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Unformatted text preview: chauhan (mmc2762) – Ch5H3 – turner – (56910) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points After the NEAR spacecraft passed Mathilde, on several occasions rocket propellant was ex pelled to adjust the spacecraft’s momentum in order to follow a path that would approach the asteroid Eros, the final destination for the mission. After getting close to Eros, further small adjustments made the momentum just right to give a circular orbit of radius 45 km around the asteroid. So much propellant had been used that the final mass was 500 kg and it of the spacecraft took 1 . 04 days to make one complete circular orbit around Eros. Cal culate what the mass of Eros must be. Use G = 6 . 67 × 10 − 11 N · m 2 / kg 2 . Correct answer: 6 . 68001 × 10 15 kg. Explanation: We begin to solve this problem by equating the net force on NEAR and the gravitational force from Eros on NEAR: vextendsingle vextendsingle vextendsingle vector F net vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vector F grav vextendsingle vextendsingle vextendsingle m  vectorv  2 R = GM m R 2 ⇒  vectorv  2 = GM R . The speed of the spacecraft is given by v = 2 π R/T . Substitute this to get an ex pression that relates the period and radius of the spacecraft’s orbit. parenleftbigg 2 π R T parenrightbigg 2 = GM R 4 π 2 R 2 T 2 = GM R ⇒ T 2 = 4 π 2 GM R 3 . The above expression is known as Kepler’s third law for circular orbits. Solving for the mass of Eros, we have ⇒ M = 4 π 2 R 3 GT 2 = 4 π 2 (45 km) 3 G (1 . 04 days) 2 = 6 . 68001 × 10 15 kg ....
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This note was uploaded on 04/03/2012 for the course PHYSICS 303K taught by Professor Antoniewicz during the Spring '11 term at University of Texas.
 Spring '11
 Antoniewicz
 Physics

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