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Unformatted text preview: Phys 341: Homework #2 Solutions 1. (a) The semimajor axis of HST’s orbit equals the radius of the Earth ( R ⊕ = 6378 km) plus the altitude above the surface (about 380 miles, or about 600 km). So let’s call it a = 7000 km. From the generalized form of Kepler’s Third Law, P = 2 πR 3 / 2 ⊕ ( GM ⊕ ) 1 / 2 = 2 π [7 × 10 8 cm] 3 / 2 [6 . 67 × 10 − 8 g − 1 cm 3 s − 2 × 5 . 97 × 10 27 g] 1 / 2 = 5800 sec = 97 min (b) To have a period of P = 1 day = 86,400 s, the semimajor axis must be a = GM ⊕ P 2 4 π 2 1 / 3 = 6 . 67 × 10 − 8 g − 1 cm 3 s − 2 × 5 . 97 × 10 27 g × (86 , 400 s) 2 4 π 2 1 / 3 = 4 . 22 × 10 9 cm = 42 , 200 km The altitude above Earth’s surface is then a − R ⊕ = 35 , 800 km. (c) Geosynchronous orbits are possible only over the equator. That’s because a satellite orbits in a plane that passes through the center of Earth. (If it is not in the equatorial plane, the orbit must have portions both north and south of the equator.) By contrast, a given point on Earth’s surface moves in a plane parallel to the equator. The only case when those planes match is when they are both the plane passing through the equator....
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 Spring '08
 Keeton
 Physics, Mass, Work

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