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Unformatted text preview: CHAPTER 9 GRAVITATION ActivPhysics can help with these problems: Activity 4.8 Section 92: 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 bodys 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 moons orbit from Appendix E to compute the moons acceleration in its circular orbit, and verify that the result is consistent with Newtons law of universal gravitation. Solution The centripetal acceleration of the moon in its orbit around the Earth is approximately a c = (2 /T ) 2 r = (2 / 27 . 3 86 , 400 s) 2 (3 . 85 10 8 m) = 2 . 73 10 3 m / s 2 (see Problem 443). (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 moons 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 moons 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) Saturns 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 N . What is their mass? Solution Newtons law of the universal gravitation (Equation 91), with m 1 = m 2 = m, gives m = radicalbigg Fr 2 G = parenleftbigg . 25 10 6 N (0 . 14 m) 2 6 . 67 10 11 N m 2 / kg 2 parenrightbigg 1 / 2 = 8 . 57 kg . (Newtons 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 planets radius....
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This note was uploaded on 09/15/2009 for the course PHY 557 taught by Professor Rijssenbeek during the Spring '08 term at Adelphi.
 Spring '08
 RIJSSENBEEK
 Mass

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