Homework Solutions Chapter 4

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Merry Lamb
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1 Homework Solutions Chapter 4 Todd B. Krause September 30, a) If joules represents the energy of a major earthquake, the energy of a -megaton bomb is smaller by a factor of joule joule 5 05 = = A major earthquake releases as much energy as five -megaton bombs. b) The annual U.S. energy consumption is about 0 0 joules, and a liter of oil yields about. 0 7 joules. Thus, the amount of oil needed to supply all the U.S. energy for a year would be 0 0 joule. 0 7 joule liter = 8 0 liter, or about 8 trillion liters of oil roughly trillion gallons). c) We can compare the Sun s annual energy output to that of the supernova by dividing: to be conservative, we use the lower number from the range for supernova energies: supernova energy Sun s annual energy output = 044 joule 0 34 joule = First, we find the U.S. energy consumption per minute by converting the annual energy consumption into units of joules per minute: 0 joules yr 0 yr 365 days day 4 hr hr min 60 joules min.

2 Next we divide this energy consumption per minute by the amount of energy available through fusion of liter of water from table 4.): joules min joules liter.7 liters min. In other words, it would take less than 3 liters of water per minute which is less than gallon per minute to meet all U.S. energy needs through nuclear fusion. This is somewhat less than the rate at which water flows from a typical kitchen faucet. So if we could simply attach a nuclear fusion reactor to your kitchen faucet, we could stop producing and importing oil, remove all the hydroelectric dams, shut down all the coal-burning power plants, and still have energy to spare. 56. Newton s version of Kepler s third law has the form p = 4π GM + M ) a3. Because the square of the period varies inversely with the sum of the masses, the orbital period itself depends on the inverse square root of the object masses: p = 4π GM + M ) a3. Thus, if we have a star four times as massive as the Sun, the the period of a planet orbiting at AU will be / 4 = / that of the Earth, or 6 months. 57. a) Using the Moon s orbital period and distance and following the method given in Mathematical Insight 4.3, we find the mass of the earth to be about M Earth = 4π a Moon ) 3 GP Moon ). Making sure that we use appropriate units, we find 4π 384, 000 km, 000 m ) 3 M Earth km ) ) m kg s days 4 day hr 3, 600 s hr = kg.

3 b) Using Io s orbital period and distance and following the method given in Mathematical Insight 4.3, we find the mass of Jupiter to be about 4π 4, 000 km, 000 m ) 3 M Jupiter km ) ) m hr 3, 600 s kg s hr = kg. We find the same answer using Europa s orbital properties; Kepler s third law does not depend on the mass of either moon because neither moon has a significant mass in comparison to the mass of Jupiter. c) We again start with Newton s version of Kepler s third law: p = 4π GM + M ) a3. We solve for the semimajor axis of the planet with a little algebra: a = 3 GM + M ) 4 π p. We convert the planet s orbital period of 63 days into seconds: 4 hr 63 days day 60 min hr 60 s min = s. Planets are much less massive than their stars are, so we can approximate M + M M star. From Appendix A, the mass of the Sun is about 0 30 kg. So we can calculate the semimajor axis: ) m 3 0 kgs 30 kg) a = 4π s) = m. Of course, this number would probably be more useful in astronomical units, so we should convert: m AU.5 0 = 0.3 AU. m The new planet is only 0.3 AU from its star, which is closer than Mercury is to the Sun. 3

4 59. In a) and c), the easiest way to find the escape velocities with the given data is by comparison to the escape velocity from earth: GM planet v R escape planet planet M planet = = REarth. v escape Earth GM Earth M Earth R planet R Earth Given that the escape velocity from Earth s surface is about km/s, this formula becomes v escape planet = km M planet REarth. s M Earth R planet a) From the surface of Mars, the escape velocity is km s = km s MMars = 5.0 km s. M Earth REarth R Mars b) From the surface of Phobos, the escape velocity is ) m kgs kg), 000 m = m s = 0. km s. c) From the surface of Jupiter, the escape velocity is km M Jupiter R Earth s M Earth R Jupiter = km s = 58.6 km s

5 d) To find the escape velocity from the solar system, starting from the Earth s orbit, we use the mass of the Sun since that is the mass we are trying to escape) and the Earth s distance from the Sun of AU, or.5 0 m: = 4, 00 m s = 4. km s m 3 kgs ) kg).5 0 m e) To find the escape velocity from the solar system starting from Saturn s orbit, we use the mass of the Sun since that is the mass we are trying to escape) and Saturn s distance from the Sun of.4 0 m: = 3, 800 m s = 3.8 km s m 3 kgs ) kg).4 0 m 5

Name Class Period. F = G m 1 m 2 d 2. G =6.67 x 10-11 Nm 2 /kg 2

Name Class Period. F = G m 1 m 2 d 2. G =6.67 x 10-11 Nm 2 /kg 2 Gravitational Forces 13.1 Newton s Law of Universal Gravity Newton discovered that gravity is universal. Everything pulls on everything else in the universe in a way that involves only mass and distance.