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Unformatted text preview: UCSB Winter 2008 Astro 1 - Homework #3 Solutions Kevin Moore 1/20/08 3.11 What is the difference between the umbra and the penumbra of a shadow? Check out fig. 3-8. The umbra is the darkest part of a shadow - the part that no light reaches - so an observer there can see no part of the light source. The penumbra is the less dark part of a shadow - some light reaches the penumbra, so an observer there can see part (but not all) of the light source. 3.12 Why doesn’t a lunar eclipse occur at every full moon, and a solar eclipse at every new moon? We don’t see both types of eclipses each month because of the 5 ◦ tilt of the plane of the Moon’s orbit with respect to the ecliptic (the plane of the Earth’s orbit). From fig. 3-7 we see that not only do we need a full or new moon to have an eclipse, but it has to fall on the line of nodes. Therefore, most months we don’t have any eclipses at all. 3.19 Which type of eclipse - lunar or solar - do you think most people have seen? Why? Considering what it takes to have an eclipse - a full or new moon lying on the line of nodes- it seems that each type of eclipse would be equally likely to occur. This is true, tables 3-1 and 3-2 show that there are 12 solar and 12 lunar eclipses between 2007 and 2011. So it must be that although each type of eclipse (solar and lunar) is equally likely, one type must be easier to see. Since the Moon is so much smaller than the Earth it easily fits inside the Earth’s shadow, so a lunar eclipse can be seen from any point on the night side of the Earth (see fig. 3-8). The Moon’s small size is also what makes solar eclipses so much harder to see. Looking at fig. 3-11, you must be in a specific part of the daytime Earth - the eclipse path - to see a solar eclipse because the Moon’s shadow is so small. So it is much easier to see a lunar eclipse than a solar eclipse, and it is reasonable to expect that more people have seen lunar eclipses than solar eclipses. 3.32 During an occultation, or ’covering up’, of Jupiter by the Moon, an astronomer notices that it takes the Moon’s edge 90 s to cover Jupiter’s disk completely. If the Moon’s motion is assumed to be uniform and the occultation was ’central’ (that is, center over center), find the angular diameter of Jupiter in arcseconds. We’re given how long it takes the moon to cover up Jupiter and need to find out how large in angular size Jupiter appears. So, to solve this problem we need to figure out how much angle is swept out by the Moon in 90s as it moves across the sky. 1 First calculate the angular speed of the moon, ω , in ◦ sec using the sidereal period (27.3 days): (remember sidereal periods are the actual orbital periods, the synodic period is the time between successive phases and is slightly longer - 29.5 days) Note: it’s fine if you approximate the period as being 30 days for ease of calculation - I just wanted to highlight the small difference between the types of periods....
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This homework help was uploaded on 04/07/2008 for the course ASTRO 1 taught by Professor Antonucci during the Winter '08 term at UCSB.
- Winter '08