ch13-p038 - 38. Energy conservation for this situation may...

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7 21 11 2.2 10 J. KKG m M rr §· =+ = × ¨¸ ©¹ (b) In this case, we require K 2 = 0 and r 2 = 8.0 × 10 6 m, and solve for K 1 : 7 12 6.9 10 J. K K GmM 38. Energy conservation for this situation may be expressed as follows: 2 2 1 2 GmM GmM KU K K += + ¡ −= . where
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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