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PHY 150 - Astronomy Homework Assignment #4 October 9, 2007 1) Rocks found on the Moon are between 3.1 and 4.6 billion years old. by contrast, the majority of the Earth’s surface is made of oceanic crust that is less than 200 million years old, and the very oldest Earth rocks are about 3 billion years old. If the Earth and Moon are essentially the same age, why is there such a disparity in the ages of rocks on the two worlds? Plate tectonics is the answer. The Earth’s crust rides on a semi-molten asthenosphere in the form of plates. New matter is continuously being brought to the surface and old rocks are being subsumed. 2) How much would an 80 kg person weigh on the Moon? How much does that person weigh on the Earth? Newton’s Universal Law of Gravitation says that F= GMm r 2 g moon = = ma = mg . This implies that g = ( 6.67 x 10− 11 )(7.348 x 1022 ) (174 . x 106 m) 2 . = 162 m s2 GM r2 . . Thus the weight of a man on the moon is (80 kg)(1.62 m/s2) = 130 N. The same man on the Earth is (80 kg)(9.8 m/s2) = 784 N, or 6 times as much. 3) The last manned spacecraft to land on the Moon was the Apollo 17 lunar module Challenger in December 1972. After lifting off from the lunar surface and returning to the command module America, in which they would return to the Earth, the Apollo 17 astronauts sent the unoccupied Challenger to crash into the lunar surface. The seismic waves from this impact were detected by seismometers left by the crews of Apollos 12, 14, 15, 16, and 17. Why was it useful to do this? Why was it not enough to detect seismic waves from naturally occurring moon quakes? The Moon has cooled off far more than the Earth owing to its much smaller volume to surface area ratio. Thus, naturally occurring moon quakes are rare. This crash was an opportunity to have the source of the seismic waves at the surface rather than deep in the interior. Plus the position could be pinpointed. Thus, the paths of the waves detected by the seismometers were different than usual and new information could be obtained about the Moon’s interior. 4) What you see when you look at the Moon is reflected light from the Sun. What is the dominant wave length of light from black body radiation from the Moon when it is full? Can you see it? The surface temperature of the sunlit side of the Moon is typically on the order of 100oC. According to Wien’s Law the radiation has a maximum at a wavelength given by λmax = .0029 .0029 m= = 7.8 × 10− 6 m = 7.8µm . 373 T( K) This radiation is in the infrared and is not visible to the naked eye. 5) Give some everyday examples of conduction, convection, and radiative diffusion. Conduction - Stick a metallic spoon in boiling water and heat is transferred from the hot water to your hand by conduction through the spoon. Convection - During the winter months, cold air from the north pole drifts southward by convective action leading to waves of cold, freezing air reaching south. Radiation - While it is true that the Sun emits a solar wind of charged particles towards the Earth, it is radiation from the Sun which heats the Earth and causes the Earth’s weather. 6) Draw a cross-sectional picture of the Sun’s interior. Describe the main physical processes that occur at various depths within the Sun. Core - The temperature is on the order of 15 million degrees Kelvin and the density is on the order of 160 times that of water. This occurs in a region from the center out to about .2 RSun. An extremely small percentage of hydrogen atoms have enough energy to undergo nuclear fussion. The energy released produces the Sun’s luminosity. Radiative zone - The temperature falls from 15 million to 1 million degrees Kelvin and the density is falling to one hundredth that of water. This region extends from the core out to about .7 RSun. This region is relatively transparent to radiation. Convective Zone - Temperature falls to thousands of degrees Kelvin and densities thousands of times less than water. This region extends out to the visible photosphere. In this region protons and electrons have recombined into hydrogen ionized hydrogen atoms which absorb radiation. Energy transfer is by way of convective motion boiling up to the surface. 7) Assuming that the current rate of hydrogen burning in the Sun remains constant, what fraction of the Sun’s mass will be converted into helium over the next 5 billion years? [Hint: consult Box 16-1.] How will this affect the overall chemical composition of the Sun? The Sun’s luminosity is 3.9 x 10 26 J/s. Luminosity has the units of energy per time. Using Einstein’s energy relationship that E = mc 2 we see that the mass conversion per time is given by ∆E 3 .9 x 1 0 26 J s L Sun ∆m kg ∆ t = = 2 = = 4 .3 3 x 1 0 9 s 2 2 ∆t c c 3 x 10 8 m s ( ) Only .7% of the hydrogen converted into helium is transformed into energy. Thus, the actual hydrogen mass M converted into helium per time is 9 kg ∆ M 4 .3 3 x 1 0 kg s = = 6 .2 x 1 0 11 s ∆t .0 0 7 5 billion years = 1.58 x 1017 s, thus the mass converted into energy in 5 billion years is kg M = 6 .2 x 1 0 11 s 1.5 8 x 1 0 17 sec = 9 .7 7 x1 0 28 kg Given that MSun = 2 x 1030 kg, this represents 5% of the total Sun’s mass. Let’s make a few simple assumptions about the composition of the Sun. A good guess is that the Sun formed with about 80% hydrogen and 20% helium. In the first 5 billion years we can assume about 5% of the hydrogen was burned. Thus, in the next 5 billion years another 5% of the hydrogen will be burned leaving the Sun about 70% hydrogen and 30% helium. This might not seem like a very big reduction in the amount of hydrogen in the Sun, but it represents most of the hydrogen in the core. ( 8) ) Do Starry Night exercise 16-60. I make the rotation to be about 29 days. This is the rate at the equator of the Sun.