homework 1 math background 20 21 answer key

Astronomy 101

Homework answer key: homework 1.

The year we use, i.e., the one which is the basis for our calendars, is the "tropical" year, so called because it is determined by the passage of the seasons. The tropical year is based on the position of the Sun with respect to the orientation of the Earth's rotation axis. Remember that it is the tilt of Earth's rotation axis that causes us to have seasons. If the rotation axis of Earth were fixed in space (i.e., with respect to the fixed stars), then the tropical and sidereal years would be identical. The Earth, however, precesses. That is, the rotation axis wobbles or changes direction. This wobble is a very precisely measurable behavior in which the direction of the rotation axis moves in a circle (as seen on the celestial sphere) just as a spinning top would wobble. This wobble is termed "precession." The rotation axis precesses very slowly, with a period of 26,000 years.

Because the Earth precesses, the orientation of the north pole with respect to the Sun is not fixed to the orbital period of Earth around the Sun. Thus, the first day of summer (or winter, etc.) "moves" to a different spot in Earth's orbit, with each successive orbit. This "first day" position moves by about 20 minutes with each orbit. (If you divide 365.256 days = 525,969 minutes by 20 minutes per year, you will get about 26,000 years.) If we used the sidereal year for our calendars, the seasons would migrate through the calendar year, with summer in the northern hemisphere eventually occurring in December (albeit in 13,000 years). Thus, instead, we use a tropical year (365 days, 5 hours, 48 minutes, 29 seconds; 365.2422 days) for our calendars, tossing in lead days as needed to make up for the difference between the non-integer number of days in the orbit and the integer number of days on our January-December calendars. This compensates for the precession and keeps the first day of summer on or about June 21.

Note that if the position of the rotation axis changes, then the plane of the celestial equator (which is perpendicular to the rotation axis) must also change. Therefore, the two "nodes," the two points at which the Sun's apparent orbit (the ecliptic) intersects the celestial equator, also changes. It is these two positions, the two nodes, that migrate around the ecliptic at the rate of 20 minutes per year. These two nodes are also used by astronomers to determine the first moment (or day) of spring and fall, called the vernal and autumnal equinoxes. Consequently, this whole process we are discussing is often referred to as the precession of the equinoxes.

A "sidereal" day (a day with respect to the stars) is the time for Earth to rotate exactly once with respect to the stars. This time period is approximately 23 hours 56 minutes. Because Earth orbits the Sun while it rotates, it moves in its (nearly) circular orbit around the Sun by about 1 degree in about one day (actually, 360 degrees in 365.25 days, or 0.986 degrees per day), the angle between Earth, the stars and the Sun changes every day and at the end of a sidereal day, the Sun is not yet directly overhead. Thus, a sidereal day would be a very inconvenient way to measure time.

• Assume you live at the equator of the Earth. How fast are you moving through space if Earth rotates? (Your final answer may be in either miles/sec or km/sec.)

[Answer] For this problem, ignore the Earth's orbital motion around the Sun (and the Sun's orbital motion in the galaxy and the galaxies motion relative to other galaxies, etc.) We assume the Earth is a sphere and we live at the equator. As the Earth spins we are carried with the surface of the Earth in a big circle. The circle is the circumference of the Earth. The velocity is the distance traveled divided by the time needed to travel that distance.

• D = distance traveled = circumference [not area] of a circle
• v = velocity = D/T
• pi = 3.1415
• R = radius of a circle (in this case the radius of the whole Earth)
• D = distance traveled = 2 x pi x R
• T = time to travel distance D
• In this case, diameter of the Earth = 2R = 12,756 km, so D = 4.01 x 10^4 km
• T = one day = 1 [day] x 24 [hr/day] x 60 [min/hr] x 60 [sec/min] = 86,400 seconds
• v = D/T = 4.01 x 10^4 km / 8.64 x 10^4 sec = 0.46 km /sec = 0.29 miles/sec
• Assume you rent a room from Santa Claus at the Earth's North Pole. How fast are you moving through space if Earth rotates?

