Translational kinetic energy vs. rotational kinetic energy

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SUMMARY

The discussion centers on the distinction between translational kinetic energy and rotational kinetic energy in the context of a moon's orbital mechanics. The participant analyzes the work done in shifting the moon's orbit, applying the conservation of energy principle. They explore the equations for kinetic energy, specifically 1/2mv² for translational kinetic energy and 1/2Iω² for rotational kinetic energy, questioning their interchangeability in this scenario. The conclusion is that both forms of kinetic energy can be applied, provided the correct relationships between linear velocity and angular velocity are maintained.

PREREQUISITES
  • Understanding of conservation of energy principles in physics
  • Familiarity with kinetic energy equations: 1/2mv² and 1/2Iω²
  • Knowledge of angular velocity and its relationship to linear velocity
  • Basic concepts of orbital mechanics and motion
NEXT STEPS
  • Study the relationship between linear and angular velocity in orbital mechanics
  • Learn about the conservation of angular momentum in celestial bodies
  • Explore the implications of rotational kinetic energy in different physical systems
  • Investigate the effects of gravitational forces on orbital motion
USEFUL FOR

Students of physics, particularly those studying mechanics and orbital dynamics, as well as educators looking to clarify concepts of kinetic energy in celestial contexts.

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Homework Statement



My question relates to a physics problem dealing with the orbit of a moon. In the problem, the orbit of the moon shrinks such that its radius to the center of the planet in question (here, Earth) is smaller. The question gives a great deal of information, but at one point, it asks how much work was done in shifting the orbit.

Homework Equations



Conservation of energy

The Attempt at a Solution



I set this problem up such that the

Total Initial Energy = Kinetic Energy at Orbit 1 - Potential Energy at Orbit 1.

Similarly, I said that

Total Final Energy = Kinetic Energy at Orbit 2 - Potential Energy at Orbit 2.

I also reasoned that Total Final Energy - Total Initial Energy = Work done.

My question is how best to conceive of kinetic energy for a problem like this. I'm having trouble distinguishing between translational kinetic energy and rotational kinetic energy...I know that a rolling cylinder has both types of energy...so I tried to reason by analogy and figured that the moon does as well due to its rotation about its own axis (rotational kinetic energy) and orbital movement (which I figured was translational kinetic energy).

So, I set the problem up such that the kinetic energy at beginning and end were found using 1/2mv^2. The periods at both positions were given, so I found the velocities using v = ωR (with omega equaling 2∏/period).

The answer key I have used the rotational kinetic energy instead of the translational kinetic energy for the kinetic energy components in the expressions above. My answer was close but just far off enough to make me wonder whether it was rounding error or an error in my strategy.

My two questions, I suppose, are:

1) Is it equally acceptable to conceive of the kinetic energy of moving planets as rotational kinetic energy or translational kinetic energy? In other words, are the expressions 1/2mv^2 and 1/2Iω^2 equal provided you correctly solve for v using ω in the first place?

And,

2) Why do we conceive of the motion of moons in an orbit to be rotational kinetic energy? Shouldn't the term be reserved only for the moon's rotation around its own axis?

Thanks!
 
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I would just say there are two kinds of rotational energy going on. And ignore the rotation of the moon on its own axis in this question unless there is something to indicate that rotation changes.

It should work out the same way whether you use Ek = ½m⋅v² or Ek = ½m⋅(rω)²
 

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