) is constant, so they can be ignored. What are m2/r2 for Jupiter and the car?

For Jupiter: m2/r2 = (2 x 1027)/(1012 x 1012) = 2 x 103 = 2000. For the car: m2/r2 = (2000)/(0.1 x 0.1) = 200,000.

The force exerted by Jupiter is 100 Bmes smaller!

















Example quesBon: gravity Suppose that on the day arer your nephew’s birth, Jupiter (whose mass is about 2 x 1027 kg) has a distance of 1012 m from the Earth. How strong is the gravitaBonal force of Jupiter on your nephew compared to the force he feels from his parents’ 2000 kg car, which is 10 cm away from him in the hospital parking lot?

A.  The gravitaBonal force exerted by Jupiter is 100 Bmes larger than that of the car.

B.  The gravitaBonal force exerted by Jupiter is 10 Bmes larger than that of the car.

C.  The gravitaBonal force exerted by Jupiter is the same as that of the car.

D.  The gravitaBonal force exerted by Jupiter is 100 Bmes smaller than that of the car.

E.  The gravitaBonal force exerted by Jupiter is 10 Bmes smaller than that of the car.

Example quesBon: unit conversion The Joule (J) is the official SI unit of energy, and the relaBonship between the energy of a photon and its frequency is E = hν, where E is in J, ν is in Hz, and h is the “Planck constant” whose value is 6.6 x 10-‐34 J Hz-‐1. Given this informaBon, what is the energy of a photon whose wavelength is 660 nm?

A.  6.6 x 10-‐19 J. B.  1 x 10-‐27 J. C.  3 x 10-‐19 J. D.  3 x 10-‐27 J. E.  6.6 x 10-‐27 J.

Remember: c = λν.

Example soluBon: unit conversion The Joule (J) is the official SI unit of energy, and the relaBonship between the energy of a photon and its frequency is E = hν, where E is in J, ν is in Hz, and h is the “Planck constant” whose value is 6.6 x 10-‐34 J Hz-‐1. Given this informaBon, what is the energy of a photon whose wavelength is 660 nm?

Remember: c = λν.

We know we need to use E = hν, and we know h. We can use ν = c/λ to determine ν. That means E = hc/λ

= (6.6 x 10-‐34 J Hz-‐1)(3.0 x 108 m/s)/(660 x 10-‐9 m) = (6.6 x 10-‐34 J Hz-‐1)(3.0 x 108 m/s)/(6.6 x 10-‐7 m) = (10-‐34 J Hz-‐1)(3.0 x 108 m/s)/(10-‐7 m) = 3.0 x 10-‐19 J


Remember: c = λν.




Remember: c = λν.




Remember: c = λν.




Remember: c = λν.




Remember: c = λν.



Example quesBon: unit conversion The Joule (J) is the official SI unit of energy, and the relaBonship between the energy of a photon and its frequency is E = hν, where E is in J, ν is in Hz, and h is the “Planck constant” whose value is 6.6 x 10-‐34 J Hz-‐1. Given this informaBon, what is the energy of a photon whose wavelength is 660 nm?

A.  6.6 x 10-‐19 J. B.  1 x 10-‐27 J. C.  3 x 10-‐19 J. D.  3 x 10-‐27 J. E.  6.6 x 10-‐27 J.

Remember: c = λν.

Example quesBon: scaling relaBons

In which of the following scenarios would the gravitaBonal force of the Earth on the Moon be the same as it is today?

A.  The Moon is two Bmes more massive than it is today, but has the same distance from the Earth as it does today.

B.  The Moon is two Bmes more massive and four Bmes farther from the Earth than it is today.

C.  The Moon is two Bmes farther from the Earth than it is today, but has the same mass as it does today.

D.  The Moon is four Bmes more massive and two Bmes farther from the Earth than it is today.

Example soluBon: scaling relaBons






Let’s say today’s force F0 = G MM ME / r2 (from cover sheet). New force will be Fnew = G (2MM) ME / r2 = 2 x F0 – nope!







Let’s say today’s force F0 = G MM ME / r2 (from cover sheet). New force will be Fnew = G (2MM) ME / (4r)2 = 1/8 x F0 – nope!






D.  The Moon is four Bmes more massive and two Bmes farther from the Earth than it is today. Let’s say today’s force F0 = G MM ME / r2 (from cover sheet).

New force will be Fnew = G MM ME / (2r)2 = 1/4 x F0 – nope!







Let’s say today’s force F0 = G MM ME / r2 (from cover sheet). New force will be Fnew = G (4MM) ME / (2r)2 = F0 – yep!

Example quesBon: scaling relaBons






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Astro&109&“Lecture”&12:& Review&for&midterm&exam&ajbaker/ph109/midterm/Lec12-Oct10-F14.pdf · Astro&109&“Lecture”&12:& Review&for&midterm&exam& ... etc.&atdiﬀerentBmes.&