To write this number in scientific notation we need to figure out how many spaces to move the decimal point to the left until we have only one digit to the left of the decimal point. Each space we move the decimal is one power of ten. So to obtain the number 2.38793387 we needed to move the decimal 5 times, or 5 powers of 10. This number to 3 significant digits is 2.39. We round the third digit up or down accordingly. So finally, the answer is:
In the first problem you can see that multiplying by a positive power of ten indicates how many decimal places to the right one must move the decimal point to obtain a normal string of digits (we went to the left to get the scientific notation, it's opposite the direction to go to obtain the normal string). So multiplying by a negative power of ten moves the decimal to the left by that power. And so we have
First off 6/3 = 2. That's easy enough. So now we are left with 104/10-6. Use the rules of multiplication and division of exponents. When dividing you subtract the exponents. So 4 - (-6) = 4 + 6 = 10. Thus the answer is
Let us remember the relationship between speed, distance, and time.
The whole point of this problem is to use ratios to get a sense of scale of space and time in the Universe. There's only one piece of math here and that's the ratio. In this scale model the Sun's diameter is 0.5 mm, let's put the numbers in meters for ease (0.5 mm x 1 m/103mm = 5 x 10-4 m). The Sun's actual size is (1.5 x 106 km) x (103 m/km) = 1.5 x 109 m. So our ratio is:
For this part, the actual average distance between stars is about 5 light-years = 5 light-years x (9.5 x 1012 km/light-year) = 4.75 x 1013 km. So let's multiply it by our ratio to find out what it is in the scale model.
Think about it, if the Sun were the size of a period located in Berkeley, the nearest star would be in San Francisco!
Let's put this number in km again. 105 light-years x (9.5 x 1012 km/light-year) = 9.5 x 1017 km. Now multiply by the ratio.
That's just a little less than the average distance between the Earth and the Moon!
Again, put the number in km. 2 x 106 light-years x (9.5 x 1012 km/light-year) = 1.9 x 1019 km. Now multiply by the ratio.
13 x 109 light-years x (9.5 x 1012 km/light-year) = 1.2 x 1023 km. Now, multiply by the ratio.
Since the scale is 41 Billion kilometers from the Sun to the most distant reaches it is clearly bigger than the diameter of Pluto's orbit (12 Billion km). The factor by which it is bigger is just
This will be the same as above but with a different ratio. Let's put both times in seconds so that we can properly compare them.
This event occurred 2000 - 1776 = 224 years ago. Let's put that in seconds. It's 224 years x (3.16 x 107 seconds/year) = 7.08 x 109 seconds ago. Now multiply by the ratio:
This occurred 2000 - 476 = 1,524 years ago. Put that in seconds: 1.524 x 103 years x (3.16 x 107 seconds/year) = 4.82 x 1010 seconds. Now multiply by the ratio...
104 years ago x (3.16 x 107 seconds/year) = 3.16 x 1011 seconds. Now multiply by the ratio...
105 years ago x (3.16 x 107 seconds/year) = 3.16 x 1012 seconds. Now multiply by the ratio...
65 x 106 years ago x (3.16 x 107 seconds/year) = 2.05 x 1015 seconds. Now multiply by the ratio...
250 x 106 years ago x (3.16 x 107 seconds/year) = 7.9 x 1015 seconds. Now multiply by the ratio...
3.8 x 109 years ago x (3.16 x 107 seconds/year) = 1.2 x 1017 seconds. Now multiply by the ratio...
In class I drew a figure similar to the one above and showed you that to do this calculation you needed only to recognize that the horizon is where your line of sight runs tangent to Earth's surface. This is of course a line that is perpendicular to the radius of the Earth, R, and thus we can identify a right triangle with legs, dH (the distance we seek), and R. Also the triangle has a hypotenuse equal to the sum of the Earth's radius and your height above it, h. Using the Pythagorean theorem for right triangles we may then write
For this part h = 1,913 ft x (1 mile/5280 ft) x (1 km/ 0.6214 miles) = 0.583 km, and thus we have
For this part h = 35,000 ft x (1 mile/5280 ft) x (1 km/ 0.6214 miles) = 10.7 km, and thus we have
Recall from class that we discussed how the Earth appears from the Moon. In that discussion I told you that Earth exhibited phases as seen from the Moon. That could be readily seen during the demonstration with the globe and the lamp. Also, recall that I told you that the phases were exactly opposite the ones people on Earth would be seeing for the Moon. When the Moon is New, Earth is Full, etc.. So, if you see Earth to be a waxing gibbous from your window, then the Moon is a waning crescent. Think about it. If the Earth is a waxing gibbous then that means it will soon be full. If it will soon be full then the Moon must be viewing it from the position in its orbit where it is between the Earth and Sun. That is when Earth will see the Moon as New. The phase just previous to New is the waning crescent.
