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CP Tidal Forces near a Black Hole. An astronaut inside a

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 88CP Chapter 13

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 88CP

CP Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030-kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer. (a) What is the tension between her ears? Would the astronaut find it difficult to keep from being torn apart by the gravitational forces? (Since her whole body orbits with the same angular velocity, one ear is moving too slowly for the radius of its orbit and the other is moving too fast. Hence her head must exert forces on her ears to keep them in their orbits.) (b) Is the center of gravity of her head at the same point as the center of mass? Explain.

Step-by-Step Solution:

Solution 88CP Introduction The tension will be equal to the difference in the gravitational force in the two ear. We have to calculate this difference. Step 1 The gravitational force at a distance r is given by F = GMm r2 Now differentiating the above equation with respect to r we get dF = 2G3m dr r dF = 2Grm dr Here M is the mass of the black hole and is 5 times the mass of the sun. Hence the mass of the black hole is M = 5 × (1.99 × 10 kg) = 9.95 × 10 kg .0 The distance is r = 120 km = 120 × 10 m,3 2 The difference in distance is dr = 12 cm = 12 × 10 m Mass of the each ear is m = 0.030 kg And the gravitational constant G = 6.67 × 10 11 N.m /kg 2 Hence the force difference is 11 2 2 30 2(6.67×10 N.m /kg )(9.95×10 kg)(0.030 kg) 2 dF = (120×10 m) 3 (12 × 10 m) = 2765 N Hence the tension between the ear is 2765 N. Since the force is very large, the astronaut will find it very difficult to keep himself from torn apart.

Step 2 of 2

Chapter 13, Problem 88CP is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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