Ch5-H3-solutions

# Ch5-H3-solutions - chauhan(mmc2762 – Ch5-H3 – turner...

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Unformatted text preview: chauhan (mmc2762) – Ch5-H3 – turner – (56910) 1 This print-out should have 10 questions. Multiple-choice 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.

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Ch5-H3-solutions - chauhan(mmc2762 – Ch5-H3 – turner...

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