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HW3 Fall09 (sol)

HW3 Fall09 (sol) - Homework 3 Solutions 1 E If M is the...

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Homework 3 Solutions 1. E If M is the mass of the Earth and R its radius, the force on a mass m at the surface has magnitude: F = G Mm R 2 ; since F = ma , the acceleration of this object would be GM/R 2 , which give us the standard value of g . If the distance from the center of the Earth is increased to 3 R (i.e., one diameter above the surface), then the acceleration will be GM/ (3 R ) 2 . Since we know g = GM/R 2 = 9 . 8 m / s 2 , it follows that the acceleration at this higher position is (9 . 8 m / s 2 ) / 9 = 1 . 1 m / s 2 . 2. B The gravitational force on M is due to the four particles of mass m . The configuration is symmetric about the origin, so the contributions to the force in the x -direction will cancel. The gravitational force on M attracts it toward the four particles of mass m , so the net force will be in the downward direction. 3. B For configurations 1, 2, and 3, we can calculate the gravitational force exactly. For 1: F 1 = GMm d 2 + GMm (2 d ) 2 = 5 4 GMm d 2 = 1 . 25 GMm d 2 . For 2: F 2 = 0, since the forces of magnitude GMm/d 2 are in opposite directions. For 3: F 3 = ( GMm d 2 ) 2 + ( GMm d 2 ) 2 = 2 GMm d 2 = 1 . 41 GMm d 2 . For 4, we don’t know the angle between the two masses, but it is less than 90 , so that the net force must be greater than F 3 . Thus, we can see that the order is: F 2 < F 1 < F 3 < F 4 . 4. D We know g = GM/R 2 = 9 . 8 m / s 2 , where M is the mass and R the radius of the Earth. To get the acceleration of a freely falling body near the surface of Venus, we can see the result must be: g v = G M v R 2 v = G (0 . 0558 M ) (0 . 381 R ) 2 = 0 . 0558 (0 . 381) 2 G M R 2 = 0 . 0558 (0 . 381) 2 g = 0 . 384(9 . 8 m / s 2 ) = 3 . 8 m / s 2 . 5. D In this situation, a mass falls from 2 R to R , for the Earth and also for some other planet with the same radius. On the other planet the final speed of the object is four times greater than on Earth; therefore its kinetic energy is 16 times larger. Thus, the change in potential energy is 16 times larger on the other planet. On Earth the change in the potential energy is: Δ U = U f - U i = - GMm R + GMm 2 R = - GMm 2 R . We see the magnitude of Δ U is proportional to M , so the mass of the other planet must be 16 times larger than the mass of Earth. 1

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6. C F g = GMm/R 2 on Earth, if M is the mass of the Earth and R its radius, where m is the mass of the man; we are told F g = 600 N. To get the gravitational force on the man on another planet, we need to replace the Earth values for M and R with those corresponding to the other planet: F other g = G (0 . 01 M ) m/ (0 . 25 R ) 2 . Comparing this to the Earth value, it follows that F other g = 0 . 16(600 N) = 96 N. 7. A One Earth radius above the surface is a distance r = 2 R from the center, if R is the radius of the Earth. We use Δ K + Δ U = 0, with the initial value of kinetic energy being K i = 0, and the initial value of potential energy being U i = - GMm/ (2 R ), where m is the mass of the object. The final value of potential energy U f = - GMM/R , so Δ U = - GMm/ (2 R ). Therefore, the final value of kinetic energy is: K f = Δ K = - Δ U = GMm/ (2 R ) = 1 2 mv 2 . Then the speed of the object just before it his the surface is v = GM/R .
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