Does Mechanical Energy of a Planet Change in an Elliptical Orbit?

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The mechanical energy (ME) of a planet in an elliptical orbit remains constant throughout its motion, as the total mechanical energy of the system is conserved. While the planet's kinetic and potential energy fluctuate during the orbit, the sum of these energies, which constitutes the mechanical energy, does not change. The precise definition of ME in a two-body system includes both kinetic energy due to the planet's velocity and gravitational potential energy relative to the star. Thus, even though the individual components of ME vary, the overall mechanical energy remains constant. Understanding this principle is essential for analyzing planetary motion in elliptical orbits.
mancity
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Homework Statement
Given an elliptical planetary orbit of a planet and a star, do:
(a) the mechanical energy of the planet change during the orbit? If so, describe the motion.
(b) the mechanical energy of the planet-star system change during the orbit? If so, describe the motion.
Relevant Equations
ME=KE+PE
Obviously the mechanical energy of the total system remains the same.

But I'm having a hard time determining of the ME of the planet is constant or if it is changing.
 
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mancity said:
Homework Statement: Given an elliptical planetary orbit of a planet and a star, do:
(a) the mechanical energy of the planet change during the orbit? If so, describe the motion.

But I'm having a hard time determining of the ME of the planet is constant or if it is changing.
What's the precise definition of the ME of a planet that is part of a two-body system?
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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