Stars Sun and Moon Research Paper

Pages: 7 (1988 words)  ·  Bibliography Sources: 7  ·  Level: Master's  ·  Topic: Astronomy

Sun, Moon and Stars

Data and results from your diameter of the sun activity:

Image diameter: .93 cm

Distance from pin-hole to image: 100 cm

.95 cm x 150000000km / 100cm

Calculated diameter of the sun: 1425000 km

How does your calculated value for the diameter of the sun compare with its actual value?

My calculated value for the diameter of the sun was larger 1,425,000km, which is larger than the actual diameter of the sun, which is 1,391,978km (Cain, 2012).

If the values are different, describe possible reasons for the difference.

There is a huge difference in order of magnitude between the figures I was using to calculate the distance and the total distance. This difference of 33022km is actually a very small percentage ( 33022km/1391978km= n/100), which is actually less than 2.4% error. That difference could simply be because I measured to two places rather than three places in my centimeter measurement.

Can you think of any improvements that could be made to the procedure? Using a graph paper with more lines, thus allowing for a more precise measurement might be enough to remove or reduce the margin of error.

2. Kepler's third law, P2 = a3, applies to any object orbiting the sun. Newton was able to derive Kepler's third law using his law of gravity. Newton's version includes the mass of both objects, P2 = a3 / (M1 + M2), and can be used for any object that orbits any astronomical body. In this formula, the masses are measured in special units called solar mass units. The mass of the sun is equal to one solar mass unit. a. If the mass of the second object is very small compared with the first mass, then, to a good approximation, P2 = a3 / M1. Solving for the mass, we get M1 = p2 / a3. Use this mass formula to determine the mass of Jupiter using data from its moon Sinope: period of orbit is 2.075 years, average orbital distance is 0.158 astronomical units.

B. calculated mass of Jupiter: .076 solar mass units

M1 = .158 AU / 2.075 years

M1 = 2.075 years / .158 AU

M1 = .076 Solar Mass Units

C. You can convert your result above into kilograms by multiplying it by the mass of the sun in kilograms: 2.00 a -- " 10-30 kg.

.076SM * 2.00 x 10^30kg =

D. calculated mass of Jupiter: 1.52*10^27kg

E. Compare your calculated mass of Jupiter (kg) to the actual value. How close did you get? Explain any difference.

The actual mass of Jupiter is 1.9 *10^27kg (Coffey, 2008). These figures are not very close. First, the mass of Sinope is not negligible; it is 8 x 1016kg, which cannot be completely disregarded (Enchanted Learning, 2010). Moreover, each of the figures was estimated, which can make a difference in calculations.

3. Research Mauna Kea Observatories -What type of telescope is used? What is the size of its mirror or radio dish? What specific wavelengths of light can be studied using this device? Why would these wavelengths of light be useful to astronomers?

There are several different types of telescopes used near the summit of Mauna Kea. Of the thirteen working telescopes on Mauna Kea, nine of them are for "optical and infrared astronomy, three of them are for submillimeter wavelength astronomy and one is for radio astronomy. They include the largest optical/infrared telescopes in the world (the Keck telescopes), the largest dedicated infrared telescope (UKIRT) and the largest submillimeter telescope in the world (the JCMT). The westernmost antenna of the Very Long Baseline Array (VLBA) is situated at a lower altitude two miles from the summit" (University of Hawaii Institute for Astronomy, 2012). The size of the UH telescope mirrors are .6m and 2.2m (University of Hawaii Institute for Astronomy, 2012). The size of the IRTF mirror is 3.0m (University of Hawaii Institute for Astronomy, 2012). The size of the CFHT mirror is 3.6m (University of Hawaii Institute for Astronomy, 2012). The size of the UKIRT is 3.8m (University of Hawaii Institute for Astronomy, 2012). The size of Keck I is 10m (University of Hawaii Institute for Astronomy, 2012). The size of Keck II is 10m (University of Hawaii Institute for Astronomy, 2012). The size of Subaru is 8m (University of Hawaii Institute for Astronomy, 2012). The size of Gemini is 8m (University of Hawaii Institute for Astronomy, 2012). The size of the JCMT is 15m (University of Hawaii Institute for Astronomy, 2012). The JCMT operates in the submillimeter wavelength range, which allows people to investigate dust and gasses from our solar system and from other galaxies (University of Hawaii Institute for Astronomy, 2012).