Since the Moon is a waning crescent it is between 90° and 0° West of the Sun as seen from Earth (Berkeley included). Thus it rises after midnight and before sunrise. It is therefore overhead in the morning sometime before noon and after sunrise.
Forever! Recall in class how I explained that the Moon always presents the same face to Earth. It spins only once per orbit about the Earth. Its solar day is the same as the synodic month. That means that if you are in a location where the Earth is in the sky, it is always there! It never rises, sets, or moves at all. The only change it goes through is the phases. So pull the shades if you want that nap.
Since the orbits are elliptical there are times when the Moon is at it's closest approach to Earth (perigee) and places where it is farthest from Earth (apogee). In these places the Moon is at its largest and smallest apparent angular sizes, respectively. Likewise the Earth's orbit brings it to a point of closest approach (perihelion) and farthest distance (aphelion). In these places the Sun is at its largest and smallest apparent angular sizes, respectively. The angular size changes in an inverse way with distance: smaller the greater the distance, bigger the smaller the distance. So when the Moon is far enough away in its orbit that its angular size is less than that of the Sun we get an annular eclipse. This is most extreme if the eclipse were to occur when the Moon was at apogee and Earth at perihelion. (NOTE: you need not have mentioned the words apogee and perihelion specifically)
They will also see a total lunar eclipse. This is true for two reasons. One is that since it's stated that the Moon is visible in both skies at the same time both places are seeing the same Moon and they will both see it completely covered by Earth's shadow. This was mentioned specifically in class. Secondly, since Earth's shadow is huge it takes the Moon longer than 12 hours to move all the way through it. Thus everyone on Earth will be able to witness at least part of the eclipse, but not the total eclipse (that only lasts for a maximum of 1 hour 40 minutes).
As mentioned numerous times in class, the angle of Polaris above the horizon is equal to your latitude. Take a look at the following figure:
In the figure, angle l is your latitude and angle p is the angle above the horizon that the North Celestial Pole makes. l + b = 90° also l + a = 90° and a + p = 90° as well. Putting those three equations together tells you that l = p. So Gilligan and company are at a latitude of 28° North.
For this calculation you need to recall that the Sun moves about the sky, covering 360° every 24 hours, or 15°/hr. Also recall that at a time of 12:00 noon PST the Sun is crossing the meridian as seen from the longitude of 120° West. Since the Sun does not cross Gilligan's meridian until 3 hours later that means that the Sun was East of the meridian at noon and hence they are West of 120° West Longitude. Since it was 3:04 PM when the Sun transited for them then they are
If you took note of your actual sign when I put up the overheads in class that's fine as well. Some of you may have found that you still had the same sign whereas others would have shifted over one sign. For example, My birthdate is September 18th, 1973. That put me just at the cusp of Virgo in the old astrologers' horoscopes and just at the other cusp in the modern constellation recognized by astronomers. But I remain a Virgo.
You will likely have found that the predictions did not agree well, at times even being contradictory. Most of these so-called predictions were also likely just advice and not a prediction at all. Any predictions that were made were likely so vague as to have any interpretation to make them appear accurate in retrospect.
Here is one example of a horoscope for Virgo on Tuesday, May 30th:
(Aug. 23 -Sept. 22)
Tackling major household projects could be your main area of focus today as things have been put off for as long as you, or the spouse. are willing to put up with. even though you are as industrious as the busiest beaver don't forget to take some time out to rest during the heat of the day.
Well that was not true at all since I didn't spend much of that day at home, and the last bit is just advice not a prediction.
Here's another for 05/30/2000:
Actions speak louder than words, so get busy. Don't wait for approval--go ahead even if the light's red.
Hmmm, I don't even know what that's supposed to mean. Certainly didn't apply to anything I did that day.
Here's one last horoscope from the New York Post for 5/30/2000:
VIRGO (Aug. 24-Sept. 23): Some think your ambitions are too lofty. Some think you are aiming too high. But according to the planets, you have what it takes to reach the top and stay there. Don't listen to those who counsel caution, because caution never got anyone anywhere. Take a few risks - they will pay off.
That pretty much can apply to anyone doing anything. It's basically just advice. Not really a prediction.
In Ptolemy's model the planets are attached to circles that are attached to circles that both rotate at the same time, the larger one is rotating about the Earth. This causes a real loop-the-loop motion in space that would appear as retrograde motion to observes on Earth. See figure 3.
In Copernicus' model the planets (including Earth) all revolve about the Sun in perfect circles. The retrograde motion is just an illusion of perspective when one planet is "lapped" by another. See figure 4
In modern scientific research we use a philosophy called Occam's razor. It states that when presented with two hypotheses that explain and predict a phenomenon equally well, we should go with the more simple. There's absolutely no reason why we should prefer the more complex explanation if the simpler does just as well at explaining it. Thus when it was seen that Copernicus' model did just as well at predicting the positions of the planets we should have chosen it for its simplicity, even though it went against our most fundamental beliefs and prejudices.
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