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equationâ ÎU = mg Îy gives a 1.0% error in the change in the potential energy. What is this height? Answer Because the surface equation assumes a constant value for g, it will give a ÎU value that is larger than the value given by the general equation, Equation 13.13. Set up a ratio reflecting a 1.0% error: Substitute the expressions for each of these changes ïU surface ï½ 1.010 ïU general gr r mg ïy ï½ i f ï½ 1.010 GM E m ï¨ ïy ri rf ï© GM E ÎU: Substitute for ri, rf, and g ï¨GM E RE 2 ï© RE ï¨ RE ï« ïy ï© GM E from Equation 13.5: Solve for Îy: ï½ RE ï« ïy ïy ï½ 1ï« ï½ 1.010 RE RE Îy = 0.010RE = 0.010(6.37 Ã 106 m) = 6.37 Ã 104 m = 63.7 km Example 13.7 Changing the Orbit of a Satellite A space transportation vehicle releases a 470-kg communications satellite while in an orbit 280 km above the surface of the Earth. A rocket engine on the satellite boosts it into a geosynchronous orbit. How much energy does the engine have to provide? SOLUTION Conceptualize Notice that the height of 280 km is much lower than that for a geosynchronous satellite, 36 000 km, as mentioned in Example 13.5. Therefore, energy must be expended to raise the satellite to this much higher position. Categorize This example is a substitution problem. Find the initial radius of ri = RE + 280 km = 6.65 Ã 106 m the satelliteâs orbit when it is still in the