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17homework14 - Husain Zeena – Homework 14 – Due Dec 2...

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Unformatted text preview: Husain, Zeena – Homework 14 – Due: Dec 2 2003, 4:00 am – Inst: H L Berk 1 This print-out should have 16 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points Two satellites A and B of the same mass are going around Earth in concentric orbits. The distance of satellite B from Earth’s center is twice that of satellite A. What is the ratio of the tangential speed of B to that of A? 1. 1 / 2 2. 2 3. 1 4. p 1 / 2 correct 5. √ 2 Explanation: For each satellite, the centripetal force is equal to the gravitational force between the satellite and Earth, and is proportional to the square of the tangential velocity and the inverse of the distance. 002 (part 1 of 1) 10 points Given: G = 6 . 6726 × 10- 11 Nm 2 / kg 2 Three m = 13 kg masses are located at points in the x-y plane as shown in the figure; x = 80 cm, y = 50 cm. x y What is the magnitude of the resultant force (caused by the other two masses) on the mass at the origin? Correct answer: 4 . 8426 × 10- 8 N. Explanation: Basic Concepts: Newton’s Law of Grav- itation: F g = G m 1 m 2 r 2 We calculate the forces one by one, and then add them using the superposition principle. The force from the mass on the right is point- ing in the x direction, and has magnitude f 1 = G mm x 2 = Gm 2 x 2 = (6 . 6726 × 10- 11 Nm 2 / kg 2 )(13 kg) 2 (0 . 8 m) 2 = 1 . 76198 × 10- 8 N . The other force is pointing in the y direction and has magnitude f 2 = G mm y 2 = Gm 2 y 2 = (6 . 6726 × 10- 11 Nm 2 / kg 2 )(13 kg) 2 (0 . 5 m) 2 = 4 . 51068 × 10- 8 N; . F f f 1 2 Now we simply add the two forces, using vector addition. Since they are at right angles to each other, however, we can use Pythago- ras’ theorem as well: F = q f 2 1 + f 2 2 = h (1 . 76198 × 10- 8 N) 2 + (4 . 51068 × 10- 8 N) 2 i 1 2 = 4 . 8426 × 10- 8 N . As you see, this force is very small. If the masses in the picture are standing on a table Husain, Zeena – Homework 14 – Due: Dec 2 2003, 4:00 am – Inst: H L Berk 2 (the view being from above), typically the force of static friction will not be overcome. In agreement with common sense, then, the masses will not move. 003 (part 1 of 1) 10 points Given: G = 6 . 67 × 10- 11 Nm 2 / kg 2 M earth = 5 . 98 × 10 24 kg An earth satellite remains in orbit at a dis- tance of r = 14884 km from the center of the earth. What speed would it have to maintain? Correct answer: 5176 . 71 m / s. Explanation: The orbiting satellite is pulled towards the earth with a force GM m r 2 , where G is the Gravitational constant, M is the mass of the earth, m is the mass of the satellite and r is the distance between the center of the earth and the satellite. This force serves as the centripetal force mv 2 r ....
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17homework14 - Husain Zeena – Homework 14 – Due Dec 2...

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