Assignment 2 PDF

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Astronomy 207 assignment 2 answers University of Calgary...

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Assignment 2 Due: 11:59pm on Sunday, September 30, 2018 To understand how points are awarded, read the Grading Policy for this assignment.

Problem 2.1

Part A

Which numbered position in the figure shows a planet at opposition?

Hint 1. Study Section S1.1 of The Cosmic Perspective, and remember that opposition means the planet is opposite the Sun in our sky (so that we see it on the meridian at midnight).

ANSWER: 1 3 4 2

Correct Notice that the planet (Jupiter) is on the opposite side of Earth from the Sun, which is why we say that it is at opposition.

Part B

Knowing that the dot near the upper center of the Sun in this photo is the planet Venus, you can conclude that the photo was taken ________.

Hint 1. Study Section S1.1 of The Cosmic Perspective.

ANSWER:

during a transit when Venus is at superior conjunction when Venus is at greatest western elongation when Venus is at greatest eastern elongation

Correct It's a transit because Venus is transiting across the face of the Sun.

Part C

On this diagram of the celestial sphere, the numbered point(s) that have declination 0º are __________.

Hint 1. Review the definition of declination in Section S1.2 of The Cosmic Perspective.

ANSWER:

1 only 1 and 3 2 only 2 and 4

Correct Point 1 is the March equinox and Point 3 is the September equinox; both are on the celestial equator, so both have declination 0º.

Part D

Among the points and circles shown on this diagram, which one(s) have varying right ascension?

Hint 1. Review the definition of right ascension in Section S1.2 of The Cosmic Perspective.

ANSWER:

the celestial equator and the ecliptic only the celestial equator Earth's equator and the celestial equator the north and south celestial poles

Correct Right ascension varies as you move east or west around the celestial sphere, which means it varies along both the celestial equator and the ecliptic.

Part E

On this map of Earth, which numbered circle represents the place where the Sun appears directly overhead at noon on the December solstice?

Hint 1. Study Section S1.2 of The Cosmic Perspective.

ANSWER:

3 4 2 1 None of the answers are correct; the Sun can be directly overhead only on the equator.

Correct Circle 3 is the Tropic of Capricorn, which has latitude 23.5ºS and hence is the latitude at which the Sun passes directly overhead at noon on the December solstice.

Part F

Suppose you go outside and see the view shown when you look due south. Knowing that the stars of Orion's Belt lie close to the celestial equator, what can you conclude?

Hint 1. Study Section S1.2 of The Cosmic Perspective.

ANSWER:

You are in the Northern Hemisphere. It is winter. In a couple of hours, Lepus will be higher in the sky than Orion. You are in the Southern Hemisphere.

Correct You can tell you are in the Northern Hemisphere because you are seeing stars along the celestial equator when you look due south. Remember that the celestial equator goes across the southern sky only for locations in the Northern Hemisphere.

Part G

The celestial coordinates of the red dot on this diagram of the celestial sphere are approximately _______.

Hint 1. Study Section S1.2 of The Cosmic Perspective.

ANSWER: Dec = -60º, RA = 12 hr Dec = +60º, RA = 6 hr Dec = -60º, RA = 6 hr Dec = +60º, RA = 12 hr

Correct The declination is +60º because the dot is 60º north of the celestial equator, and the RA of 6hr indicates that it is half or the way around the celestial sphere from the circle labeled 0hr.

Part H

This sequence of photos shows the Sun at one-hour intervals throughout the day of the June solstice; directions and times are indicated. Where was this photo taken?

Hint 1. Think about the motion of the Sun as seen from different latitudes, and study Section S1.2 of The Cosmic Perspective.

ANSWER: the equator the North Pole the Antarctic Circle the Arctic Circle

Correct You can tell it is the Arctic because the Sun skims the northern horizon at midnight.

Part I

This ancient navigational instrument is called a(n) _______.

Hint 1. Note that this figure appears in Section S1.3 of The Cosmic Perspective.

ANSWER: analemma cross-staff astrolabe celestial sphere

Correct The astrolabe was invented by the ancient Greeks and greatly improved by Islamic scholars of the Middle Ages.

