A 1 030-kg satellite orbits the Earth at a constant altitude of 99-km. A 1 030-k

Question

A 1 030-kg satellite orbits the Earth at a constant altitude of 99-km.

A 1 030-kg satellite orbits the Earth at a constant altitude of 99-km. How much energy must be added to the system to move the satellite into a circular orbit with altitude 203 km? How is the total energy of an object in circular orbit related to the potential energy? MJ What is the change in the system's kinetic energy? What is the change in the system's potential energy? What is the equation for the gravitational potential energy of a system of two objects? MJ

Explanation / Answer

a) E = -GMm/2r = -6.67*10^-11*5.97*10^24*1030/2*6.37*10^6*99000 = -3.17*10^10 J

high orbit E = -GMm/2r = -3.12*10^10 J

delta E = (-3.12-(-3.17)10^10 = 5 *10^8 J

b) delta K = -delta E = - 5*10^8 J

c) delta U = 2*delta E = 2*5*10^8 = 10*10^8 J

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