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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online