Part 2

1. Data and results from your sun position activity: (you can also find this out from sources on astronomy daily updates websites) Location of observations: Gaithersburg, MD

Observation (1) date October 31, 2012 time 6:08 P.M.

Observation (2) date November 1, 2012 time 6:07 P.M.

Observation (3) date November 2, 2012 time 6:06 P.M.

Observation (4) date November 3, 2012 time 6:05 P.M.

Observation (5) date November 4, 2012 time 5:04 P.M.

Observation (6) date November 5, 2012 time 5:03 P.M.

During the course of your observations, what was the general direction of movement of the sun's position at sunset?

The general direction of the movement's of the sun's position at sunset appears to be south of due west (Pandian, 2002).

Did the sun's position ever reverse direction? No.

Using the methods laid out for the sun position activity, estimate the total amount of angular change in the sun's position at sunset over the course of your observations.

The suns moves about one degree per day, so that over the course of six days of observation, there appears to be 6 degrees of change.

What is this amount? 6 degrees.

2. Data and results from your moon observation activity: Location of observations: Gaithersburg, MD

Observation (1) date November 2, 2012 time 8:52 P.M.

Observation (2) date November 3, 2012 time 9:45 P.M.

Observation (3) date November 4, 2012 time 9:41 P.M.

Describe any changes in the moon's appearance from observation to observation.

The moon was waning gibbous from the period from November 2 to November 4th. This means that it began relatively large and got steadily smaller.

"Describe any change in the moon's position in the sky from observation to observation.

Since there are 360 degrees in a circle, the Moon moves 360 / 27.3 or 13.2 degrees per day relative to the stars, which is just over half a degree per hour, and approximately equal to its apparent size. This means that from night to night, the Moon moves a little more than one hand-width to the East (the direction of its motion around the Earth) relative to the stars, and from hour to hour, it moves about one diameter to the East, among the stars" (Seligman, Unk.)

Using the methods laid out for the moon observation activity, estimate the daily amount of angular change in the moon's position. What is this estimate? The angular change is approximately 13 degrees per day (Seligman, Unk.).

4. Radioactive decay is one of the sources of the heat that drives Earth's geologic activity. Radioactive decay also enables us to date rocks and to determine the age of Earth and other solar system bodies. In this activity, you will simulate the Radioactive decay of 36 atoms of a rare isotope of uranium, U-235. U-235 has a half-life of 700 million years. Gather 36 coins and arrange them in a 6 a -- " 6 grid with all of the coins facing heads-up. Flip each coin once and place it back in its original location. This represents the passage of one half-life (700 million years, for this example). The coins that come up heads represent atoms that have not yet decayed; the coins that come up tails represent atoms that have decayed. Record the number of heads below. Next, flip each of the remaining heads-up coins once and place it back in its original location. 1.4 billion years have now passed (2 a -- " 700 million). Record the number of remaining heads below. Repeat this process until all coins are tails-up

36 Original number of U-235 atoms

21 Remaining number U-235 atoms after first flip

9 Remaining number U-235 atoms after second flip

5 Remaining number U-235 atoms after third flip

2

Remaining number U-235 atoms after fourth flip

0 Remaining number U-235 atoms after fifth flip

How many half-lives did it take for all of the atoms to decay?

Five

To how many years does that equate?

5*700 million = 35 million years

Do you think everyone in class will get the same answer? Why or why not?

No, because somewhere near half of the coins should have come up heads with each flip, but the number of flips it will take each student to get down to zero will vary.

Stage 3

1. Data and results from your star… [END OF PREVIEW]

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