Part J

The figure shows the equation of time; notice that the vertical axis shows "apparent solar time minus means solar time." Suppose the date is about Feb. 15 and you have a digital watch that is set to the local mean solar time. If the watch say sit is precisely noon, then _________.

Hint 1. Note that this figure appears in Section 3.1 of The Cosmic Perspective.

ANSWER:

the Sun is now crossing your meridian the Sun crossed your meridian about 15 minutes ago the Sun will reach your meridian in about 15 minutes the Sun is now at its highest point in your sky

Correct You know this because the graph shows that on February 15, your mean solar clock is ahead of apparent solar time by about 15 minutes, and apparent solar time tells you the actual position of the Sun.

Problem 2.2 Learning Goal: To understand the difference between a solar day and a sidereal day. Introduction. Study the figure , which shows this demonstration of why our 24-hour solar day is slightly longer than the sidereal day: Set an object representing the Sun on a table, and stand a few steps away to represent Earth. Point at the Sun and imagine that you are also pointing toward a distant star that lies in the same direction. If you rotate in place, you'll again point at both the Sun and the star after one full rotation. However, to show that Earth also orbits the Sun, you should take a couple of steps around the Sun while you rotate. In that case, you will again point in the direction of the distant star after one full rotation, but you need to rotate a little extra to point back at the Sun. The actual "extra" rotation needed by Earth is about 1 per day.

Part A In the demonstration, 360 of rotation (one full rotation) represents a sidereal day. You can actually measure the length of the sidereal day by measuring the time from when __________ crosses your meridian on one day (or night) until it crosses the meridian on the next day (or night). Check all that apply.

Hint 1. What does the word sidereal mean? The word sidereal means "related to __________." ANSWER: the stars the Sun rotation

ANSWER:

the Sun the star Sirius the Moon a GPS satellite the star Vega

Correct In fact, any star can be used to measure the length of the sidereal day, because the daily motion of stars across the sky is due only to Earth's daily rotation.

Part B As shown in the figure, a solar day requires about an extra 1° of rotation, or a total of about 361° of rotation for Earth. Therefore, a solar day is longer than a sidereal day by about __________.

Hint 1. How long is a solar day? On average, the length of a solar day is __________. ANSWER: 24 hours 23 hours, 56 minutes 24 hours, 10 minutes

ANSWER: 1 minute 1/360 of a minute 24 hours 1/360 of 24 hours

Correct This works out to be about 4 minutes, so the sidereal day is about 23 hours, 56 minutes.

Part C Which of the following changes would cause the solar day to be shorter (rather than longer) than the sidereal day?

Hint 1. What planet has a solar day shorter than its sidereal day? Which planet has a solar day that is shorter than its sidereal day? ANSWER: Mercury Venus Jupiter

Hint 2. How to approach the problem Be sure you have actually tried the demonstration for yourself. Note that the reason that Earth's solar day is longer than its sidereal day is that you have to rotate a bit "extra" to demonstrate the solar day. If you try each of the options offered in Part C, you will find the option that makes the solar day require less rotation (rather than "extra" rotation) than the sidereal day.

ANSWER:

Decreasing Earth's orbital speed around the Sun Increasing Earth's rotation period Increasing Earth's orbital speed around the Sun Decreasing Earth's rotation period Reversing the direction of Earth's rotation

Correct In our solar system, Venus is the only planet that rotates "backward" relative to its orbit, so it is the only planet with a solar day that is shorter than its sidereal day.

Part D Suppose that Earth orbited the Sun 10 times as fast as it actually does but kept the same rotation period it has now. Which of the following would be true?

Hint 1. How to approach the problem The key to answering this question is to recall from Part B that the solar day is about 4 minutes longer than the sidereal day because Earth moves around its orbit by about 1 per day. An orbital speed 10 times as fast would mean that Earth would move about 10 per day around its orbit. How would that affect the time difference between the solar and sidereal day?

ANSWER:

The solar day would still be 24 hours, but the sidereal day would be a little less than 23 1/2 hours. The solar day would be 10 times as long, or about 240 hours. The sidereal day would still be 23 hours, 56 minutes, but the solar day would be a little over 24 1/2 hours. The solar day would actually be a bit shorter than the sidereal day.

