chap9 - CHAPTER 9 GRAVITATION radius Thus R/RE = 1 3 = 57.7...

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CHAPTER 9 GRAVITATION ActivPhysics can help with these problems: Activity 4.8 Section 9-2: The Law of Universal Gravitation Problem 1. Space explorers land on a planet with the same mass as Earth, but they find they weigh twice as much as they would on Earth. What is the radius of the planet? Solution At rest on a uniform spherical planet, a body’s weight is proportional to the surface gravity, g = GM/R 2 . Therefore, ( g p /g E ) = ( M p /M E )( R E /R p ) 2 = 2 . Since M p /M E = 1 , R p = R E / 2 . Problem 2. Use data for the moon’s orbit from Appendix E to compute the moon’s acceleration in its circular orbit, and verify that the result is consistent with Newton’s law of universal gravitation. Solution The centripetal acceleration of the moon in its orbit around the Earth is approximately a c = (2 π/T ) 2 r = (2 π/ 27 . 3 × 86 , 400 s) 2 (3 . 85 × 10 8 m) = 2 . 73 × 10 3 m / s 2 (see Problem 4-43). (One could also use a c = v 2 /r , but the orbital speed given in Appendix E is less accurate.) The acceleration of gravity at the moon’s approximate distance from the Earth is g ( r ) = GM E /r 2 = (6 . 67 × 10 11 N · m 2 / kg 2 )(5 . 97 × 10 24 kg) ÷ (3 . 85 × 10 8 m) 2 = 2 . 69 × 10 3 m / s 2 . Since the moon’s orbit is actually elliptical (with 5.5% eccentricity), these values based on circular orbits are not inconsistent. Problem 3. To what fraction of its current radius would Earth have to be shrunk (with no change in mass) for the gravitational acceleration at its surface to triple? Solution If the surface gravity of the Earth were three times its present value, with no change in mass, then GM E /R 2 = 3 GM E /R 2 E , where R E is the present radius. Thus, R/R E = 1 / 3 = 57 . 7% gives the new, shrunken radius. Problem 4. Calculate the gravitational acceleration at the surface of (a) Mercury and (b) Saturn’s moon Titan. Solution With reference to the first two columns in Appendix E: (a) g Merc = G (0 . 330 × 10 24 kg) / (2 . 44 × 10 6 m) 2 = 3 . 70 m / s 2 , and (b) g Titan = G (0 . 135 × 10 24 kg) ÷ (2 . 58 × 10 6 m) 2 = 1 . 35 m / s 2 . Problem 5. Two identical lead spheres are 14 cm apart and attract each other with a force of 0 . 25 μ N . What is their mass? Solution Newton’s law of the universal gravitation (Equation 9-1), with m 1 = m 2 = m, gives m = radicalbigg Fr 2 G = parenleftbigg 0 . 25 × 10 6 N (0 . 14 m) 2 6 . 67 × 10 11 N · m 2 / kg 2 parenrightbigg 1 / 2 = 8 . 57 kg . (Newton’s law holds as stated, for two uniform spherical bodies, if the distance is taken between their centers.) Problem 6. The gravitational acceleration at the surface of a planet is 22 . 5 m / s 2 . Find the acceleration at a height above the surface equal to half the planet’s radius. Solution The surface gravity of a spherical planet is g = GM p /R 2 p . At a height of 1 2 R p above the surface, the distance to the center of the planet is r = 3 2 R p , so the acceleration of gravity at that height is g ( r ) = GM p /r 2 = 4 9 g = 4 9 (22 . 5 m / s 2 ) = 10 . 0 m / s 2 .
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122 CHAPTER 9 Problem 7. What is the approximate value of the gravitational force between a 67-kg astronaut and a 73,000-kg space shuttle when they’re 84 m apart?
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