Final answer:
The work-energy theorem in satellite motion often simplifies to W = -ΔU because changes in kinetic energy correspond to changes in potential energy due to gravity. Satellites maintain circular orbits due to the balance of gravitational potential energy and kinetic energy, with kinetic energy being half the magnitude of potential energy in this system. The ratio of kinetic to potential energy in orbits decreases as the orbit size increases.
Step-by-step explanation:
While applying the work-energy theorem on a satellite, it's often sufficient to equate the work done to the change in gravitational potential energy, W = -ΔU, without explicitly considering the kinetic energy because the theorem also takes into account that any change in kinetic energy has a corresponding change in potential energy when dealing with conservative forces such as gravity.
For a satellite in a circular orbit, its velocity v = √(GM/r) comes from the gravitational pull of the Earth, which acts as the centripetal force to keep the satellite in orbit. This results in a scenario where the satellite's total mechanical energy, which is the sum of its kinetic and potential energies, is negative, indicating a bound orbit.
It is true that the energy required to lift a satellite into orbit is much less than the energy needed to keep it moving, which is primarily kinetic energy. The relationship between kinetic and potential energy for circular orbits is that the magnitude of the kinetic energy is exactly one-half the magnitude of the potential energy.
As the size of the orbit increases, the total mechanical energy becomes less negative, which means the satellite becomes less tightly bound and the ratio of kinetic to potential energy decreases.