Correct The sidereal day is unchanged because Earth's rotation period is unchanged. To understand the change in the solar day, recall that a solar day is about 4 minutes longer than the sidereal day because Earth moves around its orbit by about 1 per day. Therefore, if Earth orbited 10 times as fast, it would move 10 times as far around its orbit each day, making the solar day longer than the sidereal day by about .

Part E The Moon takes about 1 month to orbit Earth (more precisely, about 27 1/3 days), and it travels in the same direction that Earth orbits the Sun. Therefore, during the time it takes the Moon to complete one orbit around Earth, Earth moves __________ around its orbit of the Sun.

Hint 1. How to approach the problem Look carefully at the figure. Notice that the reason Earth moves about 1 per day around the Sun is that there are 365 days in a year—and 1/365 is very close to 1/360, or 1 . So how far does Earth move along its orbit in one month?

ANSWER:

about 12 about 1 exactly 27.333 about 1/12 of the way

Correct Earth takes a year to orbit the Sun, and there are about 12 months in a year, so during the 1 month that it takes the Moon to orbit Earth, Earth must move about 1/12 of the way around its orbit of the Sun.

Part F Based on your answer to Part E, the time from one new moon to the next must be _________.

Hint 1. How to approach the problem This diagram shows the geometry of the Moon's orbit around Earth relative to Earth's rotation around the Sun. Note that 30 is 1/12 of 360 , so the extra 30 of rotation arises because it takes about one month (1/12 of a year) for the Moon to orbit Earth.

ANSWER: shorter than the Moon's actual orbital period by about 1/12 of the 27-day orbital period shorter than the Moon's actual orbital period by about 1/12 of a year longer than the Moon's actual orbital period by about 1/12 of a year longer than the Moon's actual orbital period by about 1/12 of the 27-day orbital period

Correct Just as the solar day is longer than the sidereal day because Earth rotates and orbits in the same direction, the fact that the Moon orbits in the same direction as Earth means that its synodic period (from new moon to new moon) must be longer than its actual orbital period. The difference is 1/12 of the 27-day orbital period, which is just over 2 days. That is why the Moon's cycle of phases lasts about 29 1/2 days on average, or about 2 days longer than the Moon's orbital period of 27 1/3 days.

Problem 2.3

Part A Define opposition, conjunction, and greatest elongation for planets both closer to and farther from the Sun than Earth. Drag the items on the left to the appropriate blanks on the right to complete the sentences. (Items can be used once or more than once.) ANSWER:

Reset

greatest elongation opposition conjunction

Opposition

Help

occurs when a planet is on the opposite side of Earth from the Sun. This can only

happen for planets located farther than Earth from the Sun. (So Venus and Mercury are never in opposition .)

Conjunction

occurs when a planet is aligned with the Sun in our sky, either between the Sun and

Earth or on the opposite side of the Sun from Earth. For planets beyond Earth, only a superior conjunction

is possible. For Mercury and Venus, which orbit closer to the Sun than we do, an inferior

conjunction

also occurs.

Greatest elongation

only occurs for planets closer to the Sun than Earth that is, for Mercury and

Venus. This is the arrangement in which the planet will appear to be as far from the Sun in our sky as it will get during an orbit.

Correct

Problem 2.4

Part A Each item following represents an amount of time. Recall from your reading that “sidereal” refers to events that are timed with respect to the distant stars, and “synodic” refers to special alignments of astronomical bodies, such as the Earth, Moon, and Sun. Rank the items from left to right in order of the amount of time they represent, from shortest time to longest time. If two items represent equal amounts of time, show this equality by dragging one on top of the other.

