1. Kepler’s Laws 1. 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) P 2 1 / P 2 2 = a 3 1 / a 3 2 Note: If P is measured in fraction of earth years, and a is measured in a.u., then P 2 = a 3 (See Table 2.1) Period is the amount of time it takes for the planet to complete one orbit

2. Kepler’s laws The 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, P 2 = a 3. As a result, a = P 2/3 = (0.241) 2/3 = 0.387 au = 0.579 x 10 11 m = 5.79 x 10 10 m Then, Distance to aphelion = a (1 - e) = 5.79 x 10 10 m (1 - 0.206) = 4.6 x 10 10 m Distance to perihelion = a (1 + e) = 5.79 x 10 10 m (1 + 0.206) = 6.98 x 10 10 m

3. Kepler’s Laws The 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, P 2 = a 3. As a result, a = P 2/3 = (163.7) 2/3 = 30.7 au = 45.93 x 10 11 m = 4.593 x 10 12 m Then, Distance to aphelion = a (1 - e) = 4.593 x 10 12 m (1 - 0.009) = 4.55 x 10 12 m Distance to perihelion = a (1 + e) = 4.593 x 10 12 m (1 + 0.009) = 4.63 x 10 12 m The 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%.

4. Isaac Newton Mass is the generator of gravity. The force of gravity exists between any objects that possess mass. The Universal Law of Gravity F = G {m 1 m 2 / r 2 } G = 6.67 x 10 -11 Nm 2 /Kg 2 The “New” Astronomy m1m1 m2m2 r r is measured center to center

5. The Universal Law of Gravity What 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 aphelion F = G (M Sun M Neptune ) / r Sun to Neptune 2 = 6.67 x 10 -11 N m 2 / kg 2 (2 x 10 30 kg ) (1 x 10 26 kg) / (4.55 x 10 12 m) 2 = 0.644 x 10 21 N = 6.44 x 10 20 N. At Perihelion, F = G (M Sun M Neptune ) / r Sun to Neptune 2 = 6.67 x 10 -11 N m 2 / kg 2 (2 x 10 30 kg ) (1 x 10 26 kg) / (4.63 x 10 12 m) 2 = 0.622 x 10 21 N = 6.22 x 10 20 N.

6. The Universal Law of Gravity The 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/sec 2. This answer would be wrong because the HST is not “near the surface of the earth.” Use a = G M Earth / (R Earth + H) 2 = 6.67 x 10 -11 N m 2 / kg 2 (6 x 10 24 kg) / (6378000 m + 600000 m) 2 = 8.21 m/sec 2. Note this acceleration is less than 9.8 m/sec 2 because the HST is not near the surface of the earth.

7. The Universal Law of Gravity What 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) g Surface of Earth = (11110 kg) (9.8 m/sec 2 ) = 1.09 x 10 5 N In orbit, W = mg = (11110 kg) g at Orbital Location = (11110 kg) (8.21 m/sec 2 ) = 0.912 x 10 5 N The HST’s weight in orbit is about 84% of it’s weight at launch.

8. The Universal Law of Gravity What 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 = R E. Therefore g = G M p / (R p + h) 2 g = G M p / (R p + R E ) 2 g = G M p / (2 R p ) 2 g = (¼) G M p / ( R p ) 2 = (¼) 9.8 m/sec 2 Therefore W = m (1/4) g = ¼ m g = ¼ (10,000 N) = 2,500 N

9. Circular Motion Using the data provided for the HST in the previous examples, what is the orbital speed of the HST? Assume a circular orbit. a = v 2 / (R Earth + H) Therefore, v = (a (R Earth + H)) 1/2 = ((8.21 m/sec 2 ) (6.9 x 10 6 m)) 1/2 = 7.53 x 10 3 m/sec = 7.53 km/sec. Note: This is about 17,000 mph.