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Astronomy 2010 Problems in Planetary Astronomy Fall_2016 Day-6 ANNOUNCEMENTS •Homework Set 5: Chapter 4 #31, 32, 33, 34, 37, 38, 41, 42, 44 •Project 1 will be due October 17. In addition to a written paper, you need to prepare a short oral presentation. Make your choice ASAP NEWTON’S 1ST LAW OF MOTION The Law of Inertia An object in straight line uniform motion will continue that motion unchanged unless some external force acts on it This was law based entirely on the work of Galileo NEWTON’S SECOND LAW The Force Law: F = ma The acceleration a body experiences is directly proportional to the net force acting on it and inversely proportional to its mass v d v a t dt Acceleration and force are actually vectors F ma VECTORS HAVE BOTH MAGNITUDE AND DIRECTION A vector can be broken into horizontal and vertical components using a few simple trigonometric relationships. These relationships come from the Pythagorean Theorem relating the sides and angles of a right triangle. NEWTON’S THIRD LAW The Action-Reaction Law For every force there is an equal and opposite reaction force USING NEWTON’S LAWS A 215 kg sailboat experiences a force from the wind of 75.0 N at 30° north of east while the current pushes it with a force of 50.0 N due south. What is the magnitude and direction of the acceleration of the boat? SOLUTION North-South Components Wind: (75.0N)sin30° = 37.5N Current: (50.0N)sin270° = -50.0N Net = 37.5N – 50.0N = -12.5N East-West Components Wind: (75.0N)cos30° = 64.45N Current: (50.0N)cos270° = 0.0N Net = 64.45N + 0.0N = 64.45N Fnet 64.45 N 12.5 N 2 2 66.14 N 12.5 N 10.48 349.5 64.45 N tan 1 Fnet ma a Fnet a 66.14 N 0.308 m s2 215kg m The acceleration is 0.31 m/s2 at 10° south of east IN ADDITION TO HIS THREE LAWS OF MOTION, NEWTON FORMULATED THE UNIVERSAL LAW OF GRAVITY FGrav m1m2 G 2 rˆ r1 2 G is the universal gravitational constant (6.67 x 10-11 ), m1 and m2 are the masses of the two objects, r1-2 is the distance between them and r̂ is the direction of the line connecting the two masses. N m2 kg 2 EASY EXAMPLE Using the information in the Appendix, determine the gravitational force of the Earth on the Moon. Which way does it act? What is the gravitational force of the Moon on the Earth? Which way does it act? SOLUTION ME = 5.97 x 1024 kg MM = 7.35 x 1022 kg rE-M = 3.844 x 105 km mE mM F G 2 rˆE M 6.67 1011 rE M 5.97 10 kg 7.35 10 3.844 10 m 24 N m kg 2 2 8 22 kg 2 1.98 1020 N Gravity is always an attractive force so the Earth pulls on the Moon: the direction is from the Moon to the Earth. The force of the Moon on the Earth is exactly the same but in the opposite direction (Newton’s 3rd Law) 1.98 x 1020 N directed from the Earth to the Moon NEWTON DERIVED KEPLER’S 3RD LAW USING UNIVERSAL GRAVITATION AND HIS LAWS OF MOTION P 4 3 r GM * 2 2 Rearranges to give 4 r M* 2 G P 2 3 Where M* is the total mass of the system. For a planet orbiting a star, this is effectively the mass of the star. For a moon or satellite orbiting a planet this is effectively the mass of the planet unless the moon is large. NEWTON’S FORM OF KEPLER’S 3RD EXAMPLE The two moons of Mars, Phobos and Deimos, orbit the planet at 9,380 km and 23,460 km respectively. If their periods are 0.32 days (Phobos) and 1.26 days (Deimos), what is the mass of Mars? The equation to use has “G” in it which has units of (Nm2/kg2) so the 1st thing to do is unit conversions: km to m and days to seconds 9380 km 1000 m km 9.38 106 m 23,460 km 1000 m km 2.346 107 m 0.32 days 86400 s day 2.7648 10 4 s 1.26 days 86400 s days 1.0886 105 s NEWTON’S FORM OF KEPLER’S 3RD EXAMPLE CONTINUED P 4 4 r M 3 r GM GP2 2 M 2 2 3 4 9.38 10 m 2 6 3 6.39 1023 kg 2.7648 10 s 4 2.346 10 m M 6.45 10 1.0886 10 s 6.67 10 6.67 10 11 Nm 2 kg 2 2 11 Nm 2 kg 2 4 7 2 3 23 5 Within round-off error of being the same. The textbook value is 6.42x1023 kg kg NEWTON’S FORM OF KEPLER’S 3RD PROBLEMS The first extra-solar planet discovered orbits the star 51 Pegasi. If the semimajor axis is 0.052 AU and the orbital period is 4.23 days, what is the mass of 51 Pegasi? Triton, the largest moon of Neptune, orbits the planet at a distance of 3.548x105 km every 5.88 days. Use this information to determine the mass of Neptune. EXAMPLE 1 EASY SOLUTION Using units of AU and years, the ratio of the cube of the semimajor axis to the square of the orbital period will give the mass in solar masses. So, we only need to convert the period from days to years. 4.23days 365.25 days year 0.01158 years Now use the ratio of A3 to P2 to find the mass in solar masses 0.052 AU A3 M solar masses 1.0484M Sun 2 2 P 0.01158 years 3 1.0484M Sun 1.98911030 kg M Sun 2.085 1030 kg EXAMPLE 1 HARD SOLUTION Since G has units of (Nm2/kg2), distances must be in meters and periods in seconds so do unit conversions first 0.052 AU 1.49598 1011 m AU 7.779 109 m 4.23days 86400 s day 365472s Next, chose the equation to use and algebraically solve it for the mass. 2 2 3 4 4 r 2 3 P r M GM GP 2 Finally, plug in numbers to solve 4 7.779 10 m 4 r 30 M 2.09 10 kg 2 2 11 Nm2 GP 6.673 10 365472s kg 2 2 2 3 9 3 EXAMPLE SOLUTION 2 Do unit conversions first 3.548 105 km 1000 m km 3.548 108 m 5.88days 86400 s day 508032s Use the same equation as the previous example 4 3.548 10 m 4 r 26 M 1.02 10 kg 2 2 11 Nm2 GP 6.673 10 508032s kg 2 2 2 3 8 3 THE BLACKBODY SPECTRUM Stefan’s Law relates the temperature to the total energy emitted per square meter F T 4 5.67 10 8 W m2 K 4 Wein’s Law relates the temperature to the wavelength of maximum intensity 0.0029 max (in meters) T BLACKBODY EXAMPLE Determine the wavelength of maximum intensity of the Earth and the total energy emitted by it. Assume it is a perfect blackbody with a temperature of 288 K and a radius of 6.378x106 m For wavelength of maximum intensity, use Wein’s law max 0.0029 K m 0.0029K m 1.01 105 m 10.1 m T 288K Next, find total energy emitted: flux times surface area Eemitted T 4 4 R 2 4 6.378 10 m 1.99 1017 W 6 5.67 10 2 8 W m2 K 4 288K 4