Hint 1. A strategy to help you The following strategy should help you complete this task: Three of the items on the list are related to Earth’s daily rotation, which means they are close in length to a day, while the other four items all are related to the Moon’s orbit around Earth, which means they are close in length to a month. So the first step in completing the task is to identify which three items are close to a day and which four are close to a month. Next, review the definitions of various sidereal and synodic periods on the list (open other hints if you need help). You will find that other items on the list are equal to each of these periods; be sure you remember to show an equality between two items by dragging one on top of the other. Hint 2. How do we measure a sidereal day? A sidereal day is __________. ANSWER: the time it takes for any star to go from being on the meridian one day until it is again on the meridian the next day the time it takes for the Sun to go from being on the meridian one day until it is again on the meridian the next day the time it takes Earth to complete one orbit of the Sun

Hint 3. How do we measure a synodic month? A synodic month is __________. ANSWER:

equal to a sidereal month the actual orbital period of the Moon the time from one new moon to the next new moon

ANSWER:

Reset

sidereal day

one Moon orbit around Earth

Help

one full cycle of moon phases

solar day one rotation of Earth on its axis

sidereal month

synodic month

Correct The first three times relate to Earth’s rotation: A sidereal day is equal to Earth’s rotation period of about 23 hours and 56 minutes; our 24-hour solar day is slightly longer because Earth is moving around its orbit at the same time that it rotates. The last four items are related to the Moon’s orbit around Earth, The Moon’s true orbital period is the sidereal month, which is about 27 1/2 days. The synodic month, which is equal to the time from one new moon to the next, is longer (about 29 1/2 days) because Earth and the Moon are moving around the Sun at the same time that the Moon orbits Earth.

Problem 2.5

Part A For what planets do we sometimes observe a transit? Check all that apply. ANSWER:

Mars Pluto Jupiter Mercury Neptune Saturn Venus Uranus

Correct

Part B Why we observe a transit? Drag the items on the left to the appropriate blanks on the right to complete the sentences. (Not all terms will be used.) ANSWER:

Reset

conjunction tilted when a planet is on the opposite side of Earth from

Help

A transit occurs when a planet appears to pass in front of the Sun's disk as seen from Earth , so it can occur only when the planet is in inferior conjunction . Moreover, because Mercury and Venus's orbits are tilted

relative to Earth's orbit, the planets usually are slightly north or south of the

Sun at inferior conjunction . However, when the planet's orbit lines up just right, we get a transit.

the Sun opposition not tilted when a planet appears to pass in front of the Sun's disk as seen from Earth

Correct

Problem 2.6

Part A Listed following are observable characteristics of equinoxes and solstices in the continental United States (which means temperate latitudes in the Northern Hemisphere). Match each characteristic to the corresponding equinox or solstice.

Hint 1. What is the Sun’s path when its declination is 0°? Which of the following is not true when the Sun has declination = 0°? ANSWER:

The Sun rises due east. The Sun sets due west. It is the summer solstice. The Sun is located on the celestial equator.

ANSWER:

Reset

Help

the noontime Sun casts the longest shadows the Sun rises due east today, but will rise slightly north of due east tomorrow

the noontime Sun reaches its highest point of the year

the Sun has declination 0 degree today, but will have a negative declination tomorrow

sunset occurs at its farthest point south of due west for the year

longest day (most daylight) of the year the Sun crosses the meridian 23.5 degrees lower in altitude than the celestial equator

Correct

Problem 2.7

Part A Match the words in the left-hand column to the appropriate blank in the sentences in the right-hand column. Use each word only once.

Hint 1. What apparent solar time is Apparent solar time is time measured by the actual position of the Sun in your local sky, defined so that noon is when the Sun is on the meridian. Hint 2. What mean solar time is TMean solar time is time measured by the average position of the Sun in your local sky over the course of the year. Hint 3. What standard time is Standard time is time measured according to the internationally recognized time zones. Hint 4. What daylight saving time is Daylight saving time is standard time plus one hour, so that the Sun appears on the meridian around 1 p.m. rather than around noon. Hint 5. What universal time is Universal time is standard time in Greenwich (or anywhere on the prime meridian).

Hint 6. What a leap year is A leap year is a calendar year with 366 rather than 365 day.; Our current calendar (the Gregorian calendar) has a leap year every 4 years (by adding February 29), except in century years that are not divisible by 400.

ANSWER:

Reset

Help

1. When the Sun casts the shortest shadows of the day, it is noon according to apparent solar time . 2. The length of the solar day i...