A 1 042kg satellite orbits the Earth at a constant altitude

A 1 042-kg satellite orbits the Earth at a constant altitude of 107-km. How much energy must be added to the system to move the satellite into a circular orbit with altitude 207 km? 937357.5 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?

Solution

a) Using

Energy added = Change in Potenital Energy + Change in Kinetic Energy

Initially

Velocity = sqrt(Gm/r)

= sqrt((6.676*10^-11*5.94*10^24)/(6478*10^3))

= 7820.52 m/sec

Therefore

Kinetic Energy + potential Energy

= 0.5*1042*7820.52^2 - ((6.673*10^-11*5.94*10^24*1042)/(6478*10^3))

= -3.189331232*10^10 J

Finally

V\' = sqrt((6.673*10^-11*5.94*10^24)/(6578*10^3))

= 7762.6 m/sec

Therefore

Total Energy = 0.5*1042*7762.6^2 - ((6.673*10^-11*1042*5.94*10^24)/(6578*10^3))

= - 3.139429312*10^10 J

So Energy Added =  3.189331232*10^10 - 3139429312*10^10

= 4.9902*10^8 J

b) Change in Kinetic Energy = - Change in Total Energy

= Energy Added

= 4.9902*10^8 = 499.02 MJ

c) Change in potential Energy

= + ((6.673*10^-11*5.94*10^24*1042)/(6478*10^3)) - ((6.673*10^-11*1042*5.94*10^24)/(6578*10^3))

= 9.6926*10^8 J

= 969.26 MJ