HW10 Solutions

# Gm m r and is directed from the origin toward m n so

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Gm m r and is directed from the origin toward m n , so that it is conveniently written as 2 3 ˆ ˆ ˆ ˆ = i + j = i + j . n n n n n n n n n n n Gm m x y Gm m F x y r r r r Consequently, the vector addition to obtain the net force on m becomes 3 3 3 9 7 net 3 3 =1 1 1 ˆ ˆ ˆ ˆ = i j ( 9.3 10 N)i (3.2 10 N)j n n n n n n n n n n m x m y F F Gm r r   . Therefore, we find the net force magnitude is 7 net 3.2 10 N F . 4. The acceleration due to gravity is given by a g = GM/r 2 , where M is the mass of Earth and r is the distance from Earth’s center. We substitute r = R + h , where R is the radius of Earth and h is the altitude, to obtain 2 2 ( ) g E GM GM a r R h . We solve for h and obtain / g E h GM a R . From Appendix C, R E = 6.37 10 6 m and M = 5.98 10 24 kg, so  11 3 2 24 6 6 2 6.67 10 m /s kg 5.98 10 kg 6.37 10 m 2.6 10 m. 4.9 m/s h Note: We may rewrite a g as 2 2 2 2 / (1 / ) (1 / ) E g E E GM R GM g a r h R h R where 2 9.83 m/s g is the gravitational acceleration on the surface of the Earth. The plot below depicts how a g decreases with increasing altitude. 5. (a) The gravitational potential energy is  11 3 2 11 6.67 10 m /s kg 5.2 kg 2.4 kg = = = 4.4 10 J. 19 m GMm U r

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(b) Since the change in potential energy is 11 11 2 = = 4.4 10 J = 2.9 10 J, 3 3 GMm GMm U r r   the work done by the gravitational force is W = U = 2.9 10 11 J. (c) The work done by you is = U = 2.9 10 11 J. 6. Let m = 0.020 kg and d = 0.600 m (the original edge-length, in terms of which the final edge- length is d /3). The total initial gravitational potential energy (using Eq. 13-21 and some elementary trigonometry) is U i = 4 Gm 2 d 2 Gm 2 2 d . Since U is inversely proportional to r then reducing the size by 1/3 means increasing the magnitude of the potential energy by a factor of 3, so U f = 3 U i U = 2 U i = 2(4 + 2 ) Gm 2 d = 4.82 10 13 J .
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