81. Energy conservation for this situation may be expressed as follows: 22112 21212GmMGmMKU KUmvmvrr+=+¡−=−whereM= 5.98 ×1024kg, r1= R= 6.37 ×106m and v1= 10000 m/s. Setting v2= 0 to find the maximum of its trajectory, we solve the above equation (noting that
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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.