Question: When it is back at Point 1, the system's potential energy is again UG =-GMm/R = 2E. So the new kinetic energy at Point 1

When it is back at Point 1, the system's potential energy is again UG =-GMm/R = 2E. So the new kinetic energy at Point 1 has to be 4 4 K,tr = Etr - Uf = E - 2E; = Ei=K,i 3 3 (1) Equation (1) gives us a factor by which the kinetic energy of the satellite must be increased. Using equation (1), by what factor must the speed be increased to produce this result? Enter your factor into the box in the simulator. You can enter math text directly into this box. For example, if you came up with a ratio of , you could enter the text "sqrt(1/2)" and it will calculate the value for you. Press the "Submit" button, then click the blue "Set Factor" button in the simulator. Note that the simulation must be stopped to do this. You will know that you've succeeded if you find that the apoapsis of the satellite's orbit now reaches exactly to the green dashed circle. (Also, the simulator will not allow you to proceed until you enter the correct value). Once you have successfully initiated this change in velocit

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