Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Kepler’s Laws1. The orbits of the planets are ellipses, with the sun at one focus of the ellipse.2. The line joining the planet to the sun sweeps out equal areas in equal times as the plates travels around the ellipse.3. The ratio of the squares of the revolutionary periods (P) of two planets is equal to the ratio of the cubes of their semimajor axes (a)P21 / P22 = a31 / a32 Note: If P is measured in fraction of earth years, and a is measured in a.u., then P2 = a3 (See Table 2.1)Period is the amount of time it takes for the planet to complete one orbitKepler’s lawsThe period of the orbit of Mercury is 0.241 years. The eccentricity of the orbit is 0.206. What is the distance to Mercury’s aphelion? What is the distance to Mercury’s perihelion?According to the Kepler’s Second Law, P2 = a3 . As a result,a = P2/3 = (0.241)2/3 = 0.387 au = 0.579 x 1011 m = 5.79 x 1010 mThen,Distance to aphelion = a (1 - e) = 5.79 x 1010 m (1 - 0.206) = 4.6 x 1010 mDistance to perihelion = a (1 + e) = 5.79 x 1010 m (1 + 0.206) = 6.98 x 1010 mKepler’s LawsThe period of the orbit of Neptune is 163.7 years. The eccentricity of the orbit is 0.009. What is the distance to Neptune’s aphelion? What is the distance to Neptune’s perihelion?According to the Kepler’s Second Law, P2 = a3 . As a result,a = P2/3 = (163.7)2/3 = 30.7 au = 45.93 x 1011 m = 4.593 x 1012 mThen,Distance to aphelion = a (1 - e) = 4.593 x 1012 m (1 - 0.009) = 4.55 x 1012 mDistance to perihelion = a (1 + e) = 4.593 x 1012 m (1 + 0.009) = 4.63 x 1012 mThe orbit of Neptune is not a perfect circle, but it’s low eccentricity indicates that the orbit is close to a circle. The distance to aphelion and perihelion differ by only (approximately) 1.7%.Isaac NewtonMass is the generator of gravity.The force of gravity exists between any objects that possess mass. The Universal Law of GravityF = G {m1 m2 / r2 }G = 6.67 x 10-11 Nm2/Kg2 The “New” Astronomym1m2rr is measured center to centerThe Universal Law of GravityWhat is the force of gravity exerted by the sun on Neptune at aphelion? What is the force of gravity exerted by the sun on Neptune at perihelion?According to the Universal Law of Gravity, at aphelionF = G (MSun MNeptune ) / rSun to Neptune 2 = 6.67 x 10-11 N m2 / kg2 (2 x 1030 kg ) (1 x 1026 kg) / (4.55 x 1012 m)2 = 0.644 x 1021 N = 6.44 x 1020 N.At Perihelion,F = G (MSun MNeptune ) / rSun to Neptune 2 = 6.67 x 10-11 N m2 / kg2 (2 x 1030 kg ) (1 x 1026 kg) / (4.63 x 1012 m)2 = 0.622 x 1021 N = 6.22 x 1020 N.The Universal Law of GravityThe mass of the Hubble Space Telescope (HST) is 11110 kg. It orbits the earth at a distance for 600 km from the surface of the eath. What is the acceleration of gravity of the HST?Potential answer: 9.8 m/sec2 . This answer would be wrong because the HST is not “near the surface of the earth.”Usea = G MEarth / (REarth + H)2 = 6.67 x 10-11 N m2 / kg2 (6 x 1024 kg) / (6378000 m + 600000 m)2 = 8.21 m/sec2 .Note this acceleration is less than 9.8 m/sec2 because the HST is not near the surface of the earth.The Universal Law of GravityWhat was the weight of the HST at launch? What is the weight of the HST in orbit? Assume the mass of the HST is 11110 kg.At launch, that is, on the surface of the earth,W = mg = (11110 kg) gSurface of Earth = (11110 kg) (9.8 m/sec2 ) = 1.09 x 105 N In orbit,W = mg = (11110 kg) gat Orbital Location = (11110 kg) (8.21 m/sec2 ) = 0.912 x 105 NThe HST’s weight in orbit is about 84% of it’s weight at launch.The Universal Law of GravityWhat is the force of gravity on a satellite that is one Earth radius above the surface of the Earth? The weight of the satellite on Earth is 10,000 N?At one Earth radius above the surface of the Earth, h = RE. Thereforeg = G Mp / (Rp + h)2 g = G Mp / (Rp + RE )2 g = G Mp / (2 Rp )2 g = (¼) G Mp / ( Rp )2 = (¼) 9.8 m/sec2Therefore W = m (1/4) g = ¼ m g = ¼ (10,000 N) = 2,500 NCircular MotionUsing the data provided for the HST in the previous examples, what is the orbital speed of the HST? Assume a circular orbit.a = v2 / (REarth + H) Therefore,v = (a (REarth + H))1/2 = ((8.21 m/sec2) (6.9 x 106 m))1/2 = 7.53 x 103 m/sec = 7.53 km/sec. Note: This is about 17,000
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