[Answer] Now, we have the same problem to solve except that we live at the north pole. At the north pole, we are standing on the Earth's rotation axis so as the Earth spins, we rotate in a circle but that circle has no size - the radius of the circle is zero. So we don't move any distance. (Our distance from Earth's rotation axis is zero). Thus, T = one day = 24 hours (no change) but R = 0 and so v = D/T = 0/1 day = 0 km/sec  

• A toughie, for the amusement of those who want a bigger challenge: Assume you live in Nashville, TN. How fast are you moving through space if Earth rotates? (Note: the northern latitude of Nashville is 36 degrees.

Clearly, our velocity changes depending on whether we are at the equator or at the pole. We would expect that the velocity at Nashville would be somewhere between v = 0 and v = 0.46 km/sec.

T still equals one day

but if v is different (as we predict), then the distance traveled must be different.

For D to be different, R must be different. So the question boils down to asking "what is R?" Well, R is the radius of the circle traveled and must be the distance along a perpendicular line connecting Nashville to the Earth's rotation axis, not the distance of Nashville from the center of the Earth. To compute this requires some trigonometry.

Draw a right triangle. Label the hypotenuse as the radius of the Earth. The short (horizontal) side of the triangle is "R"; the second (vertical) side of the triangle is "Earth's rotation axis" pointing from the center of the Earth to the north. "Y" is the angle between two lines from the center of the Earth, on e pointing toward a point on the surface of the Earth at the equator, one pointing toward Nashville.

__________ @ Nashville |*******Y*/ |********/ |*******/ |******/radius of Earth |*****/ |****/ |***/ |**/ |*/ |/_Y__________________ equator of Earth @ center of the Earth

Y = Nashville's latitude = 36 degrees. Simple geometry tells us that the small angle of the triangle at the Nashville apex is also Y degrees.

R = 6378 km x cos(latitude) = 6378 km x cos (36), or R = 6378 km x sin(54) R = 5160 km

so for the distance traveled we have

D = 2 x pi x R = 2 x 3.14 x (5,160 km) = 3.24 x 10^4 km

and v = D/T = 3.24 x 10^4 km / 8.64 x 10^4 sec = 0.37 km/sec = 0.23 miles/sec. 4. Assume for simplicity that Earth's orbit around the Sun is a perfect circle. How fast (at what speed or velocity) is Earth moving through space as it orbits the Sun? (Your final answer may be in either miles/sec or km/sec.)

The velocity of the Earth is the distance traveled divided by the time needed to travel that distance.

If the Earth's orbit is a circle, then the distance traveled by the Earth is the circumference of the circle, where the radius is the (average) distance of the Earth to the Sun (the Astronomical Unit). Note that this is not the radius of the Sun. Also note that you can essentially think of the AU as the distance from the center of the Sun to the center of the Earth.

So, we calculate:

v = velocity pi = 3.1415 D = distance traveled = 2 x pi x R R = 1 AU = 1.5 x 10^8 km T = time for Earth to orbit the sun = 365.25 days = 365.25 days x 8.64 x 10^4 sec/day = 3.15 x 10^7 sec v = D/T = 2 x pi x 1.5 x 10^8 km / 3.15 x 10^7 sec v = 29.86 km/sec = 18.56 miles/sec  

Twice per orbit (about every 6 months), because the orbit of the Earth would eventually carry the "line of nodes" into the Earth-Sun line. This would be true no matter what the tilt of Earth's rotation axis.

Home > CC1 > Chapter 3 > Lesson 3.1.1 > Problem 3-20

Study the dot pattern below.   

Can you see a pattern in the completed figures? Use this pattern to complete figures 3 and 5.

Use the eTool below to create the patterns for Fig. 3 and Fig. 5. Click the link at right for the full version of the eTool: 3-20 HW eTool

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